I ’ R M

Transcription

I ’ R M

INSTRUCTOR’S RESOURCE
MANUAL
M ATHEMATICS FOR
E LEMENTARY T EACHERS
WITH A CTIVITIES
Sybilla Beckmann
University of Georgia
Boston Columbus Indianapolis New York San Francisco Upper Saddle River
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson from electronic files supplied by the author.
Copyright © 2014, 2011, 2008 Pearson Education, Inc.
Publishing as Pearson, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-83673-1
ISBN-10: 0-321-83673-1
1 2 3 4 5 6 OPM 16 15 14 13 12
www.pearsonhighered.com
Contents
0 General Advice
0.1 General Advice on Teaching Courses for Elementary Teachers
0.2 Comments and Advice on the Class Activities . . . . . . . . .
0.3 Comments and Advice on the Practice Exercises and Problems
0.4 Grading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0.5 Sample Syllabi . . . . . . . . . . . . . . . . . . . . . . . . . .
0.6 Advice to Give Students Who Are Struggling . . . . . . . . .
0.7 How Is Teaching from this Book Different? . . . . . . . . . . .
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1 Numbers and the Base-Ten System
1.1 The Counting Numbers . . . . . . . . . . . . . . . . . . . .
The Counting Numbers as a List . . . . . . . . . . . . . .
Connecting Counting Numbers as a List with Cardinality .
How Many Are There? . . . . . . . . . . . . . . . . . . . .
1.2 Decimals and Negative Numbers . . . . . . . . . . . . . . .
Representing Decimals with Bundled Objects . . . . . . .
Representing Decimals as Lengths . . . . . . . . . . . . . .
Zooming In on Number Lines . . . . . . . . . . . . . . . .
Numbers Plotted on Number Lines . . . . . . . . . . . . .
Negative Numbers on Number Lines . . . . . . . . . . . .
1.3 Comparing Numbers in Base Ten . . . . . . . . . . . . . .
Misconceptions in Comparing Decimals . . . . . . . . . . .
Finding Smaller and Smaller Decimals . . . . . . . . . . .
Finding Decimals Between Decimals . . . . . . . . . . . . .
Decimals Between Decimals on Number Lines . . . . . . .
“Greater Than” and “Less Than” with Negative Numbers
1.4 Rounding Numbers . . . . . . . . . . . . . . . . . . . . . .
Explaining Rounding . . . . . . . . . . . . . . . . . . . . .
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CONTENTS
Rounding with Number Lines . . . . . . . . . . . . . . . . . . . . . .
Can We Round This Way? . . . . . . . . . . . . . . . . . . . . . . . .
What Does “Population 120,000” Really Mean? . . . . . . . . . . . .
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2 Fractions and Problem Solving
2.1 Solving Problems and Explaining Solutions . . . . . . . . . . . . . . .
2.2 Defining and Reasoning About Fractions . . . . . . . . . . . . . . . .
Fractions of Things . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Relating Fractions to Wholes . . . . . . . . . . . . . . . . . . . . . .
Comparing Quantities with Fractions . . . . . . . . . . . . . . . . . .
Fractions of Non-Contiguous Wholes . . . . . . . . . . . . . . . . . .
Is the Meaning of Equal Parts Always Clear? . . . . . . . . . . . . . .
Number Line Errors with Fractions . . . . . . . . . . . . . . . . . . .
Fractions on Number Lines, Part 1 . . . . . . . . . . . . . . . . . . .
Improper Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Equivalent Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equivalent Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . .
Misconceptions About Fraction Equivalence . . . . . . . . . . . . . .
Common Denominators . . . . . . . . . . . . . . . . . . . . . . . . . .
Solving Problems by Changing Denominators . . . . . . . . . . . . .
Fractions on Number Lines, Part 2 . . . . . . . . . . . . . . . . . . .
Simplifying Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Comparing Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . .
What is another way to Compare these Fractions? . . . . . . . . . . .
Comparing Fractions by Reasoning . . . . . . . . . . . . . . . . . . .
Can We Reason this Way? . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Percent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Math Drawings, Percentages, and Fractions . . . . . . . . . . . . . .
Reasoning About Percent Tables to Solve “Portion Unknown” Percent
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Reasoning about Percent Tables . . . . . . . . . . . . . . . . . . . . .
Equivalent Fractions and Percent . . . . . . . . . . . . . . . . . . . .
Percent Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Addition and Subtraction
3.1 Interpretations of Addition and Subtraction . . . . . . . . . . . . . .
Relating Addition and Subtraction:The Shopkeepers’ Method of Making Change . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Writing Add To and Take From Problems . . . . . . . . . . . . . . .
Writing Put Together/Take Apart and Compare Problems . . . . . .
Identifying Problem Types and Difficult Language . . . . . . . . . . .
Properties of Addition and Mental Math . . . . . . . . . . . . . . . .
Mental Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Children’s Learning Paths for Single-Digit Addition . . . . . . . . . .
Children’s Learning Paths for Single-Digit Subtraction . . . . . . . .
Reasoning to Add and Subtract . . . . . . . . . . . . . . . . . . . . .
Algorithms for Addition and Subtraction . . . . . . . . . . . . . . . .
Adding and Subtracting with Base-Ten Math Drawings . . . . . . . .
Understanding the Standard Addition Algorithm . . . . . . . . . . .
Understanding the Standard Subtraction Algorithm . . . . . . . . . .
A Third Grader’s Method of Subtraction . . . . . . . . . . . . . . . .
Regrouping in Base 12 . . . . . . . . . . . . . . . . . . . . . . . . . .
Regrouping in Base 60 . . . . . . . . . . . . . . . . . . . . . . . . . .
Adding And Subtracting Fractions . . . . . . . . . . . . . . . . . . .
Why Do We Add and Subtract Fractions the Way We Do? . . . . . .
Adding and Subtracting Mixed Numbers . . . . . . . . . . . . . . . .
Are These Word Problems for 12 + 31 ? . . . . . . . . . . . . . . . . . .
Are These Word Problems for 12 − 13 ? . . . . . . . . . . . . . . . . . .
What Fraction is Shaded? . . . . . . . . . . . . . . . . . . . . . . . .
Addition with Whole Numbers, Decimals, Fractions, and Mixed Numbers: What Are Common Ideas? . . . . . . . . . . . . . . . . .
Adding and Subtracting Negative Numbers . . . . . . . . . . . . . . .
Word Problems and Rules for Adding and Subtracting with Negative
Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 Multiplication
4.1 Interpretations of Multiplication . . . . . . . . . . . . . . . . . . . . .
Showing Multiplicative Structure . . . . . . . . . . . . . . . . . . . .
Writing Multiplication Word Problems . . . . . . . . . . . . . . . . .
Problems About Pairs . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Why Multiplying by 10 Is Special in Base Ten . . . . . . . . . . . . .
Multiplying by 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Commutative, Associative Properties, Area, Volume . . . . . . . . . .
Explaining and Applying the Commutative Property of Multiplication
Multiplication, Areas of Rectangles, and the Commutative Property .
Ways to Describe the Volume of a Box with Multiplication . . . . . .
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3.5
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CONTENTS
4.4
4.5
4.6
Explaining the Associative Property . . . . . . . . . . . . . . . . . .
Using the Associative and Commutative Properties of Multiplication
Using Multiplication to Estimate How Many . . . . . . . . . . . . . .
The Distributive Property . . . . . . . . . . . . . . . . . . . . . . . .
Explaining the Distributive Property . . . . . . . . . . . . . . . . . .
Applying the Distributive Property . . . . . . . . . . . . . . . . . . .
Relating Multiplication Problems . . . . . . . . . . . . . . . . . . . .
Properties of Arithmetic, Mental Math, Basic Facts . . . . . . . . . .
Using Properties of Arithmetic to Aid the Learning of Basic Multiplication Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solving Arithmetic Problems Mentally . . . . . . . . . . . . . . . . .
Writing Equations that Correspond to a Method of Calculation . . .
Showing the Algebra in Mental Math . . . . . . . . . . . . . . . . . .
Why Algorithms for Multiplying Work . . . . . . . . . . . . . . . . .
The Standard Versus the Partial Products Multiplication Algorithm .
Why the Multiplication Algorithms Work, Part I . . . . . . . . . . .
Why the Multiplication Algorithms Work, Part II . . . . . . . . . . .
The Standard Multiplication Algorithm Right Side Up and Upside Down
5 Multiplying Fractions, Decimals, Negatives
5.1 Multiplying Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . .
Extending the Definition of Multiplication to Fractions . . . . . . . .
Explaining Why the Procedure for Multiplying Fractions Is Valid . .
When Do We Multiply Fractions? . . . . . . . . . . . . . . . . . . . .
What Fraction Is Shaded? . . . . . . . . . . . . . . . . . . . . . . . .
Multiplying Mixed Numbers . . . . . . . . . . . . . . . . . . . . . . .
5.2 Multiplying Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . .
Multiplying Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . .
Explaining Why We Place the Decimal Point Where We Do When We
Multiply Decimals . . . . . . . . . . . . . . . . . . . . . . . .
Decimal Multiplication And Areas of Rectangles . . . . . . . . . . . .
5.3 Multiplying Negative Numbers . . . . . . . . . . . . . . . . . . . . . .
Patterns With Multiplication and Negative Numbers . . . . . . . . .
Explaining Multiplication with Negative Numbers (and 0) . . . . . .
5.4 Powers and Scientific Notation . . . . . . . . . . . . . . . . . . . . . .
Multiplying Powers of 10 . . . . . . . . . . . . . . . . . . . . . . . . .
Scientific Notation Versus Ordinary Decimal Notation . . . . . . . . .
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6 Division
6.1 Interpretations of Division . . . . . . . . . . . . . . . . . . . . . . . .
What Does Division Mean? . . . . . . . . . . . . . . . . . . . . . . .
Division Word Problems . . . . . . . . . . . . . . . . . . . . . . . . .
Why Can’t We Divide by Zero? . . . . . . . . . . . . . . . . . . . . .
6.2 Division and Fractions and Division with Remainder . . . . . . . . .
Relating Fractions and Division . . . . . . . . . . . . . . . . . . . . .
What to Do With the Remainder? . . . . . . . . . . . . . . . . . . .
6.3 Why Division Algorithms Work . . . . . . . . . . . . . . . . . . . . .
Dividing Without Using a Calculator or Long Division . . . . . . . .
Why the Scaffold Method of Division Works . . . . . . . . . . . . . .
Interpreting the Standard Division Algorithm as Dividing Bundled
Toothpicks . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Interpreting the Standard Division Algorithm in Terms of Area . . .
Student Errors in Using the Division Algorithm . . . . . . . . . . . .
Interpreting the Calculation of Decimal Answers to Whole Number
Division Problems in Terms of Money . . . . . . . . . . . . . .
Errors in Decimal Answers to Division Problems . . . . . . . . . . . .
6.4 Fraction Division, How Many Groups? . . . . . . . . . . . . . . . . .
“How Many Groups?” Fraction Division Problems . . . . . . . . . . .
Dividing Fractions by Dividing the Numerators and Dividing the Denominators . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 Fraction Division, How Many in One Group? . . . . . . . . . . . . .
“How Many in One Group?” Fraction Division Problems . . . . . . .
Are These Division Problems? . . . . . . . . . . . . . . . . . . . . . .
6.6 Dividing Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Quick Tricks and Estimation with Decimal Division . . . . . . . . . .
Decimal Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7 Ratio and Proportional Relationships
7.1 Motivating and Defining Ratio and Proportional Relationships . . .
Comparing Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . .
Walking Times and Distances . . . . . . . . . . . . . . . . . . . . .
Using strip diagrams to represent ratios as “fixed numbers of parts”
7.2 Reasoning with Multiplication and Division . . . . . . . . . . . . .
Using Strip Diagrams to Solve Ratio Problems . . . . . . . . . . . .
Ratio Problem Solving with Strip Diagrams . . . . . . . . . . . . .
Multiplicative Relationships in Ratio Tables . . . . . . . . . . . . .
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7.3
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7.5
Solving Proportions by Reasoning About How Quantities Compare .
Going Through 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
More Ratio Problem Solving . . . . . . . . . . . . . . . . . . . . . . .
Unit Rates and the Values of a Ratio . . . . . . . . . . . . . . . . . .
Unit Rates and Multiplicative Comparisons Associated with a Ratio .
Solving Proportions by Cross-Multiplying Fractions . . . . . . . . . .
Proportional Versus Inversely Proportional . . . . . . . . . . . . . . .
Can You Use a Proportion or Not? . . . . . . . . . . . . . . . . . . .
A Proportional Relationship Versus an Inversely Proportional Relationship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Who Says You Can’t Do Rocket Science? . . . . . . . . . . . . . . . .
Percent Revisited: Percent Increase and Decrease . . . . . . . . . . .
How Should We Describe the Change? . . . . . . . . . . . . . . . . .
Calculating Percent Increase and Decrease . . . . . . . . . . . . . . .
Calculating Amounts from a Percent Increase or Decrease . . . . . . .
Can We Solve it This Way? . . . . . . . . . . . . . . . . . . . . . . .
Percent Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . .
Using the Commutative Property of Multiplication . . . . . . . . . .
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8 Number Theory
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8.1 Factors and Multiples . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
Factors and Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . 132
Finding All Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
Do Factors Always Come in Pairs? . . . . . . . . . . . . . . . . . . . 132
8.2 Even and Odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
Why Can We Check the Ones Digit to Determine if a Number is Even
or Odd? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
Questions About Even and Odd Numbers . . . . . . . . . . . . . . . 133
Extending the Definitions of Even and Odd . . . . . . . . . . . . . . 134
8.3 Divisibility Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
The Divisibility Test for 3 . . . . . . . . . . . . . . . . . . . . . . . . 135
8.4 Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
The Sieve of Eratosthenes . . . . . . . . . . . . . . . . . . . . . . . . 135
The Trial Division Method for Determining whether a Number Is Prime135
Factoring into Products of Primes . . . . . . . . . . . . . . . . . . . . 136
8.5 Greatest Common Factor and Least Common Multiple . . . . . . . . 136
The “Slide Method” . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
GCF and LCM with Products of Powers of Primes . . . . . . . . . . 138
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8.6
8.7
Problems Involving GCFs and LCMs . . . . . .
Spirograph Flower Designs . . . . . . . . . . . .
Rational and Irrational Numbers . . . . . . . .
Decimal Representations of Fractions . . . . . .
Writing Terminating and Repeating Decimals as
What is 0.9999. . . ? . . . . . . . . . . . . . . . .
The Square Root of 2 . . . . . . . . . . . . . . .
Pattern Tiles and the Irrationality of the Square
Looking Back at the Number Systems . . . . . .
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Fractions
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9 Algebra
9.1 Numerical Expressions . . . . . . . . . . . . . . . . . . . . . . . . .
Writing Expressions for Dot, Star, and Stick Designs . . . . . . . .
How Many High-Fives? . . . . . . . . . . . . . . . . . . . . . . . . .
Sums of Counting Numbers . . . . . . . . . . . . . . . . . . . . . .
Sums of Odd Numbers . . . . . . . . . . . . . . . . . . . . . . . . .
Expressions with Fractions . . . . . . . . . . . . . . . . . . . . . . .
Evaluating Expressions With Fractions Efficiently and Correctly . .
9.2 Expressions with Variables . . . . . . . . . . . . . . . . . . . . . . .
Describing the Structure of an Expression . . . . . . . . . . . . . .
Defining Variables and Writing Expressions . . . . . . . . . . . . . .
Equivalent Expressions . . . . . . . . . . . . . . . . . . . . . . . . .
Expressions for Quantities . . . . . . . . . . . . . . . . . . . . . . .
Encoding and Decoding Secret Numbers . . . . . . . . . . . . . . .
Rocket Science Reasoning . . . . . . . . . . . . . . . . . . . . . . .
9.3 Equations for Different Purposes . . . . . . . . . . . . . . . . . . .
Different Ways We Use Equations . . . . . . . . . . . . . . . . . . .
Identities Arising from Rectangular Designs . . . . . . . . . . . . .
Equations About Quantities That Vary Together . . . . . . . . . .
Formulating Equations for Story Problems . . . . . . . . . . . . . .
9.4 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solving Equations by Reasoning About Expressions . . . . . . . . .
Two Ways of Reasoning to Solve “One Step” Equations . . . . . . .
Solving Equations Algebraically and With a Pan Balance . . . . . .
What Are the Solutions of These Equations? . . . . . . . . . . . . .
9.5 Solving Problems with Strip Diagrams . . . . . . . . . . . . . . . .
Solving Word Problems with Strip Diagrams and with Equations .
Solving Word Problems in Multiple Ways and Modifying Problems
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9.6
9.7
9.8
Solving Word Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 163
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
Repeating Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
Solving Problems Using Repeating Patterns . . . . . . . . . . . . . . 164
Arithmetic Sequences of Numbers Corresponding to Sequences of Figures166
How Are Expressions for Arithmetic Sequences Related to the Way
Sequences Start and Grow? . . . . . . . . . . . . . . . . . . . 166
Explaining Expressions for Arithmetic Sequences . . . . . . . . . . . 167
Comparing and Contrasting Sequences . . . . . . . . . . . . . . . . . 168
What’s the Next Entry? . . . . . . . . . . . . . . . . . . . . . . . . . 169
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
What Does the Shape of a Graph Tell Us About a Function? . . . . . 170
Graphs and Stories . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
How Does Braking Distance Depend on Speed? . . . . . . . . . . . . 173
Reasoning About Equations for Functions . . . . . . . . . . . . . . . 173
Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
Proportional Relationships as Functions . . . . . . . . . . . . . . . . 174
Arithmetic Sequences as Functions . . . . . . . . . . . . . . . . . . . 175
Different Equations for Linear Functions . . . . . . . . . . . . . . . . 175
Linear Versus Inversely Proportional Relationships . . . . . . . . . . 176
Is it Linear or Not? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
What Kind of Relationship Is It? . . . . . . . . . . . . . . . . . . . . 177
10 Geometry
10.1 Visualization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Visualizing Lines and Planes . . . . . . . . . . . . . . . . . . . . . . .
The Rotation of the Earth . . . . . . . . . . . . . . . . . . . . . . . .
Explaining the Phases of the Moon . . . . . . . . . . . . . . . . . . .
10.2 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Angles Formed by Two Lines . . . . . . . . . . . . . . . . . . . . . .
Angles Formed when a Line Crosses Two Parallel Lines . . . . . . . .
Seeing That the Angles in a Triangle Add to 180◦ . . . . . . . . . . .
Using the Parallel Postulate to Prove that the Angles in a Triangle
Add to 180◦ . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Explaining Why the Angles in a Triangle Add to 180◦ by Walking and
Turning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Angle Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Students’ Ideas and Questions About Angles . . . . . . . . . . . . . .
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186
CONTENTS
xi
10.3 Angles and Phenomena in the World . . . . . . . . . . . . . . . . . . 186
Eratosthenes’s Method for Determining the Circumference of the Earth 186
How Big Is the Reflection of Your Face in a Mirror? . . . . . . . . . . 187
Why Do Spoons Reflect Upside Down? . . . . . . . . . . . . . . . . . 187
10.4 Circles and Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
Using Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
The Global Positioning System (GPS) . . . . . . . . . . . . . . . . . 189
Circle Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
10.5 Triangles, Quadrilaterals, and Other Polygons . . . . . . . . . . . . . 189
What Shape Is It? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
What Properties Do These Shapes Have? . . . . . . . . . . . . . . . . 189
Classify Shapes into Categories Based on Their Properties [core] . . . 190
Classify Triangles Based on Their Properties . . . . . . . . . . . . . . 191
Venn Diagrams Relating Quadrilaterals . . . . . . . . . . . . . . . . . 191
Using a Compass to Construct Triangles and Quadrilaterals . . . . . 192
Making Shapes by Folding Paper . . . . . . . . . . . . . . . . . . . . 193
Making Shapes by Walking and Turning Along Routes . . . . . . . . 193
11 Measurement
11.1 Fundamentals of Measurement . . . . . . . . . . . . . . . . . . . .
The Biggest Tree in the World . . . . . . . . . . . . . . . . . . . .
What Does “6 Square Inches” Mean? . . . . . . . . . . . . . . . .
Using a Ruler . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 Length, Area, Volume, and Dimension . . . . . . . . . . . . . . .
Dimension and Size . . . . . . . . . . . . . . . . . . . . . . . . . .
11.3 Error and Precision in Measurements . . . . . . . . . . . . . . . .
Reporting and Interpreting Measurements . . . . . . . . . . . . .
11.4 Converting from One Unit of Measurement to Another . . . . . .
Conversions: When Do We Multiply? When Do We Divide? . . .
Conversion Problems . . . . . . . . . . . . . . . . . . . . . . . . .
Using Dimensional Analysis to Convert Measurements . . . . . .
Area and Volume Conversions . . . . . . . . . . . . . . . . . . . .
Area and Volume Conversions: Which Are Correct and Which Are
Problem Solving with Conversions . . . . . . . . . . . . . . . . . .
195
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Not?200
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12 Area of Shapes
203
12.1 What Area Is . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
Units of Length and Area in the Area Formula for Rectangles . . . . 204
xii
CONTENTS
12.2 The Moving and Additivity Principles About Area . . . . . . . . .
Different Shapes with the Same Area . . . . . . . . . . . . . . . . .
Using the Moving and Additivity Principles . . . . . . . . . . . . .
12.3 Areas of Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Determining Areas of Triangles in Progressively Sophisticated Ways
Choosing the Base and Height of Triangles . . . . . . . . . . . . . .
Explaining Why the Area Formula for Triangles Is Valid . . . . . .
Area Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . .
12.4 Areas of Parallelograms and Other Polygons . . . . . . . . . . . . .
Do Side Lengths Determine the Area of a Parallelogram? . . . . . .
Explaining Why the Area Formula for Parallelograms Is Valid . . .
Finding and Explaining a Trapezoid Area Formula . . . . . . . . . .
12.5 Shearing: Changing Shapes Without Changing Area . . . . . . . . .
Is This Shearing? . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Shearing Parallelograms and Triangles . . . . . . . . . . . . . . . .
12.6 Areas of Circles and the Number Pi . . . . . . . . . . . . . . . . . .
How Big Is the Number π? . . . . . . . . . . . . . . . . . . . . . . .
A Hands-On Method to Approximate the Value of π . . . . . . . .
Why the Area Formula for Circles Makes Sense . . . . . . . . . . .
Area Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.7 Approximating Areas of Irregular Shapes . . . . . . . . . . . . . . .
Determining the Area of an Irregular Shape . . . . . . . . . . . . .
12.8 Perimeter Versus Area . . . . . . . . . . . . . . . . . . . . . . . . .
Perimeter Misconceptions . . . . . . . . . . . . . . . . . . . . . . .
How Are Perimeter and Area Related for Rectangles? . . . . . . . .
How Are Perimeter and Area Related for All Shapes? . . . . . . . .
12.9 The Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . .
Side Lengths of Squares Inside Squares . . . . . . . . . . . . . . . .
A Proof of the Pythagorean Theorem . . . . . . . . . . . . . . . . .
13 Solid Shapes
13.1 Polyhedra and Other Solid Shapes . . . .
Making Prisms and Pyramids . . . . . .
Analyzing Prisms and Pyramids . . . . .
What’s Inside the Magic 8 Ball? . . . . .
Making Platonic Solids . . . . . . . . . .
13.2 Patterns and Surface Area . . . . . . . .
What Shapes Do These Patterns Make?
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226
CONTENTS
Patterns and Surface Area for Prisms and Pyramids . . . . . . . . . .
Patterns and Surface Area for Cylinders . . . . . . . . . . . . . . . .
Patterns and Surface Area for Cones . . . . . . . . . . . . . . . . . .
Cross-Sections of a Pyramid . . . . . . . . . . . . . . . . . . . . . . .
Cross-Sections of a Long Rectangular Prism . . . . . . . . . . . . . .
13.3 Volumes of Solid Shapes . . . . . . . . . . . . . . . . . . . . . . . . .
Why the Volume Formula for Prisms and Cylinders Makes Sense . . .
Comparing the Volume of a Pyramid With the Volume of a Rectangular
Prism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
The 3 in the Volume Formula for Pyramids andCones . . . . . . . . .
Volume Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . .
Deriving the Volume of a Sphere . . . . . . . . . . . . . . . . . . . . .
Volume Versus Surface Area and Height . . . . . . . . . . . . . . . .
13.4 Volume Versus Weight . . . . . . . . . . . . . . . . . . . . . . . . . .
Underwater Volume Problems . . . . . . . . . . . . . . . . . . . . . .
Floating Versus Sinking: Archimedes’s Principle . . . . . . . . . . . .
xiii
227
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232
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234
14 Geometry of Motion and Change
235
14.1 Reflections, Translations, and Rotations . . . . . . . . . . . . . . . . 236
Exploring Reflections, Rotations, and Translations with Transparencies 236
Reflections, Rotations, and Translations in a Coordinate Plane . . . . 237
Which Transformation Is It? . . . . . . . . . . . . . . . . . . . . . . . 238
14.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
Checking for Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . 239
Traditional Quilt Designs . . . . . . . . . . . . . . . . . . . . . . . . . 239
Creating Interlocking Symmetrical Designs . . . . . . . . . . . . . . . 239
14.3 Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
Motivating a Definition of Congruence . . . . . . . . . . . . . . . . . 240
Triangles and Quadrilaterals of Specified Side Lengths . . . . . . . . . 240
What Information Specifies a Triangle? . . . . . . . . . . . . . . . . . 240
Explaining Why Isosceles Triangles and Rhombuses Decompose into
Congruent Right Triangles . . . . . . . . . . . . . . . . . . . . 241
Sewing Boxes and Congruence . . . . . . . . . . . . . . . . . . . . . . 242
14.4 Constructions With Straightedge and Compass . . . . . . . . . . . . . 242
Relating the Constructions to Properties of Rhombuses . . . . . . . . 242
Constructing a Square and an Octagon with Straightedge and Compass 242
14.5 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
Mathematical Similarity Versus Similarity in Everyday Language . . 244
xiv
CONTENTS
A First Look at Solving Scaling Problems . . . . . . . . . . .
Using the Scale Factor and Internal Factor Methods . . . . . .
Addressing a Common Misconception About Scaling . . . . .
Measuring Distances by “Sighting” . . . . . . . . . . . . . . .
Using a Shadow or a Mirror to Determine the Height of a Tree
14.6 Areas, Volumes, and Scaling . . . . . . . . . . . . . . . . . . .
Surface Areas and Volumes of Similar Boxes . . . . . . . . . .
Determining Surface Areas and Volumes of Similar Objects . .
A Scaling Proof of the Pythagorean Theorem . . . . . . . . .
Area and Volume Problem Solving . . . . . . . . . . . . . . .
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15 Statistics
251
15.1 Formulating Questions and Gathering Data . . . . . . . . . . . . . . 252
Statistical Questions Versus Other Questions . . . . . . . . . . . . . . 252
Choosing a Sample . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
Using Random Samples . . . . . . . . . . . . . . . . . . . . . . . . . . 252
Using Random Samples to Estimate Population Size by Marking (CaptureRecapture) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
15.2 Displaying Data and Interpreting Data Displays . . . . . . . . . . . . 255
What Is Wrong With These Displays or Their Interpretation? . . . . 255
Three Levels of Questions about Graphs . . . . . . . . . . . . . . . . 256
Display These Data about Pets . . . . . . . . . . . . . . . . . . . . . 257
Investigating Small Bags of Candies . . . . . . . . . . . . . . . . . . . 257
The Length of a Pendulum and the Time It Takes to Swing . . . . . 257
Balancing a Mobile . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
15.3 The Center of Data: Mean, Median, and Mode . . . . . . . . . . . . . 258
The Mean as “Making Even” or “Leveling Out” . . . . . . . . . . . . 258
Solving Problems about the Mean . . . . . . . . . . . . . . . . . . . . 259
The Average as “Balance Point” . . . . . . . . . . . . . . . . . . . . . 259
Same Median, Different Average . . . . . . . . . . . . . . . . . . . . . 260
Can More than Half Be Above Average? . . . . . . . . . . . . . . . . 260
Errors with the Mean and the Median . . . . . . . . . . . . . . . . . 261
15.4 Summarizing, Describing, and Comparing Data Distributions . . . . . 261
What Does the Shape of a Data Distribution Tell Us About the Data? 261
Distributions of Random Samples . . . . . . . . . . . . . . . . . . . . 262
Comparing Distributions: Mercury in Fish . . . . . . . . . . . . . . . 262
Using Medians and Interquartile Ranges to Compare Data . . . . . . 263
Using Box Plots to Compare Data . . . . . . . . . . . . . . . . . . . . 264
CONTENTS
xv
Percentiles Versus Percent Correct . . . . . . . . . . . . . . . . . . . . 265
Comparing Paper Airplanes . . . . . . . . . . . . . . . . . . . . . . . 265
Using Dot Paper to Estimate Area . . . . . . . . . . . . . . . . . . . 266
16 Probability
267
16.1 Basic Principles of Probability . . . . . . . . . . . . . . . . . . . . . . 268
Probabilities with Spinners . . . . . . . . . . . . . . . . . . . . . . . . 268
Some Probability Misconceptions . . . . . . . . . . . . . . . . . . . . 268
Empirical Versus Theoretical Probability: Picking Cubes From a Bag 269
Using Empirical Probability to Make Predictions . . . . . . . . . . . 269
If You Flip 10 Pennies Should Half Come Up Heads? . . . . . . . . . 269
16.2 Counting the Number of Outcomes . . . . . . . . . . . . . . . . . . . 270
How Many Keys Are There? . . . . . . . . . . . . . . . . . . . . . . . 270
Counting Outcomes: Independent Versus Dependent . . . . . . . . . 270
16.3 Calculating Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . 270
Number Cube Rolling Game . . . . . . . . . . . . . . . . . . . . . . . 271
Picking Two Marbles From a Bag of 1 Black and 3 Red Marbles . . . 271
More Probability Misconceptions . . . . . . . . . . . . . . . . . . . . 273
Expected Earnings from the Fall Festival . . . . . . . . . . . . . . . . 273
16.4 Using Fraction Arithmetic to Calculate Probabilities . . . . . . . . . 274
Using the Meaning of Fraction Multiplication to Calculate a Probability274
Using Fraction Multiplication and Addition to Calculate a Probability 274
A Blackline Masters
277
xvi
CONTENTS
Chapter 0
General Advice on Teaching, Class
Activities, Grading, and Syllabi
1
2
CHAPTER 0
This manual contains solutions to the Class Activities as well as my advice and
recommendations on teaching the material. I hope the manual will help you teach
successful, enjoyable courses using my book. Teaching prospective elementary teachers can be extremely satisfying work. I look forward to going to class every day and
hope that you will too!
0.1
General Advice on Teaching Courses for Elementary Teachers
It’s almost too obvious to say, but it helps me to keep the following key point in
mind when developing and teaching mathematics courses for prospective teachers:
These courses should help prepare teachers to teach mathematics. Therefore such
courses should be designed and taught to “travel into the classroom.” In other words,
teachers should be able to draw on what they have learned in their mathematics
courses (including mathematics content courses, not just methods courses) to teach
mathematics effectively. Thus mathematics courses for prospective teachers should
focus on the mathematics that the teachers will teach, but should also help teachers
understand how this mathematics will develop and be built upon in subsequent grades.
I wrote this book for use in courses that focus on explaining and making sense
of the mathematics that teachers will teach, therefore I encourage this focus in your
courses. Most prospective elementary teachers entering my courses do not know that
procedures such as the standard multiplication algorithm can actually be explained.
They often seem to view the procedures and formulas of math as “given.” So expecting
students to explain why these procedures and formulas work in terms of fundamental
principles and concepts requires a shift in students’ thinking about mathematics. The
payoff is that students will develop a more coherent understanding of mathematics
because they will know how to link procedures and formulas to underlying concepts.
Students will need time and practice to help them learn what qualifies as a mathematical explanation. If you ask students to explain why a formula or a procedure
is valid, many will give mnemonic or other devices for remembering the formula, or
they will describe the procedure instead of explaining why it is valid. For example,
to explain why we place the decimal point where we do when we multiply decimals, a
student might say that we “took out” some decimal places so we have to “put back”
the same number of decimal places. Such a statement is not sufficiently specific or
complete to qualify as a valid mathematical explanation. You will need to keep pressing students to seek logical explanations that draw on central mathematical ideas.
c 2014 Pearson Education, Inc.
Copyright SECTION 0.2
3
In the decimal multiplication example you might ask what we do mathematically in
order to “take out” a decimal point (we multiply by 10 a suitable number of times to
move the decimal point or to shift the digits of the number).
The ideas of elementary school math are surprisingly deep and subtle. It is therefore likely that you and your students will encounter some mathematics in the book
that you will find surprising, unfamiliar, and in fact, difficult. A natural reaction to
such an encounter is to say “if this is hard for me, then it’s too hard for my students.”
Such a reaction can be followed by the dreaded question, “why do we need to know
this?”. This is a legitimate question, so I urge you to take it seriously and treat it
respectfully. (I actually rarely get this question any more.) It might also help to comment that young students are just as smart as we are, they are just less experienced.
Therefore young students’ ways of thinking about mathematical ideas may actually
be less rigid than ours.
To help your students see the relevance of what they are learning to their future
work, you may wish to point them to the mathematics standards in your state. Many
states are now expecting students to make sense of math and to discuss, reason about,
and explain mathematical ideas rather than simply carry out procedures without
understanding.
Some students may have their heart set on teaching a certain grade and may
resist thinking about material that is taught at other grade levels. If so, remind the
students that they will be certified to teach a range of grades and that they must be
prepared to teach the math at all of those grade levels. Also, to teach 5th grade, say,
one must be thinking ahead to the material students will be learning at 6th and 7th
grades.
Even if you do not have experience with elementary school teaching, you can refer to the carefully thought out recommendations of respected national groups and
organizations. The Common Core State Standards for Mathematics were written by
national experts. The National Council of Teachers of Mathematics (NCTM) has
developed principles and standards for the teaching of mathematics in schools (Pre-K
– 12) and Curriculum Focal Points for Prekindergarten through Grade 8 Mathematics. The Conference Board of the Mathematical Sciences (CBMS) has developed
recommendations for the preparations of teachers. If we take these recommendations seriously, then teachers will need to develop a much deeper understanding of
elementary school mathematics than has been traditional.
Copyright 4
0.2
CHAPTER 0
Comments and Advice on the Class Activities
The Class Activities were developed in tandem with the text as a means to help
students engage actively with the ideas during class. They are intended to be used
flexibly to meet the needs of your students and to give you a choice of things to do in
class. I also hope they will inspire your own inventiveness and enthusiasm for teaching
the material.
It is unlikely that you will have time to do all the Class Activities in class in the
sections you cover, so most likely, you will have to pick and choose. I have labeled
the activities that I view as most central and important with the “core” symbol to
help you. Even so, please don’t feel that you must use all these core activities. It’s
also fine not to do all the problems in a Class Activity. In many cases, there will be
more problems in an activity than you have time for, especially if you have vigorous
class discussions. Also, the Class Activities are not all of the same length. Some will
be quick and won’t require a lot of discussion, others will require more thought and
discussion and therefore more time.
It usually makes sense to precede a Class Activity with a short lecture on the
underlying principles or concepts that will be addressed in the activity. After a short
introduction, have the students begin to work on one or several problems in the Class
Activity. I like to use a “think, pair, share” approach. First ask students to think
individually about a problem in an activity (or perhaps the activity as a whole), then,
when the students are ready, or after a predetermined amount of time, ask students to
share their results or questions with a neighbor. Finally, have a whole-class discussion
about the problem. I like to proceed through the activities one or two problems at
a time rather than having students do the whole activity before we discuss it. By
proceeding a problem or two at a time, students are less likely to get bogged down
and are more likely to spend class time productively. At the end of the activity,
summarize the findings and draw students’ attention to key points.
You can also assign Class Activities to do outside of class and then discuss students’ solutions and questions in class. I find that this makes efficient use of class
time.
How can you lead a discussion on a problem? You can begin such a discussion by
asking for a volunteer to present a solution (either at the board or orally). Rather
than stating whether the solution is correct or not, you can ask the other students
if they have any questions or comments about the presentation. Or you can have
several students put their solutions on the board (especially if you believe they have
somewhat different approaches) and then ask the class to discuss the collection of
solutions. In my experience, some of the most fruitful discussions occur when several
5
different explanations can be examined, compared, and contrasted.
It will be important for you to set a classroom climate in which students are
expected and encouraged to discuss and critique each other’s work and feel comfortable doing so. You will need to actively promote and encourage such discussion, for
example, by stating that in class, initial correctness is not important, and that the
opportunity to discuss proposed solutions and to deliberate whether or not they are
correct offers valuable learning opportunities. (It’s a good idea to warn students that
you do expect to see correct arguments on tests.) Explain that teachers must be good
at discussing mathematics and at evaluating solutions proposed by their students, so
that for students who intend to become teachers, active participation in class discussions and careful listening to proposed solutions is especially important. Regularly
encourage your students to think carefully about whether proposed arguments make
sense or not and give them time to think, decide, and comment.
Needless to say, it’s important never to belittle students’ understanding of mathematics and to assure students that it’s ok to make errors during class discussions.
When students have finished asking questions and making comments about a proposed solution, you can either ask your own questions and make your own comments,
or you might first ask if anyone had a different way of solving the problem or explaining their solution. A major benefit of class discussions is that you will learn a lot
about how your students are thinking about the material and you will be better able
to gauge how you might push their thinking further.
A word of warning: even if your class has had an excellent discussion and you
pulled the activity together at the end by summarizing the findings and drawing
students’ attention to key points, do not assume that everything is crystal clear in all
the students’ minds. In my experience, follow-up reading, written homework, periodic
quizzes or tests, and a comprehensive final exam are invaluable in helping students to
solidify their understanding. Should you assign the reading before or after you do the
related activities? I think that you can do it either way, but I prefer to have students
do the activities before I assign the related reading. For this reason I have generally
placed the activities before the related portions of the text. I think the text is best
used to help students solidify their ideas after they have begun to explore the ideas
in class.
Although my colleagues and I never took mathematics classes that used “think,
pair, share” work on class activities, we have been successful in teaching this way by
using these Class Activities. If you are used to lecturing (as I had been), then it may
take some time to get used to “think, pair, share” and class discussions. But I think
you will agree with me that discussing mathematics helps students learn mathematics,
and that class discussions and the opportunity for students to express their thinking
Copyright 6
CHAPTER 0
will help you monitor your students’ progress.
One final comment about the Class Activities: please don’t be a slave to them!
You are welcome to modify them, to use them as a starting point in your instruction,
or as an inspiration for something different. The goal is to help the students learn
the material, not to “cover” activities. I think it’s important for instructors to be
actively engaged in their teaching and to be thinking about how to make the course
material interesting for their students and themselves. No set of activities, no matter
how good, can substitute for your active engagement in and enthusiasm for helping
your students learn!
0.3
Comments and Advice on the Practice Exercises and Problems
Most students have little or no experience explaining why mathematics problems are
solved the way they are, and therefore require time, patience, and effort to learn to
do this. As the instructor, you must be patient and persevere. I think it works well to
give frequent (e.g., daily) small assignments with prompt feedback. You may need to
assign fewer problems to be turned in as homework than you are used to, otherwise
the grading can become overwhelming. I like to hold students responsible for doing
the practice exercises by giving closed-book quizzes and tests with similar problems.
Remind students that the answers to the practice problems provide them with many
examples of good explanations of the sort they should learn to write, so that studying
these explanations will help them improve their own.
0.4
Grading
Recently I’ve been grading students’ work on a 10 point scale. I give extra points for
truly exemplary work. I describe the scores to my students as follows:
7
# of points description
10.5
exemplary
10
very good
9
8
competent
7
6
basic
4
emerging
0
no credit
characteristics
work that could serve as a model for other students
correct work that is carefully thought out and thorough
work that contains only a minor flaw
good, solid work that is largely correct
work that has merit but also has some shortcomings
work that has merit but also has significant shortcomings
work that shows effort but is seriously flawed
no work submitted or no serious effort shown
I determine the score on assignments and tests by the extent to which the work
meets the following criteria, which I give to students:
• The work is factually correct, or nearly so, with only minor, inconsequential
flaws.
• The work addresses the specific question or problem that was posed. It is focused, detailed, and precise. Key points are emphasized. There are no irrelevant
or distracting points.
• The work could be used to teach a student: either a child or another college
student, whichever is most appropriate. Thus points that are difficult or likely
to be confusing are attended to with sufficient care and elaboration.
• The work is clear, convincing, and logical. An explanation should be convincing
to a skeptic and should not require the reader to make a leap of faith.
• Clear, complete sentences are used.
• Mathematical terms and symbols are used correctly and as needed.
• If applicable, supporting pictures, diagrams, or equations are used appropriately
and as needed.
• The work is coherent.
I’m a firm believer in the importance of a comprehensive final exam, which I think
is essential for helping students to solidify their understanding.
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0.5
CHAPTER 0
Sample Syllabi
A Three Semester Sequence
At the University of Georgia, we use this book for a 3 semester sequence:
Semester 1 A course on numbers and operations covering chapters 1 – 5 and sections
6.1 – 6.3. This is a lot of material and I find I have to go more quickly than I
would like to.
Semester 2 A course on geometry and geometric measurement covering chapters 10
– 14.
Semester 3 A course on algebra, number theory, probability and statistics covering
sections 6.4 – 6.6, chapter 7, most of chapter 8 (usually omitting section 8.6,
except for Class Activity 12U on Pattern Tiles and the Irrationality of the
Square Root of 3), chapter 9, and chapters 15, 16.
A Four Semester Sequence
If you have 4 semesters, you will have time to cover numbers and operations to the
depth they deserve and you will have more time for problem solving. I suggest the
following:
Semester 1 Course 1 on numbers and operations covering chapters 1 – 5.
Semester 2 Course 2 on numbers and operations covering chapters 6 – 8.
Semester 3 A course on geometry and geometric measurement covering chapters 10
– 14.
Semester 4 A course on algebra, probability, and statistics covering chapters 9, 15,
and 16.
A One Semester Course in Numbers and Operations
If you only have one semester in which you must cover the material of chapters 1 – 8,
then the first thing I recommend is to lobby for an additional course! One semester is
not enough for all this important material, which elementary teachers need to know
well. Here is what I recommend:
9
• Chapter 1: Do sections 1.1 and 1.2 only.
• Chapter 2
• Chapter 3: Skip section 3.5 on adding and subtracting negative numbers.
• Chapter 4
• Chapter 5: Do section 5.1 only.
• Chapter 6: Do sections 6.1 – 6.3 only.
• Chapter 7: Do sections 7.1, 7.2 only.
• Skip chapter 8.
A Course on Proportional Reasoning
If you want to teach a course that focuses on ratios and proportions as they apply in
a range of settings, consider using the following sections:
• Section 2.5, Percent
• If time: Section 5.1, Multiplying Fractions, preceded by a brief discussion of the
(general) definition of multiplication in Section 4.1
• If time: Sections 6.4, 6.5 on Dividing Fractions, preceded by a brief discussion
of the (general) definitions of division in Section 6.1.
• If there is not time for Sections 5.1 and 6.4, then discuss the definitions of
multiplication and division in sections 4.1 and 6.1 briefly. It is essential to
understand how to reason with these definitions since “proportional reasoning”
is essentially reasoning about multiplication and division.
• Chapter 7
• Section 14.4 on Similarity.
• If time: Section 14.5, Areas, Volumes, and Scaling
• The part of Section 15.1 on using random samples.
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CHAPTER 0
Whole Number Arithmetic First, Then Fraction Arithmetic
If you prefer doing whole number arithmetic before you get into fractions, I suggest
the following sequence. It’s too much material for one semester, but can be turned
into a two semester sequence by adding material from chapters 9, 15, and 16 or by
spending more time on problem solving or on pedagogy.
1. Whole number arithmetic and number theory:
• Chapter 1, Numbers and the Decimal System: Section 1.1.
• Chapter 3, Addition and Subtraction: Sections 3.1 – 3.3.
• Chapter 4, Multiplication: All sections, 4.1 – 4.6.
• Chapter 6, Division: Sections 6.1 and 6.3.
• If time: Chapter 8, Number Theory: Sections 8.1 – 8.5.
2. Fraction and decimal arithmetic:
• Chapter 1, Numbers and the Decimal System: Sections 1.2 and 1.3. Include section 1.4 if there will be time.
• Chapter 2, Fractions: All sections.
• Chapter 3, Addition and Subtraction: Section 3.4.
• Chapter 5, Multiplying Fractions and Decimals: Sections 5.1 and 5.2.
• Chapter 6, Division: Sections 6.4 – 6.6.
• If time: Chapter 7 on Proportional Reasoning.
0.6
Advice to Give Students Who Are Struggling
Some students who are used to simply following and memorizing step by step processes
in mathematics have a hard time learning to give explanations. Here are some things
you might discuss with these students.
It’s important for prospective teachers to realize that today’s expectations for
mathematics in elementary school and for teaching to today’s standards call for a
focus on reasoning and sense-making and for communicating that reasoning. Teachers
will be responsible not just for showing students how to carry out mathematical
procedures but also for helping students understand the underlying mathematical
ideas and why the procedures make sense. There is a much greater conceptual focus
11
and teachers will need to be able to lead discussions about the ideas and reasoning
that is involved in mathematics. It’s this same kind of conceptual focus on reasoning
and sense making that we are after in mathematics courses for teachers. Students
need to realize that their old approaches to mathematics may no longer be adequate if
they had previously only focused on copying what the teacher did rather than trying
to really understand the ideas. Such students need the recognition, willingness, and
patience to do the hard work it takes to learn the reasoning behind the mathematics
and to focus on understanding.
Here are some specific steps I recommend for students:
• The student must take responsibility for his or her own learning. She has to
convincer herself that she’s the one who has to actually understand the ideas.
There are no shortcuts to this. It takes work, time, commitment, and patience!
• The student must actively engage in the class discussions and class work and
should use class time to think about the ideas and the reasoning being discussed.
Most instructors who use this book structure the class time so that students
will spend time thinking through how to solve a problem and discussing their
ideas with a small group or one other student. This provides a valuable learning
opportunity—if students take it. Some students make the mistake of not taking
advantage of this time to really think, but instead they wait for someone to tell
them the answer. This will not lead to success in understanding the ideas. And
the student then feels they haven’t been taught the material.
• The student should read the relevant sections of the textbook in depth for
understanding and monitor comprehension. He might look into “comprehension
monitoring” strategies. One that might help is to repeatedly ask “why?” while
reading. “Retrieval practice” is another effective study method.
• The student should do all the practice exercises at the end of the relevant
sections. After working out the problem, the student should then look at the
solutions and compare critically.
• When writing up homework, the student should write a draft, read it critically,
and revise. Repeat if necessary!
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0.7
CHAPTER 0
How Is Teaching from this Book Different from
Teaching Standard Math Major Courses?
If you are used to teaching standard math major courses, teaching a math course for
prospective elementary teachers will probably require some adjustments on your part.
I think that it is more valuable for prospective teachers to learn to give clear “explanatory arguments” as opposed to fully rigorous, complete proofs that fit within
an axiomatic development. For example, whereas in more advanced mathematical
treatments we usually take the standard procedure for fraction multiplication as a
definition, and we check that this definition makes sense because it agrees with multiplication for the integers and turns the rational numbers into a field, a different
treatment of fraction multiplication will be more useful for prospective elementary
teachers. These teachers should be able to give examples of word problems for a
given fraction multiplication problem, and they should be able to explain, in the
context of a story or scenario, why the answer provided by the standard fraction
multiplication procedure agrees with what one expects. Thus the focus here is on
giving legitimate mathematical arguments that focus on sense-making rather than on
a rigorous axiomatic development of the mathematics.
Another difference between the explanations that this book intends to encourage
and proofs developed in more advanced mathematics courses is that teachers should
ideally know several different kinds of explanations. Teachers should know multiple
explanations because they are charged with bringing diverse students to high levels
of learning. A teacher who only knows one way to think about the material will not
be well prepared for this difficult task.
Copyright Chapter 1
Numbers and the Base-Ten System
13
14
CHAPTER 1
1.1
The Counting Numbers
Go online to My Math Lab for another activity for this section:
Show the Values of Base-Ten Units.
Class Activity 1A: The Counting Numbers as a List
1. Common errors that children make (and are shown here) are: putting numbers
in the wrong order, skipping one or more numbers, repeating numbers.
In the correct list each counting number appears exactly once and in the correct
order.
2. You could use this question as a lead in to the next activity, which answers it.
Class Activity 1B: Connecting Counting Numbers as a List
with Cardinality
1. No, see the next two parts.
2. You can introduce the terminology “one-to-one correspondence” when you discuss this part. When objects are counted correctly, a one-to-one correspondence
is made between the objects and the relevant part of the list of counting numbers.
3. Although we don’t know for sure, Child 2 may not understand that the last
number word said also indicates how many objects there are. It’s usually surprising to adults that this is a separate piece of knowledge!
5. To be fully fluent, as demonstrated by Child 3, children must be able to go back
and forth between the “list view” of the counting numbers and cardinality. After
establishing that there were 6 bears initially, the child must be able to think
about the number 6 in the list, realize that the next number in the list is 7, then
go back to the cardinality point of view to understand that there are 7 bears.
15
Class Activity 1C: How Many Are There?
Includes IMAP video opportunity
I recommend leading into this activity by raising the issue about how to represent
ever larger numbers as discussed in the text.
Students should gradually come to see that we can determine how many toothpicks
there are by bundling them into tens, and then bundling bundles of 10 into hundreds
(and perhaps even bundling bundles of hundreds into thousands). Most students find
it interesting to “rediscover” the base-ten system.
Instead of working through parts 1 through 4 step by step, you can just pass out
a bunch of toothpicks and ask the students to come up with ways of determining
how many toothpicks there are without counting them one by one. You might let
them discuss their ideas in pairs first, then ask them to discuss as whole group. Let
them present all their ideas, then gradually steer the students toward grouping the
toothpicks. When the time seems right, give a short lecture on the base-ten system
and place value.
Parts 5 and 6 might be best done after a brief lecture on the decimal system. For
5, it’s better to say “2 tens” because that draws attention to a ten as a single unit and
that the digit 2 really does stand for 2 of something (just not 2 individual counters,
but 2 bundles of ten counters). Then you might like to show and discuss IMAP video
(number 5 on the DVD), Zenaida, who answers questions on how many ones and tens
there are in various numbers. In particular, she notes, correctly, that there are 32
ones in 32. So asking “how many ones are in 32?” is different from asking “what is
in the ones place of 32?”.
For part 6, students should show 1 bundle of 100, 3 bundles of 10, and 7 individual
toothpicks. Although their pictures may not show this detail, a bundle of 100 should
be thought of as 10 bundles of 10, in keeping with the structure of the decimal system.
At this point, you could watch a segment of the IMAP video (number 7 on the
DVD), Talecia, to emphasize the importance of understanding the value of each place
as 10 of the previous place’s value. The video shows a student trying to add two
3-digit numbers. The student is confused about what 10 hundred is. To save time,
you might want to start watching the video at time 2:24, when the student adds 600
and 400 and gets “10 hundred.”
For part 7, the problem is that there are more than 9 items in each “place”. We
must bundle 10 of the individual toothpicks to make another bundle of 10 and then
bundle 10 of the bundles of 10 to make a bundle of 100, making a total of 1 bundle
of 100, 3 bundles of 10, and 5 individual toothpicks, which is 135 toothpicks.
Follow this activity with a discussion of the place value drawings shown in the
Copyright 16
CHAPTER 1
text. Perhaps also show some drawings such as the ones in Figure 1.1.
For 2-digit
numbers:
OR:
This is too cumbersome, but
it is a natural starting place.
OR:
These two types of representations have
helped first graders learn to add and subtract
with regrouping.
For numbers with 2 or more digits:
100
10
10
Figure 1.1: Math Drawings to Represent Numbers in the Decimal System
1.2
Decimals and Negative Numbers
Class Activity 1D: Representing Decimals with Bundled Objects
IMAP Video opportunity
In the second half of the IMAP video (number 9 on the DVD) , starting at around
1:04, Megan and Donna are asked to let a “long” base ten block (ten cubes joined
together lengthwise) stand for 1 and are asked to represent 1.8. They do so by using
a small cube to represent a decimal point and use 8 more longs to represent the 8
17
tenths. This representation does not fit with or show the structure of the decimal
system, because tenths are not represented in a way that 10 of them make 1.
1. For 0.034 draw 4 individual paper clips and 3 bundles of 10.
For 0.134 add a bundle of 100.
For 0.13 drop the 4 individual paper clips.
2. For 0.0028 draw 8 individual beads and 2 bundles of 10.
For 0.012 drop the 8 individual beads and add a bundle of 100.
3. 137 if a single toothpick represents 1 (note that whole numbers are also “decimals”)
13.7 if a single toothpick represents
1.37 if a single toothpick represents
1
.
10
1
.
100
1370 if a single toothpick represents 10. Other answers are possible, of course.
Class Activity 1E: Representing Decimals as Lengths
The idea behind this activity is that the number 1.234 can be physically represented
by one 1 unit strip, two 0.1 unit strips, three 0.01 unit strips, and four 0.001 unit
strips. The remaining numbers are represented in a similiar manner. Emphasize that
the value of each place in a decimal number is ten times the value of the place to its
right and that we can see this relationship in the lengths of the strips.
Instead of having students cut the strips out of the back of the manual, you might
prefer to make sets of strips out of card stock that students could use in pairs. You
could also make a large set to stick up on the board (with tacky clay). Or you could
have lengths of plastic tubing cut (and use washers for the thousandths) as described
in the reference given in the text.
Class Activity 1F: Zooming In on Number Lines
There is a blackline master on page 278 for use with zooming in. You might want
to copy it onto a transparency to demonstrate several examples of zooming in. Or
make enough copies for groups of students to show examples of zooming in. Then
have some groups present their examples to the class on the overhead.
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CHAPTER 1
The issues on saying and writing decimals that are discussed in the text will
be good ones to bring up in the context of this and the next activity, or before
starting these activities. In particular, discuss that we can always put extra zeros
after numbers to the right of the decimal point. For example, 3.2 = 3.20 = 3.200.
1. See Figure 1.2.
3
3.2
3.29
3.299
3.1
3.21
3.291
3.2
3.3
etc.
3.9
4
3.22
3.23
etc.
3.29
3.3
3.292
3.293
etc.
3.2991 3.2992 3.2993
etc.
3.299
3.30
3.2999
3.300
Figure 1.2: Zooming In
2. When observing the location of 3.2996 on each number line, point out that every
number line has lots and lots of decimals on it, including those with many digits
down to small decimal places, even if we don’t see the relevant tick marks on
the particular number line.
19
The number 3.2996 lies between the whole numbers 3 and 4, between the tenths
3.2 and 3.3, between the hundredths 3.29 and 3.30, between the thousandths
3.299 and 3.300. Discuss with the students that the decimal places farther
and farther to the right give more and more detail about where, specifically, a
number is located.
3. See Figure 1.3. If students get stuck, tell them to think about “zooming in”.
Students may make the error shown in the next problem.
4.8
4.81
4.82
4.83
etc.
4.89
4.9
4.89
4.891
4.892
4.893
etc.
4.899
4.9
4.8991 4.8992 4.8993
etc.
4.8999 4.9
4.899
Figure 1.3: Labeled Number Lines
4. The problem with this labeling is that to fit with the structure of the decimal
system, if the short tick marks are tenths, then the long tick marks should be
whole numbers, or if the long tick marks are tenths, then the short tick marks
should be hundredths. The long and short tick marks should fit with the way
the value of each decimal place is divided into 10 equal parts in the decimal
system.
5. See Figure 1.4.
Class Activity 1G: Numbers Plotted on Number Lines
1. The point labeled A was positioned by the computer to be at 1.4263. Of course
it’s not possible to tell by eye, or even by measuring that the point is exactly at
this location. Students should be able to tell the the decimal representation is
close to 1.4 because it is between 1 and 2 and less than 1.5. They might even be
able to tell that it’s roughly 1.42 or 1.43, but in any case less than 1.45. So the
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CHAPTER 1
4.9
4.90
4.91
4.92
4.93
4.901
4.902 4.903
4.900 4.9001 4.9002 4.9003
etc.
etc.
etc.
4.99
5.0
4.909 4.91
4.9009
4.901
Figure 1.4: Labeled Number Lines
decimal representation must be of the form 1.3. . . ... or 1.4. . . . The numbers
1.3915, 1.4, 1.4263, and 1.43 are all plausible. The number 1.18 is too small to
be A. The numbers 1.861, and 1.644 are too large to be A.
2. The first number line should be labeled 23 = 23.0, 23.1, 23.2, . . . 24 = 24.0.
The second should be labeled 0.03 = 0.030, 0.031, 0.032, . . . 0.039, 0.04 = 0.040.
The third should be labeled 0.0 = 0.00, 0.01, 0.02, . . . 0.09, 0.1 = 0.10.
The last should be labeled 7 = 7.0, 7.1, 7.2, . . . 7.9, 8 = 8.0. Note that on the
last number line, 7.0095 will be almost indistinguishable from 7.
3. First number line: the long tick marks are 9.616 and 9.617.
Second number line: the long tick marks are 9.6163 and 9.6164.
Third number line: the long tick marks are 9.61 and 9.62.
Fourth number line: the long tick marks are 9 and 10.
Fifth number line: the long tick marks are 9.6 and 9.7.
Class Activity 1H: Negative Numbers on Number Lines
1. Katie’s and Matt’s numbers are decreasing to the right, not increasing. A
negative number such as −7 is 7 units away from 0, but to the left of 0. So
−8 and −7.2 have to be farther away from 0 than −7. It might help to review
the location of numbers such as −1 and −2 on a number line that also shows
0 on it. Parna’s labeling does show −8 as farther from 0 than −7, but her
21
labeling doesn’t fit with the decimal structure that is indicated by the heavier
and lighter tick marks. The left-most tick mark could be labeled as −8 or as
−7.1 (or as −7.01 or −7.001 etc.). In the latter cases, it would make sense to
rewrite −7 as −7.0 (or as −7.00 or −7.000 etc.).
2. The heavy tick mark on the far left is −1. The numbers between 0 and 1 are
all of the form 0. ∗ ∗ ∗ ∗ . . . and the numbers between 0 and −1 are all of the
form −0. ∗ ∗ ∗ ∗ ∗ . . ..
1.3
Comparing Numbers in Base Ten
Class Activity 1I: Misconceptions in Comparing Decimals
In the IMAP video (number 9 on the DVD), up to about time 0:45 on the clip,
Megan and Donna seem to know how to compare decimals at first, saying that 4.7
and 4.70 are equal because you can add a 0 to 4.7. But then, when they are told it’s
“you can’t add a zero day,” they think 4.70 is bigger than 4.7.
See also IMAP video 8 (on the DVD), in which a third grader and a fifth grader
add decimals and also compare decimals. Both children seem to have the “whole
number thinking” misconception about comparing decimals.
2. correct order: 3.05, 3.25, 3.251, 3.3, 3.4
Whole number thinking: 3.3, 3.4, 3.05, 3.25, 3.251
Column overflow thinking: 3.05, 3.3, 3.4, 3.25, 3.251
Denominator focused: 3.251, 3.05, 3.25, 3.3, 3.4
Reciprocal thinking: 3.251, 3.25, 3.05, 3.4, 3.3
Money thinking: Either 3.05, 3.251, 3.25, 3.3, 3.4 or 3.05, 3.25, 3.251, 3.3, 3.4
3. Answers will vary. Notice that including pairs like the following can be useful:
3.25 and 3.3
3.05 and 3.3
3.05 and 3.25
3.251 and 3.05
3.26 and 3.251
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CHAPTER 1
Students who get the first and second one wrong may be using whole number
thinking.
Students who get the first one wrong and the second one right may be using
column overflow thinking.
Students who get the first two right and the third and fourth one wrong may
be using reciprocal thinking.
Students who get the first three right and the fourth one wrong may be using
denominator focused thinking.
Students who get the first four right and the fifth one wrong maybe be using
money thinking.
Class Activity 1J: Finding Smaller and Smaller Decimals
Students should come to see that no matter what decimal is listed that is greater
than 2, there is always another decimal that is smaller than it. We can find such a
decimal simply by putting 0s out beyond the first non-zero digit that occurs in the
candidate decimal and then putting a 1 in some place after that.
Expect some students to come up with the description given in part 4. This does
not describe a decimal because there is no specific decimal place in which the digit 1
occurs.
Class Activity 1K: Finding Decimals Between Decimals
If students are stumped by a pair such as 1.2 and 1.3 ask them to consider hundredths.
Hopefully they will see that they can write these as 1.20 and 1.30. If not, remind
them of this.
Students should come to see that no matter what pair of decimals is given, there
is always another decimal in between.
Be prepared for the question: “What about 0.9999999. . . and 1?” The fact that
these numbers are equal is explained in the chapter on number theory. You may wish
to discuss that material if there is time.
Be prepared for someone to talk about a number that is “right next to” another
number (e.g., the number that is “right next to 1.1.” Ask what “right next to” would
mean. The number would have to have some decimal representation. What would
that be?
23
Class Activity 1L: Decimals Between Decimals on Number
Lines
You might want to remind students that they can always put extra zeros after numbers
to the right of the decimal point. For example, 3.4 = 3.40 = 3.400.
1. See Figure 1.5.
1.6
1.65
1.7
Figure 1.5: A Labeled Number Line
We can think of 1.6 and 1.7 as represented by $1.60 and $1.70 respectively.
$1.65 is in between.
2. See Figure 1.6.
12.540
12.5405
12.541
3. See Figure 1.7.
2.781
2.7342
2.74
2.75
2.76
2.77
2.78
4. See Figure 1.8.
5. The number −1.91 is one example. Going to the left, consecutive long tick
marks could be −1.9 then −2 = −2.0, then −2.1, then −2.2 etc. The number
−1.91 is one small tick mark to the left of −1.9.
You could ask students to write inequalities comparing the three numbers. They
could write −2 < −1.91 < −1.9.
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CHAPTER 1
23.995
23.99
24 =
24.00
6. The number −3.4891 is one example. Going to the left, consecutive long tick
marks could be −3.489, then −3.490 = −3.49, then −3.491, then −3.492, etc.
The first small tick mark to the left of −3.489 is then −3.4891, the second is
−3.4892, etc.
You could ask students to write inequalities comparing the three numbers. They
could write −3.49 < −3.4891 < −3.489.
Class Activity 1M: “Greater Than” and “Less Than” with
Negative Numbers
1. For one explanation you can use a number line, for another, view negative
numbers as amounts owed.
2. Use a number line or view the negative numbers as amounts owed.
3. (a) −6 < −5.9; (b) −4.777 > −4.8; (c) −0.555 < −0.1. These can be explained
with a number line, or by determining how the numbers compare without the
negative signs and then reversing the comparison (because of the mirror symmetry across 0).
1.4
Rounding Numbers
Go online to My Math Lab for two additional activities for this section:
Why Do We Round,
Explaining Rounding by Zooming Out.
Class Activity 1N: Explaining Rounding
1. A rounds to 7300 because that is the hundred it is closest to. B rounds to 7400
because that is the hundred it is closest to.
25
The base-ten representation of A could be 7334 or thereabouts. The base-ten
representation of B could be 7385 or thereabouts.
2. Numbers on the left half of the interval between 7300 and 7400. These numbers
are characterized by having a 0, 1, 2, 3, or 4 in the tens place.
3. Numbers on the right half of the interval between 7300 and 7400. These numbers
are characterized by having a 5, 6, 7, 8, or 9 in the tens place. By (usual)
convention, the midpoint 7350 rounds up to 7400.
4. The digit in the tens place tells us whether the number is on the left or right
half of the interval between two hundreds, and therefore it tells us which way
to round.
Class Activity 1O: Rounding with Number Lines
1. The number 38721 is between 38000 and 39000 and closer to 39000, so rounds
to 39000.
2. The number 5643 is between 5640 and 5650 and closer to 5640, so rounds to
5640. The remaining tick marks should show 5640 and 5650 (in addition to two
other tens, such as 5630 on the left and 5660 on the right).
3. The number 2.349 is between 2.34 and 2.35 and closer to 2.35, so rounds to
2.35.
4. The number 2.349 is between 2.3 and 2.4 and closer to 2.3, so rounds to 2.3.
5. Label the tick marks 100, 200, 300, . . . , 54700, 54800, 54900.
6. Label the tick marks 0.01, 0.02, 0.03, . . . 16.92, 16.93, 16.94.
7. Label the tick marks 0.1, 0.2, 0.3, . . . , 16.8, 16.9, 17.0.
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CHAPTER 1
Class Activity 1P: Can We Round This Way?
Maureen’s method is not a valid way to round even though it does give the correct
answer in this case. If Maureen used her method to round 1.2487 to the nearest tenth
she would give the answer 1.3. But the correct answer is 1.2 since 1.2487 is closer to
1.2 than to 1.3 because it is less than 1.25.
As an extension, you might ask students to determine when Maureen’s method
will give the correct answer and when it won’t.
Any time there is a 4 followed by a digit 5 or greater in the place to the right of
the place to be rounded, Maureen’s method will fail.
Class Activity 1Q: What Does “Population 120,000” Really
Mean?
1. These numbers are between 115,000 and 125,000 (up to, but not including the
latter). On the number line, they are the numbers in the interval that starts
half way between 110,000 and 120,000 and goes to half way between 120,000
and 130,000.
2. We should assume that the number has been rounded to the nearest tenthousand because there are zeros below that place. Therefore the population
could be any whole number between 115,000 and 125,000 (but not including
the latter, so up to 124,999) based on part 1.
3. We should assume that the number has been rounded to the nearest hundred
thousand because there are zeros below that place. The actual population
could therefore be any whole number between 2,650,000 and 2,750,000 (but not
including the latter).
Copyright Chapter 2
Fractions and Problem Solving
27
28
CHAPTER 2
2.1
Solving Problems and Explaining Solutions
You might like to assign this section as reading but weave a discussion of this section
into the next section. Although the norms of mathematical reasoning and explanation are certainly based in our natural capacity for logical thought, these norms are
culturally transmitted, and students must learn what qualifies as valid mathematical
reasoning. This takes many examples as well as guidance from instructors.
2.2
Defining and Reasoning About Fractions
The Whole Associated With a Fraction
Class Activity 2A: Defining Fractions
I like to have students do part 1 before I even give our definition of fraction. Once
they see the limitation of this “A out of B” definition, they are interested in learning
a different definition.
1. It doesn’t make sense to talk about 4 parts out of 3 equal parts because there
are only 3 parts making the whole. Also, this “A out of B” language seems
to cause some students to focus on the remaining part of the whole when the
fractional part is “taken out of the whole,” i.e., removed from the whole. This
misconception is discussed in IMAP Video 12 (on the DVD; 345 on the MML
list, I think). In particular, the discussion at the end explains why the “A out
of B” language may cause problems for students.
2. To explain why an amount is 13 of the whole (or unit amount), students can
explain that 3 copies of the amount makes the whole, or that the amount was
obtained by dividing the whole into 3 equal parts.
Emphasize that students should describe 23 as 2 parts, each of which is 13 of
the whole (or unit amount). Students should use similar wording for the other
non-unit fractions.
29
Class Activity 2B: Fractions of Things
During the discussion of these problems, be sure students see that an amount can
be described with different fractions depending on what the unit amount (whole) is
taken to be. This issue is explored further in the activity after this one.
1. Show 43 of the piece of paper because you want to show 3 parts, given that the
piece of paper is 4 parts.
2. Benton should use 31 of his dough, which he can get by cutting the dough into
3 equal parts either horizontally or vertically. He needs 13 of the dough because
the 34 cup of butter consists of 3 parts (each of which is 41 cup) but he only
wants 1 of those parts.
3. Because 32 is 2 parts, each of which is 31 of the other whole design, the hexagon
should be thought of as 2 parts. Two trapezoids make the hexagon, so a trapezoid is 13 of the other design. Therefore the other design’s area can be made from
3 trapezoids (which can also be made from various combinations of triangles,
rhombuses, and trapezoids).
If the hexagon is 32 of the area of another design, then it is 3 parts, each of
which is 21 of another design. Three rhombuses make the hexagon, so the other
design’s area can be made from 2 rhombuses (or 4 triangles, rhombuses or a
trapezoid and a triangle).
As an extension, ask students to make a design, then make a fraction of the
design and give it to another group as a challenge to make their original design.
Class Activity 2C: Relating Fractions to Wholes
Note that identifying the whole that a fraction is “of” will be especially important
later in analyzing fraction multiplication and division (in chapters 5 and 6) because
in fraction multiplication and division word problems, different fractions in a problem
can refer to different wholes. So during the discussion of this activity you might want
to tell students that they will be encountering the “what’s the whole?” question later
on as well.
We have to specify the whole in order to interpret unambiguously pictures that
represent fractional amounts. Students will see this point in this activity; it is especially important when it comes to improper fractions and mixed numbers. In your
Copyright 30
CHAPTER 2
class discussions, you might encourage students to draw a separate picture of the
whole and to label the whole as such.
Encourage students to talk about fractions as “of a whole” rather than just as
1
“parts out of parts.” For example, encourage them to say “ 12
of the park” (in addition
to saying “1 out of 12 equal parts”).
You may wish to use the term “unit amount” in addition to “whole” to help
students with the idea that there can be different wholes within one situation. Note
that the Common Core State Standards use the term “whole” in relation to fractions.
1
1. The swing area consists of 12
of the park because it is 1 out of 12 equal parts
1
(after suitably subdividing the park). Note that the whole for this 12
and for
1
1
the 3 is the area of the entire neighborhood park. The whole for 4 is the area
of the playground.
For part (b), the swings’ area can be described both as
1
of the area of the park.
playground and as 12
1
4
of the area of the
You don’t need to bring this up now, but some students might observe that the
swing problem is a word problem for 14 · 13 .
2. Ben should use 12 of the oil in the bottle. Note that the whole for this 21 is the
amount of oil in the bottle. But the whole for the 31 and the 23 is a cup of oil.
For part (b), the
bottle.
1
3
of a cup of oil can also be described as
1
2
of the oil in the
You don’t need to bring this up now, but some students might observe that the
oil problem is a word problem for 31 ÷ 23 .
Class Activity 2D: Comparing Quantities with Fractions
1. Nate raised 32 as much as Tyler. The whole for this 23 is the amount Tyler raised.
Tyler raised 32 as much as Nate. The whole for this 32 is the amount Nate raised.
It is also correct to say that Tyler raised 1 21 times as much as Nate, or even
that Tyler raised 50% more than Nate (although the latter answer doesn’t fit
with the “as much as” wording).
2. The strip for Company A will consist of 4 equal pieces and the one for Company
B will consist of 5 equal pieces of the same size. Company B sells 45 as much as
Company A. It is also correct to say that Company B sells 1 14 times as much
as Company A and even that Company B sells 25 % more than Company A.
31
3. Students should notice that in each case, the two fractions we use to describe the
relationship between two quantities are inverses (or reciprocals) of each other.
A B
This previews the relationship B
· A = 1.
Class Activity 2E: Fractions of Non-Contiguous Wholes
2. Mariah’s method is not correct because the 10 parts are not equal in area. If
the two plots did have the same area, then her reasoning would be correct.
3. Yes, Aysah’s picture can be used. It shows that the shaded portion is 15 of
Peter’s garden because 5 copies of the shaded portion can be joined to make
the whole garden. Or in other words, picture shows that the garden is divided
into 5 parts and one of those parts is shaded (note that a “part” can consist of
several distinct pieces).
4. Matt’s reasoning is not valid because the two 51 each refer to a different whole
and the question is about Peter’s entire garden.
If the two plots were each 1 acre, then the shaded pieces would in fact be
two parts, each of which is 51 of an acre, and therefore the two parts together
would be 25 of an acre. So yes, Matt’s reasoning can be used to make a correct
statement. As always, attending to the whole is crucial when working with
fractions!
At this point, you might like to show IMAP video 14 on the DVD, in which
Felisha divides 2 cookies equally among 5 people. There is some confusion about
2
how to describe each person’s share. Is it 10
or 25 ? The answer depends on what
we take the whole to be. (This video is an even better fit for the discussion of
the relationship between division and fractions in chapter 6.)
Class Activity 2F: Is the Meaning of Equal Parts Always
Clear?
You can use this activity to point out that “messy” shapes or situations add a layer
of difficulty when working with fractions. For this reason, fractions are usually introduced to children with length-based models, such as fraction strips.
Copyright 32
CHAPTER 2
1. In most situations, it’s clear what “equal parts” means. But when we have
different sized objects like these, it’s not clear if all the marbles can be considered
equal.
2. Arianna did divide the shape into 4 parts of equal area even though the 4 parts
are not congruent.
3.
of the shapes used in the design are triangles. 41 of the area of the design
is made of triangles. For (a) the equal parts were the 13 different shapes that
make up the design. For (b) we can divide each rhombus into two triangles that
are congruent to the triangles in the outer ring of the design. The hexagon can
be divided into six more congruent triangles. The outer six triangles are now 41
of the 24 triangles in the design.
6
13
Class Activity 2G: Number Line Errors with Fractions
1. (a) Eric may understand that making fourths involves dividing a unit into 4
pieces, but he may think he can accomplish that by inserting 4 tick marks
between 0 and 1 instead of dividing the interval from 0 to 1 into 4 equal
pieces. Eric has actually divided the interval from 0 to 1 into 5 equal
pieces.
(b) Kristin may simply be counting tick marks. She may not understand that
the interval between 0 and 1 must be divided into 4 equal pieces. Her
number line doesn’t show 0. Although not incorrect, it probably means
she is not attending to the distance from 0.
2. Tyler may have thought of 34 as 3 intervals out of the 4 intervals drawn on the
number line. Clearly he is confused about exactly what is the whole.
3. See text for the idea of circling intervals. Students may think of other ways of
highlighting the intervals.
Class Activity 2H: Fractions on Number Lines, Part 1
1. Be sure students talk about dividing the interval between 0 and 1 into 4 equallength segments, and in particular, focus on length. It will be worth noting that
dividing the interval from 0 to 1 does not involve making 4 new tick marks. See
33
text for full discussion of where to locate fractions on number lines and see the
answer to the practice problem in this section. You might encourage students
to put a strip labeled “1 whole” above the interval from 0 to 1.
2. As the tick marks should be 14 apart, draw 3 equally spaced tick marks between
the tick marks for 0 and 1 (so as to divide the interval between 0 and 1 into 4
equal pieces) and then continue on.
the tick marks for 0 and 23 , so as to subdivide the interval between 0 and 23 into
3 equal pieces, and then continue on. Note that students must recognize that
3
consists of 3 pieces, each of length 12 .
2
the tick marks for 0 and 43 and then continue on.
the tick marks for 0 and 54 and then continue on.
Class Activity 2I: Improper Fractions
1. If the two large rectangles together are taken as the whole or unit amount, then
the shaded portion is 85 of the whole. To interpret the shaded region as 45 we
must state that the whole consists of one large rectangle. You could suggest to
students that they draw another rectangle and label it with “1 whole” in order
to make clear what the whole is.
2. You might discuss with Enrico that we can use copies of parts. You could
emphasize our definition of fractions, 5 parts (or copies of parts), each of which
is 41 of the whole. You could show him that it makes sense to have 45 cups of
water, or a string of length 45 of an inch.
3. Meili might think that the 6 pieces together must make the whole since it is
larger than the 5 pieces that make the strip that is labeled as the whole. Some
questions to ask Meili are why she changed from fifths to sixths and whether
the whole or unit amount has changed or not. What if we decide not to change
the unit amount? Then what will that last piece be?
Copyright 34
CHAPTER 2
Class Activity 2J: Improper Fraction Problem Solving with
Pattern Tiles
1. Because 54 means 5 parts, each of which is one fourth (of the original design),
we need to find how to break Design A into 5 equal parts. We can do this by
making the design with 5 green triangles. Then 4 of those green triangles will
have the area of the original design (because 4 fourths make the whole). This
area can also be made with a trapezoid and 1 triangle or with 2 blue rhombuses.
2. Because 35 means 5 parts, each of which is one third (of the original design),
we need to find how to break Design B into 5 equal parts. We can do this by
making the area of the design with 5 trapezoids (even though we can’t make
the design itself that way). A design whose area is the same as the area of 3
trapezoids is what we are looking for (because 3 thirds make the whole).
3. Because 97 means 9 parts, each of which is one seventh (of the original design),
we need to find how to break Design C into 9 equal parts. We can do this by
making the area of the design with 9 blue rhombuses (even though we can’t
make the actual design that way). Then a design whose area is the same as the
area of 7 blue rhombuses is what we are looking for (because 7 sevenths make
the whole).
2.3
Equivalent Fractions
Before you have students work on the next Class Activity, you might have them work
with fraction strips first. Give each student 3 long strips of paper (e.g., from a piece
of paper cut into strips lengthwise). Ask students how they could fold the paper
to show fourths of the strip (fold in half and then in half again while still folded).
Have the students fold each strip to make fourths and then unfold and shade 3 of the
fourths to show 43 of each strip.
Next have students fold the second strip back into fourths and then fold in half
again. Ask students to predict the size of the parts (eighths) and ask them to predict
how they will be able to describe the shaded part in terms of those new parts (as 6
eighths).
Have students fold the third strip back up and have them fold and third strips back
into fourths, and then fold in half again, and then again. Ask them to predict the size
of the parts (sixteenths) and how many parts will be shaded (12 sixteenths). Have
students write equations that correspond with what they have found and ask them
35
how the numerators and denominators are related to the numerator and denominator
of 34 .
Class Activity 2K: Equivalent Fractions
Before you do the activity, you might ask students if they can think of ways to explain
why the two fractions are equal. Some students will think of the idea of multiplying
by 1 in the form of 44 . Although this is a nice way to explain fraction equivalence,
it is only suitable for students who have already studied fraction multiplication. In
the Common Core State Standards, students study equivalent fractions (in Grade 4)
before they complete their study of fraction multiplication. Also, standard 4.NF.1
asks students to explain fraction equivalence with visual fraction models (i.e., as in the
activity). Note that a brief comment is made about explaining fraction equivalence
by multiplying by 1 in the text.
1. See text for an explanation using different numbers. You may wish to present an
explanation to students first using different numbers. Then have them explain
this example to each other. Press students to explain the multiplication of the
8
.
numerator and denominator and not just to say that 32 = 12
2. The issue addressed in this part may already come up in your discussion of part
1. Note that it makes sense that when we divide each piece into 4 equal pieces
there will be 4 times as many pieces (in all, and shaded). So the pieces become
smaller and to compensate, there are more of them. The pieces themselves are
divided, but the number of pieces is multiplied. In the picture we are working
with the pieces themselves, but in the numerical work we are working with the
number of pieces.
3. In the math drawing, each piece should be divided into 5 equal pieces.
A
4. Repeat the above with 23 replaced by B
and 4 (or 5) replaced by N. If there
isn’t time to work all the way through this, you could instead remark that there
wasn’t anything special about the numbers 2, 3, 4, 5 used in the previous explanations and that the very same reasoning will apply no matter what counting
numbers are involved.
Copyright 36
CHAPTER 2
Class Activity 2L: Misconceptions About Fraction Equivalence
1. Even though Anna did “do the same thing to the top and bottom of the fraction,” she doesn’t get equivalent fractions this way, which we can see immediately by drawing pictures. When we teach about equivalent fractions, we should
be more specific about how to get equivalent fractions than just saying “do the
same thing to the top and the bottom.”
2. Both fractions are “one part away from a whole” but 17ths and 12ths are different size parts.
3. Both fractions are equal to 1. Peter may not realize that when the whole is
divided into more parts, each part becomes smaller.
A similar issue comes up again in the activity “Can We Reason this Way?” in
the section on comparing fractions.
Class Activity 2M: Common Denominators
1. The two smallest common denominators are 12 and 24. When giving these
fractions common denominators we are dividing the strips and number lines
into like parts.
2. Let the distance between adjacent tick marks equal
mark represent 23 = 64 .
1
6
and let the leftmost tick
99
= 12
.
mark represent 33
4
1
12
and let the leftmost tick
most tick mark represent 0.7 = 14
.
20
1
20
= 0.05 and let the left-
Class Activity 2N: Solving Problems by Changing Denominators
1.
of the larger (imagined) piece of paper is formed by 14 of the given piece of
paper. Since 23 = 64 , and since we want 1 sixth and we have 4 sixths, we must
1
6
37
show 1 out of 4 equal parts of the given piece of paper. In solving this problem,
2
also appears as 46 . We can make 4 equal parts horizontally or vertically, which
3
look different.
2. Using a math drawing, we see that Jean needs 21 = 36 cups of butter but only
has 13 = 26 cups of butter, which means she has 2 out of the 3 parts she needs,
so she has 23 of what she needs. In solving the problem the fractions appear as
equivalent fractions with denominator 6.
You could also ask students to discuss the different wholes that are used in
the problem. The whole associated with 21 and 31 is a cup of butter, but the
whole associated with 23 is the amount of butter Jean needs, namely the 12 cup
of butter.
3. Using a math drawing, we see that Joey should eat 83 of a cup of cereal. In the
drawing, we turn 34 into 86 in order to solve the problem.
You could also ask students to discuss the different wholes that are used in the
problem. The whole associated with the 34 in the problem is a cup of cereal.
The whole associated with the answer, 38 , is also a cup of cereal. But the whole
associated with 12 is a serving of cereal.
Class Activity 2O: Fractions on Number Lines, Part 2
In this activity, students will need to give fractions common denominators to solve
the problems and they will need to focus on the meaning of fractions in terms of
number lines.
8
9
and 34 = 12
, so break the interval
1. Using the common denominator 12, 32 = 12
2
between 0 and 3 into 8 equal parts by drawing seven equally spaced tick marks
between the tick marks for 0 and 32 . The distance between adjacent tick marks
1
equals 12
. If we extend these tick marks past 23 the tick mark for 43 is the first
one past 23 .
2. Draw five equally spaced tick marks between the tick marks for 0 and 21 . The
1
distance between adjacent tick marks equals 10
. If we extend these tick marks
1
3
1
past 2 the tick mark for 5 is the first one past 2
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CHAPTER 2
Class Activity 2P: Simplifying Fractions
1. See practice exercise 10 and its solution for a similar problem. During the
process, the total number of squares (or cubes) does not change and the amount
of colored squares (or cubes) does not change. Only the number of parts and
the size of the parts into which the squares (or cubes) are grouped changes.
2. See a similar figure in the text and the text discussion.
3.
2.4
12
18
=
2·6
3·6
=
2
3
12
18
=
6·2
9·2
=
6
9
=
2·3
3·3
=
2
3
Comparing Fractions
Can We Compare Fractions This Way?.
I like to begin this section by putting two fractions on the board (such as 23 and
and asking students what methods they know for determining which of those two
fractions is greater. We then discuss the general methods and why they work.
3
)
5
Class Activity 2Q: What is another way to Compare these
Fractions?
Use this activity to discuss the method of comparing fractions that have the same
numerator by considering the denominators. At the end of this activity, you could
mention that even if two fractions don’t have the same numerator to start with, we
can give the fractions common numerators to compare them. In some cases this is
easier than giving fractions common denominators.
1. The second fraction is bigger because when you divide an object into 39 equal
pieces, each piece is slightly bigger than if you divided the object into 49 equal
pieces.
2. Use the same reasoning.
39
Class Activity 2R: Comparing Fractions by Reasoning
27
43
>
26
45
because 27 > 26 and forty-thirds are greater than forty-fifths.
34
> 70
because the first fraction is greater than 21 but the second fraction is less
than 21 .
17
19
< 20
because each fraction is one piece less than a whole but 20ths are smaller
18
than 19ths, so the second fraction is closer to a whole than the first.
51
65
< 67
because both fractions are 2 pieces less than a whole but sixty-sevenths
53
are smaller than fifty-thirds, so the second fraction is closer to a whole than the first.
9
12
9
< 44
because 40
< 41 < 12
.
40
44
13
5
<
because
both
fractions
are just over 12 but 85 is one eighth more than 12
25
8
13
is half of a twenty-fifth more than 12 .
whereas 25
13
25
Class Activity 2S: Can We Reason this Way?
In this case, Claire reaches the correct conclusion but her reasoning is not valid
because even though 4 pieces is more than 3 pieces, 9ths are smaller than 8ths. So the
numerators and denominators are actually working against each other in this case.
Ask students to think about other examples and whether Claire will always get
the right answer using her approach. If they can’t find any examples where Claire
4
and 38 and ask
will get the wrong answer, give them the example of comparing 11
what Claire would predict and whether her prediction would be correct. Claire would
most likely predict that
4
3
>
11
8
which is false.
You may wish to show or assign IMAP Video 11 in which Ally talks about changing digits in order to determine which fraction is larger. It’s not necessarily the same
reasoning that Claire and Conrad are using in this activity, but is in the same style.
You could also show or assign Video 12 where prospective teachers discuss a misconception in comparing fractions that many noticed when they interviewed students.
2.5
Percent
You might introduce this section by asking students to jot down what they think
“percent” means and why they think we have percents. You could then discuss the
material at the beginning of the section. You might then ask students why we use
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CHAPTER 2
100 as the denominator for percentages instead of 10, say, or 1000. The denominator
of 100 allows percentages to be expressed with 2 digits, which is short enough to be
quickly grasped yet provides enough information to distinguish and separate many
cases.
Class Activity 2T: Math Drawings, Percentages, and Fractions
This activity gives students a chance to see various percentages as arising from combining other common “benchmark” percentages. This will be useful in mental percent
calculations.
1. Diagram 1: 95%
Diagram 2: 80% (viewed as 75% plus
1
5
of 25%)
Diagram 3: 45%
Diagram 4: 12.5% (half of 25%)
Diagram 5: 87.5% (75% plus half of 25%)
Class Activity 2U: Reasoning About Percent Tables to Solve
“Portion Unknown” Percent Problems
Most students find the “percent tables” to be really helpful for developing a feel for
percentages. (If you used a previous edition of this book, note that I changed the
name from “percent diagram” to “percent table” in order to make clearer later on that
a percent table is really a form of ratio table. And a percent table can also be viewed
as a table for a function.) When we use percent tables, we engage in proportional
reasoning.
1.
1
10
of 80,000 is 8000. So half of 8000, namely 4000, is 5% of 80,000. If we take
4,000 away from 80,000 that will be 95% of 80,000. So the answer is 76,000.
Percent diagram:
100%
10%
5%
95%
−→ 80, 000
−→
8000
−→
4000
−→ 76, 000
41
Some students may also calculate 9 times 8000 and then add 4000 to that, so
it’s a good idea to ask for their reasoning, not just their percent diagrams. Have
students recognize that this is another valid way to compute 95% of 80,000.
2.
1
10
of 6500 is 650. So half of that, namely 325 is 5% of 6500. Since 10% + 5% =
15%, we have that 650 + 325 = 975 is 15% of 6500.
3. 1% of $25 is $.25. Hence 7% of $25 is $1.75.
4.
1
2
of 810 is 405. 10% of 810 is 81. So 60% of 810 is 405 + 81 = 486.
5. 20% of 810 is 162. 40% of 810 is 162 + 162 = 324. Hence 60% of 162 is
324 + 162 = 486.
6.
1
2
of 180 is 90. 10% of that is 9, which is also 5% of 180 (or you can find 5%
of 180 by taking 10% of 180 first and then take half of that). So 55% of 180 is
90 + 9 = 99.
Class Activity 2V: Reasoning about Percent Tables
1. See Figure 2.1 for a picture.
2 boxes are in each of the 3 shaded rectangles,
so there are 20 boxes total
Figure 2.1: Boxes of Paper
Using a percent table:
30% → 6
10% → 2
100% → 20
You could also ask students to reason about equivalent fractions (without cross
multiplying) to solve the problem. To do so, they should solve
30
6
=
100
?
which they can do this way: Since
30
100
=
5·6
5·20
=
6
20
his order must have been 20.
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CHAPTER 2
2. 7% → $2.10
1% → $.30
100% → $30
3. As seen in Figure 2.2, last year, 3 out of the 5 parts of normal rainfall fell, so
3
= 60% of normal rainfall fell.
5
1 inch of
rain
normal
last year
Figure 2.2: Rainfall
Using a percent table, students can reason this way:
100% → 5 eighths
20% → 1 eighth
60% → 3 eighths
Notice that this is another version of “going through 1” because it is going
through 1 eighth.
1
4. Since 14 cup gives 100% of your full daily value of clacium, 12
gives you 33 13 %
4
of your daily value. Hence 13 = 12
of a cup gives you 133 13 % of your daily value.
Class Activity 2W: Equivalent Fractions and Percent
1.
60
400
2.
6
25
3.
6
100
=
=
=
6
40
6·4
25·4
=
=
6÷4
100÷4
3
20
=
24
100
=
15
100
= 15%.
= 24%
1.5
25
= 1.5% With a percent table:
100% → 25
1% → 0.25
6% → 1.50
4.
225
500
=
43
9·25
20·25
=
9
20
=
45
100
= 45%.
An easier way is to get a denominator of 1000 first:
225
500
=
450
1000
=
45
100
= 45%.
You might ask students to solve this using a percent table too. They might see
225 as 25 less than 250, which is 50% of 500. Since 50 is 10% of 500, then 25 is
5%, and so 225 is 50% − 5% = 45%.
5.
12
75
=
12÷3
75÷3
=
4
25
=
16
100
= 16%
6. You may need to discuss with students that just as we can have improper
= 32 =
fractions, we can also have percentages that are greater than 100%. 24
16
150
= 150%.
100
7.
16
24
=
2
3
=
0.666...
1
=
66.66...
100
= 66.66 . . . %. Note that this used “going through 1.”
Class Activity 2X: Percent Problem Solving
1. With a percent table:
40% −→ 30
20% −→ 15
100% −→ 75
With equivalent fractions: we want to solve
30
40
=
100
?
40
4
2
30
=
= =
100
10
5
75
2. Draw two strips to represent the running distances, making Marcie’s 10 pieces
long and Andrew’s 4 pieces long. From the comparison, we see that Marcie’s
distance is 250% as long as Andrew’s.
3. The 8 extra female bugs are as much as 10% of the male bugs. So there are
10 · 8 = 80 male bugs and 88 female bugs. This makes 168 bugs all together.
Using a percent table we might say
10% −→ 8
100% −→ 80
but this 100% refers to 100% of the male bugs. It would be easy to think
incorrectly that this 100% refers to all the bugs.
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CHAPTER 2
4. Draw two strips, one for the dogs, one for the cats. Make the dog strip 25%
longer than the cat strip by making the cat strip 4 pieces long and the dog strip
one piece longer, so 5 pieces long. Then the cats and dogs together are “made
of” a total of 9 pieces and the cats are 4 of those 9 pieces. So the percentage of
cats is 94 , which is about 44%.
Copyright Chapter 3
Addition and Subtraction
45
46
3.1
CHAPTER 3
Interpretations of Addition and Subtraction
Using Strip Diagrams to Decide Whether to Add or Subtract to Solve a Word
Problem.
Class Activity 3A: Relating Addition and Subtraction: The
Shopkeepers’ Method of Making Change
If the patron gives the shopkeeper $A for a purchase of $B, then the shopkeeper must
give the patron $C in change, where $C = $A − $B. When we use the shopkeeper’s
method of making change, we are viewing $C as the amount we add to $B in order
to make $A, in other words we use the equation
$B + $C = $A
which is an addition equation that is equivalent to the subtraction equation
$C = $A − $B
Class Activity 3B: Writing Add To and Take From Problems
Most students will automatically write Add To and Take From problems, which involve change over time, but be on the lookout for other types in case they do occur.
If so, you can point out then that it’s a different type of problem. See the text for
examples.
Class Activity 3C: Writing Put Together/Take Apart and Compare Problems
See the text for examples.
Class Activity 3D: Identifying Problem Types and Difficult
Language
If you will want to extend this activity and have students think about the ways
young children might solve these problems, then you might prefer to do this activity
47
in conjunction with the activities in the next section on children’s learning paths for
single-digit addition and subtraction.
1. Problem type: Add To; Equation: 3+? = 12; Keywords: “more bears.”
2. Problem type: Put Together/Take Apart; Equation: 3+? = 12; Keyword:
“altogether”
IMAP Video clip 2 shows Dillon, a 2nd grader, incorrectly solving problem 2,
which is a similar problem.
3. Problem type: Add To; Equation: ? + 3 = 12; Keywords: “got 3 more bugs,”
“altogether”
IMAP Video clip 2 shows Dillon, a 2nd grader, incorrectly solving problem 3,
which is a similar problem.
4. Problem type: Compare; Equation: ? + 3 = 12 or 12 − 3 =?; Keywords: “3
more red bugs”
5. Problem type: Compare; Equation: 3+? = 12 or 12 − 3 =?; Keywords: “more
red triangles”
6. Problem type: Take From; Equation: ? − 5 = 9; Keywords: “gave away,” “left”
7. Problem type: Compare; Equation: ? − 5 = 9; Keywords: “fewer”
3.2
The Commutative and Associative Properties
of Addition, Mental Math, and Single-Digit
Facts
Go online to My Math Lab for additional activities for this section:
Solving Addition and Subtraction Word Problems: Easier and Harder Sub-Types,
Using Properties of Addition in Mental Math, Writing Correct Equations,
Writing Equations that Correspond to a Method of Calculation.
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CHAPTER 3
Class Activity 3E: Mental Math
I like to use this activity as a springboard for discussing the commutative and associative properties of addition. I have the students talk about how they solved the
problem mentally, and then I write equations for the properties and show that they
actually used these equations even if they didn’t consciously think about it.
One thing to watch out for is that some students may think the commutative
property applies only when 3 terms are added because the examples here use 3 terms.
For adults, it’s really only in the situation of 3 or more terms where the commutative
property is useful. But when you study the “basic addition facts” next in this section,
students will see that young children use the commutative property with only 2 terms
when they “count on from larger.”
1. Combine the 1 with the 7999 to make 8000, then add 857 to make 8857.
2. Combine 98 and 2 to make 100, then add 367 + 100 = 467.
3. Combine the 7 with the 153 to make 160, then add on 19 to make 179.
4. Combine 6.95 and .05 to make 7.00, then add 7.89 to make 14.89.
Class Activity 3F: Children’s Learning Paths for Single-Digit
Addition
2. The “count on from larger” method relies on the commutative property of addition. Children use the fact that 3 + 9 = 9 + 3 (even if they don’t know this
equation) when they replace 3 + 9 with 9 + 3 in order to count on 3 from 9
instead of counting on 9 from 3.
3. The commutative property of addition helps children lighten their learning load,
cutting it nearly in half because once they know 9 + 3, they also know 3 + 9 by
the commutative property.
4. The make-a-ten method relies on the associative property of addition (even if
children are not consciously aware of it). We can see this from equations such
as these:
8 + 7 = 8 + (2 + 5) = (8 + 2) + 5 = 10 + 5 = 15
The associative property was used at the 2nd equal sign.
49
5. The make-a-ten method is especially important because we bundle by tens in
the decimal system. The method emphasizes 10 as a unit.
6. The doubles plus 1 method relies on the associative property of addition:
6 + 7 = 6 + (6 + 1) = (6 + 6) + 1 = 12 + 1 = 13
The associative property was used at the 2nd equal sign.
7. In the example shown, the child had to know the partner to 10 of 8, had to
break 7 into 2 + 5, and had to understand 15 as 10 + 5. Notice that being able
to break 7 into 2 + 5 is developed by Put Together/Take Apart, Both Addends
Unknown problems, as described in the previous section.
Class Activity 3G: Children’s Learning Paths for Single-Digit
Subtraction
The first two make-a-ten methods are especially nice to use when the subtrahend
is relatively large, thus leaving only a small ten-partner to add to the ones part of
the minuend.
The third make-a-ten method is easy when the subtrahend is small.
As an extension, ask students to think about the difference in how they calculate
53 − 2 and 53 − 51 mentally. To calculate 53 − 2 it is easier to think about taking
away 2 from 53 (like the third strategy at Level 3), but to calculate 53 − 51 it is
easier to think about the problem as an unknown addend problem 51+? = 53 (like
the first strategy at Level 3). Thank you to Sarah Donaldson for pointing out this
nice extension to me!
You could show IMAP Video clip 1 on the DVD, in which 3 children solve an
unknown addend problem. The first child solves 13 − 6 with a “direct modeling”
method, the second with a “count on” method, and the third with a derived fact
strategy. This activity only refers to the make-a-ten methods, but see the addition
activity for the “doubles ±1” method.
Class Activity 3H: Reasoning to Add and Subtract
Before doing part 3 of this activity, either show students several examples of how
to write correct strings of equations or have them do the activities “Writing Correct
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CHAPTER 3
Equations” and “Writing Equations that Correspond to a Method of Calculation”
which can be found online in My Math Lab.
1. (a) Anne is right. See Figure 3.1.
move back 100 units
253 - 100
253
153
253 - 99
move back 99 units
Figure 3.1: Subtracting Using a Number Line
(b) Since taking away 100 takes away 1 more than you want to take away, you
must add 1 back to make the answer correct.
2. (a) As Figure 3.2 indicates, the distance between the numbers remains the
same because each number is shifted over by the same amount.
Notice that this is taking a Compare perspective on subtraction, whereas
the number line for the previous part took a Take Away perspective (which
fits with the way we defined addition and subtraction on number lines).
view 253 - 99 as the distance between 253 and 99
99 100
253 254
view 254 - 100 as the distance between 254 and 100
Figure 3.2: Subtracting Using a Number Line
(b) When we add 1 to both numbers we are really adding 1 and subtracting 1
(because we are subtracting an extra 1 from what we had been subtracting).
51
(c) Yes, in this case you can add 2 to both numbers (which really adds 2
and subtracts 2) and replace the problem with the equivalent problem
326 − 300 = 26. You can add the same amount to each number in a
subtraction problem and create an equivalent subtraction problem.
3. (a) 183 + 99 = 183 + 100 − 1 = 283 − 1 = 282
(b) 268 + 52 = 268 + 2 + 50 = 270 + 50 = 270 + 30 + 20 = 300 + 20 = 320
(c) 600 − 199 = 600 − 200 + 1 = 401
(d) 70 + 30 = 100, 100 + 64 = 164, so 164 − 70 = 30 + 64 = 94
(e) 999 + 9999 = 1000 − 1 + 10, 000 − 1 = 10, 000 + 1000 − 2 = 10, 998
(f) $10.00 − $2.99 = $10.00 − $3.00 + 1 = $7.01
(g) Compute 3 + 4 + 2 + 5 − 0.04 = 14 − .04 = 13.96
3.3
Why the Standard Algorithms for Adding and
Subtracting Numbers in the Base-Ten System
Work
Subtracting Across Zeros.
Class Activity 3I: Adding and Subtracting with Base-Ten Math
Drawings
1. Student 1 and 2 use the standard addition but regroup in a slightly different
way. Student 1 computes 7 + 8 = 7 + 3 + 5 = 10 + 5. Student 2 computes
7 + 8 = 5 + 2 + 5 + 3 = 5 + 5 + 2 + 3 = 10 + 5. Student 3 counts out the answer.
The work of student 4 suggests that a make-a-ten method was used. Namely,
to subtract 5 from 12, the child viewed 12 as 10 and 2, subtracted 5 from 10,
and added the remaining 5 to the 2 to make 7. See activities of the previous
section.
2. When adding the numbers in the one’s place student 1 would compute 6 + 7 =
6 + 4 + 3 = 10 + 3 and would then carry the 10 to the top of the ten’s column.
Student 2 would compute 6 + 7 = 5 + 1 + 5 + 2 = 5 + 5 + 1 + 2 = 10 + 3 and
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CHAPTER 3
would then carry the 10 to the bottom of the ten’s column. Student 3 would
first count the tens, getting 50, then get 56, and then count 7 more from 56.
Student 4 would regroup 1 ten from the tens place in 43 as 10 ones and would
then compute 10−8+3 = 5 for the one’s place of the answer (by first computing
10 − 8 = 2 and then adding 3).
Class Activity 3J: Understanding the Standard Addition Algorithm
At the end of this activity you could talk about extending the addition algorithm to
decimals and you could show video clip 8 in which a third and fifth grader both apply
the addition algorithm correctly for adding decimals (even though the third grader
has not learned the algorithm for decimals, only for whole numbers) but neither child
knows how to compare decimals correctly. This would seem to indicate that knowing
the mechanics of the addition algorithm doesn’t necessarily lead to understanding
place value.
1.
11
14 7
+195
342
2. & 3. When students work with toothpicks their inclination will probably be to add
the hundreds first, then the tens (adjusting the hundreds), and then the ones
(adjusting the tens). Point out that this is a perfectly valid method of addition,
but is not quite the same as the standard addition algorithm.
Before you ask students to draw pictures, you might want to show the class
some simple drawings, such as the “ten-structured pictures” in Figure 3.3, so
that they don’t waste time making elaborate drawings. To speed up the drawing
of a hundred, students can just draw a big loop without drawing all the tens
inside (and perhaps write 100 inside). Students can then circle 10 of the tens
and 10 of the ones to show regrouping taking place.
53
Figure 3.3: Adding 147 + 195
4. Here regrouping is shown from right to left but it could also be done from left to
1(100) + 4(10) + 7(1)
+1(100) + 9(10) + 5(1)
right, as students might have done in part 2. 2(100) + 13(10) + 12(1)
2(100) + 14(10) + 2(1)
3(100) + 4(10) + 2(1)
Class Activity 3K: Understanding the Standard Subtraction
Algorithm
You might like to show IMAP video clip 3 in which Gretchen, a 2nd grader
attempts to calculate 70 − 23. She is convinced that the common algorithm that she
uses (incorrectly) must be correct even though it is in conflict with the answer she
gets when working with base ten blocks. When Gretchen attempts to use the common
algorithm, she makes the classic error of “taking the smaller from the larger” when
subtracting (taking 0 from 3 at the first step), rather than recognizing that regrouping
is needed. It’s interesting to see how persistent she is in trying to figure out what’s
wrong and how willing she is to stick with the problem rather than to simply give
up. It demonstrates that children do want to make sense of math!
Note also that there are two problems in the text on other subtraction algorithms
that you could also choose to discuss in class.
Copyright 54
CHAPTER 3
1.
1115
1/ 2/ 5/
−68
57
2. & 3. Show 1 bundle of 100, 2 bundles of 10, and 5 individual toothpicks. Now there
are several ways you could proceed.
You could unbundle 1 bundle of 100 as 10 bundles of 10 and unbundle 1 bundle of
10 as 10 individual toothpicks to make 11 bundles of 10 and 15 individuals from
which you can take away 6 bundles of 10 and 8 individual toothpicks. Notice
that this approach works from left to right, unlike the standard algorithm.
Instead of unbundling everything first and then taking away 6 bundles of 10 and
8 individuals, you could take away 6 bundles of 10 once you have unbundled
the 100 as 10 bundles of 10.
To demonstrate how the common algorithm works, by first unbundle 1 bundle
of 10 to make a total of 15 individual toothpicks, then take away 8 to leave 7,
then unbundle the bundle of 100 to make 10 bundles of 10, for a total of 11
bundles of 10, then take away 6 bundles of 10 to leave 5 bundles of 10.
It’s a good idea to discuss with students that these are all valid ways to subtract,
even though they may only be familiar with the common subtraction algorithm.
4.
125 = 1(100) + 2(10) + 5(1)
= 1(100) + 1(10) + 15(1)
= 0(100) + 11(10) + 15(1)
11(10) + 15(1)
−[6(10) + 8(1)]
5(10) + 7(1)
6. Because there is a 0 in the tens place of 104, the 1 hundred must first be
unbundled into 10 tens, and then 1 of those tens can be unbundled into 10
ones, to make 9 tens and 14 ones. Now 7 ones can be subtracted from 14 ones,
55
leaving 7 ones, and 4 tens can be subtracted from the 9 tens leaving 5 tens. So
the result is 57.
The problem can be solved mentally by subtracting 50 from 104 and then adding
on 3 (because 3 too many were subtracted).
As a follow up, you might ask students to explain another example of subtracting
across zeros, such as 1006 − 78.
Class Activity 3L: A Third Grader’s Method of Subtraction
Students are often surprised that a third grader could come up with this method.
And they find it a very cool method. You could show IMAP video clip 4, in which
Andrew solves 47 − 39 in a similar way although without writing as much as Pat did
or explaining it quite as clearly.
1. Yes, Pat’s method is legitimate. We can show Pat’s method with expanded
forms:
4(100) + 2(10) + 3(1)
−[1(100) + 5(10) + 7(1)]
3(100) − 3(10) − 4(1)
= 266
In terms of “horizontal” equations, we can write
4(100) + 2(10) + 3(1) − 1(100) − 5(10) − 7(10) = 3(100) − 3(10) − 4(1) = 266
2. Pat might write the following
2−
70−
100
28
3(100) + 1(10) + 7(1)
−[2(100) + 8(10) + 9(1)]
1(100) − 7(10) − 2(1)
= 28
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CHAPTER 3
Class Activity 3M: Regrouping in Base 12
This activity essentially involves base 12. It’s an opportunity for students to focus
on the idea of regrouping in bases other than 10 without getting bogged down in the
notation of arithmetic in other bases.
We want to calculate
5(boxes)
- [2(boxes)
+ 4(bags)
+ 9(bags)
+ 3(individual)
+ 7(individual)]
To do so, let’s first unbundle some boxes and bags:
5(boxes)
= 4(boxes)
= 4(boxes)
+
+
+
4(bags) + 3(individual)
Why 16 bags and not 14? Because 1 box is 12 bags, so when we unbundle 1 box
and add the bags to the 4 bags, that makes 16 bags. Similarly, when we unbundle
a bag we get 12 individual favors, combining them with the 3 individuals makes 15
individuals.
Now we can subtract:
4(boxes)
- [2(boxes)
2(boxes)
+ 12(bags)
+ 9(bags)
+ 3(bags)
+ 15(individual)
+ 7(individual)]
+ 8(individual)
So 2 boxes, 3 bags, and 8 individual favors were sold.
Class Activity 3N: Regrouping in Base 60
This activity essentially involves base 60. It’s an opportunity for students to focus
on the idea of regrouping in bases other than 10 without getting bogged down in the
notation of arithmetic in other bases.
1(hour)
+ 43(minutes) + 38(seconds)
+1(hour) + 48(minutes) + 29(seconds)
2(hours)
+ 91(minutes) + 67(seconds)
= 3(hours) + 32(minutes) + 7(seconds)
3.4
57
Adding And Subtracting Fractions
How Do We Find a Suitable Common Denominator for Adding or Subtracting
Fractions?,
Mixed Numbers and Improper Fractions,
Are These Problems for 12 + 31 or 12 − 13 ?,
Writing a Fraction as a Sum of Fractions.
Class Activity 3O: Why Do We Add and Subtract Fractions
the Way We Do?
You could show IMAP video clip 15 in which a 2nd grader correctly adds 34 + 12
by reasoning about pictures even though she has not been taught the procedure for
adding fractions.
1. (a) In short, Patti’s method is not valid because she is using two different wholes.
When we add fractions, we must use the same whole for each summand and for
the answer. See the text for a discussion relevant to part (b).
2. (b) Students can fold back the unused portion of a strip. So for example, to
show 32 , students can fold back one third to leave two of the thirds showing.
The lengths 21 + 13 and 23 + 21 can be shown by placing appropriate strip portions
end to end. The length 32 − 12 can be shown by placing the 12 piece on top of the
2
piece. The difference is the portion of the 23 piece that is not covered.
3
(c) We need common denominators in order to express a sum or difference as a
fraction. When we have different fractional parts in a sum or a difference, such
as halves and thirds, the parts aren’t the same size. But to express a length as
a fraction, we need to describe it as a certain number of parts, all of which have
the same size.
Students should relate their strips to equations like this:
3 2
3+2
5
1 1
+ = + =
=
2 3
6 6
6
6
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CHAPTER 3
Class Activity 3P: Adding and Subtracting Mixed Numbers
You might like to show video clip 10, which is about the decimal subtraction
problem 4−.7. The student uses the kinds of pictures we often use for mixed numbers,
and also writes the answer as a mixed number, so this ties in nicely here.
You could also show video clip 25 in which a student correctly identifies 1 32 as
equal to 53 , but her argument is not logical.
from 3 but should have borrowed 44 from 3, and
1. (a) Student 1 borrowed 10
4
then done the subtraction 54 − 34 . Student 2 has a correct method and only
needs to record the 14 and the 1 resulting from 2 − 1 below the line to show
the answer.
(b) This student correctly counted by thirds from 2 13 to 3, but should have
answered 23 instead of 2. The student may have been thinking “2 more
thirds.”
(c) To complete the problem compute
4
4
+ 43 .
(d) The 3 completely shaded rectangles were correctly added but when the
2
shaded rectangles were added the whole became two squares instead of
3
one.
2. Both mixed numbers could be converted to improper fractions with a common
denominator and then the numerators could be subtracted. As an alternative,
first compute 13 − 12 = 62 − 36 = − 61 . Then compute 7 − 4 − 16 .
Class Activity 3Q: Are These Word Problems for
1
2
+ 13 ?
If you are feeling pressed for time, you could look at only problems 4 and 5, since
these are usually the most difficult.
Please make clear to the prospective teachers that this is an activity that is specifically designed to addressed teacher knowledge and is not intended for their students.
As teachers, they must be able to write word problems for their students and they
will need to be aware that small changes in the wording of a problem can have a big
impact on the meaning of the problem and the kind of operation(s) needed to solve
the problem.
Note the need to attend carefully to the whole (referent unit), as in chapter 2.
1. Yes.
59
2. No, the problem is not clearly stated. It says “Tom pours in another 13 ” but
does that mean 13 of a cup or does it mean another 31 of the amount that is
there? It’s not clear.
3. We can’s solve this because we don’t know the relative sizes of Heeltoe County
and Toejoint County. These are different “wholes”.
4. No, we still can’t solve this problem by adding 12 + 13 because even though Heeltoe
and Toejoint Counties have the same area, the two-county region has twice the
area and is a different whole. We can solve the problem though: Heeltoe’s forest
is 14 of the land of the two-county region and Toejoint’s forest is 16 of the land
5
of the two-county region. Together, the forested land is 14 + 61 = 12
of the land
of the two-county region.
If students still have a hard time understanding why we can’t add 21 + 13 to
solve this problem, then you could ask them what would happen if the fraction
of land in each county that is covered with forest is 21 . If you could add the
fractions, then you’d predict that the whole land area of the two-county region
is covered with forest. But clearly that isn’t right.
5. We can’t solve this because some of the kids may like both pizza and hamburger.
6. This is not solved by adding
1
2
+
1
3
but it is solved by adding
1
2
Class Activity 3R: Are These Word Problems for
+
1
6
= 23 .
1
2
− 31 ?
Please make clear to the prospective teachers that this is an activity that is specifically
designed to addressed teacher knowledge and is not intended for their students. As
teachers, they must be able to write word problems for their students and they will
need to be aware that small changes in the wording of a problem can have a big
impact on the meaning of the problem and the kind of operation(s) needed to solve
the problem.
1. No, we can’t solve this because what does “Zelha pours out 13 ” mean? Does it
mean she pours out 13 cup of water or 13 of what is in the bowl?
2. Yes.
3. No, but we can solve this by subtracting
1
2
−
1
6
= 13 .
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CHAPTER 3
4. Yes.
5. Yes, but only because there is 12 of a pizza left today. It is really a story problem
for (1 − 12 ) − 31 , but it can still be solved by subtracting 13 from 12 .
6. No, but we can solve this by subtracting
1
2
−
1
6
= 13 .
Class Activity 3S: What Fraction is Shaded?
There is a blackline master on page 279 that you may wish to print out for students
to write and experiment on.
Students usually enjoy thinking about these problems and find them an interesting
challenge and change of pace. I like to advertise these problems as an opportunity to
“think outside the box” (by thinking inside the box!).
Since the area of the rectangle at the top of square 1 is 14 the area of the square
1
and the area of the triangle at the botton is 12
the area of the square, the area of the
1
1
of
shaded region is 3 the area of the square. We can see that the triangle’s area is 12
the area of the square by making the square from 12 copies of the triangle, although
some of the triangles will have to be cut in half to do this.
1
Since the area of the triangle in the lower left hand corner of square 2 is 18
the
area of the area of the square and the area of triangle in the upper right hand corner
1
is 81 of the area of the square, the area of the shaded region is 1 − 18
− 81 = 59
of the
72
area of square.
Since the area of each of the 4 unshaded triangles in square 3 is 41 of the area of
the shaded region is 1 − 34 = 14 of the area of the square.
Since the areas of the unshaded triangles in the lower left and upper right corners
of square 4 are both 41 and the area of the triangle in the lower right corner is 18 the
area of the shaded region is 1 − 14 − 41 − 18 = 83 of the area of the square.
Class Activity 3T: Addition with Whole Numbers, Decimals,
Fractions, and Mixed Numbers: What Are Common Ideas?
Some important common ideas are: The need for a common whole throughout; working with like parts (we add tens to tens, tenths to tenths, fifths to fifths, ones to
ones); regrouping with mixed numbers is like regrouping in the base-ten system; the
operations solve the same kind of problem.
3.5
61
Adding and Subtracting Negative Numbers
Class Activity 3U: Word Problems and Rules for Adding and
Subtracting with Negative Numbers
You can use this activity as a springboard for the upcoming material in the text
and in particular for discussing addition and subtraction with negative numbers on
number lines.
1. (a) It was −5◦ C. If the temperature goes up 5◦ C, what will the temperature
be?
(b) (−N) + N = 0
2. (a) At midnight the temperature was 2 degrees below zero. By 6:00 A.M.
the temperature was 5 degrees colder. What was the temperature at 6:00
A.M.?
(b) Paul has −2 dollars in his bank account, meaning that his bank account
is overdrawn by $2. Marci has $5 less in her bank account. How much
money does Marci have in her bank account?
(c) (−A) − B = −(A + B)
3. (a) Paul’s bank account is overdrawn by $5 (so Paul has −5 dollars in his bank
account). Marci has $2 in her bank account. How much more money does
Marci have in her bank account than Paul has in his?
(b) A − (−B) = A + B
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Multiplication
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4.1
Interpretations of Multiplication
I like to introduce this section by asking students how they think about multiplication.
How would they tell children what 3 × 5 means? Some will say it means 5 added 3
times, or 5 + 5 + 5, others may say it means 3 + 3 + 3 + 3 + 3. Then I ask, “what
about 21 × 12 ? It doesn’t make sense to talk about this multiplication problem in terms
of repeated addition. You can then lead into a brief lecture about the definition of
multiplication given in the text and follow that with the next activity.
Class Activity 4A: Showing Multiplicative Structure
Ask students to use the definition of multiplication consistently and to interpret the
first factor as the number of groups and the second factor as the number in each
group.
Note that multiplicative comparisons are revisited in section 7.3 in the chapter on
ratio and proportional relationships and ordered pair problems arise again in section
16.2 in the chapter on probability.
1. We can think of the ladybugs as made of 5 groups, where each group is a row
of 8 ladybugs. Of course you can also view the rug as made of 8 groups, where
each group is a column of 5 ladybugs.
This example illustrates the commutative property of multiplication, which will
be discussed in an upcoming section.
You could also ask students how else children could determine the number of
ladybugs, other than just counting one by one. Children could determine the
number of ladybugs with repeated addition, by adding 8 five times or adding 5
eight times. Thus multiplication is linked to repeated addition.
2. For each pair of pants there are 4 options for the shirts. This makes 3 groups
of 4 outfits, as can be seen by displaying the outfits in an array (where each
row consists of the outfits for one fixed pair of pants, for example), with an
organized list, or with a tree diagram. You can use this problem to bring up
these ways of displaying pairs if students don’t come up with them.
3. Show a strip representing Eva’s puppy’s weight. For Micah’s puppy’s weight,
show another strip that consists of 4 smaller strips joined together, each smaller
strip of the same size as the strip for Eva’s puppy.
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Ask students: if a story problem contains the wording “4 times as much as” or
“4 times as many as” does that automatically mean you will multiply by 4 to
solve the problem?
4. Eva’s puppy weighs 14 as much as Micah’s puppy. Eva’s puppy weighs 9 pounds.
How much does Micah’s puppy weigh?
5. Micah’s puppy weighs 4 times as much as Eva’s puppy. If Micah’s puppy weighs
36 pounds, how much does Eva’s puppy weigh?
Class Activity 4B: Writing Multiplication Word Problems
See the text for examples of array, ordered pair, and multiplicative comparison word
problems.
In part 2, students should see that by switching from organizing groups by row
to organizing them by column (or vice versa), that the problem could also be solved
by multiplying 7 × 3. Students may also bring up that the problem can be solved by
repeated addition, either with 7 + 7 + 7 or with 3 + 3 + 3 + 3 + 3 + 3 + 3.
Be sure that students use “5 times as much as” wording (or a variation such as
“5 times as long as”) in part 4. In part 5, the problem statement should involve “ 15
as much as” wording (or a variation such as “ 51 as wide as”).
Class Activity 4C: Problems About Pairs
If you have collections of small objects, such as snap cubes, marbles, tiles, or countingbears in 6 different colors, or labeled 1 to 6, you might like to have the students use
these objects to help them distinguish the different problems. (It’s ok to use the same
collection of small objects for all the problems.)
The point of this activity is for students to see that there different types of pairing
problems and that subtle differences in the way a problem is worded makes a big
difference in how it is solved and how hard or easy the problem is.
1. You can organize these pairs of 2-digit numbers into 6 groups of 5. Here’s why:
for each of the 6 numbers you can pick first, there are 5 numbers you can pick
second (these 5 are all the remaining numbers that weren’t picked). So there
are 6 × 5 = 30 tile-pairs.
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2. As with the first problem, there would be 6 groups of 5 pairs if the pairs were
considered to be ordered. But in this case we are only interested in the two
things in the pair, not which one is first and which is second. So the 30 ordered
pairs “double count” the unordered pairs because, for example, red-blue is the
same pair as blue-red. Therefore we divide by 2 to get the number of possible
pairs, which is 15.
Students will see this problem again if they do the the “high five” activity (also
known as the “handshake problem”) in the Algebra chapter.
3. In the tile-pair problem we are working with ordered pairs, whereas in the cubepair problem we are working with unordered pairs.
4. The bear-pair problem is just a division problem: 6 ÷ 2 = 3. There are 3 pairs
of bears. In this case we are not looking for all possible ways to pair the bears,
we are just pairing them up in one way all at once and we want to know how
many pairs that makes.
Notice that this problem is not asking for all possible ways of pairing the bears
(which is a much harder problem!), it’s just asking how many pairs there will
be.
The bear-pair problem is suitable even for very young children, but the tile-pair
and cube-pair problems provide a good problem solving opportunity for 4th or
5th graders.
4.2
Why Multiplying by 10 Is Special in Base Ten
Go online to My Math Lab for an “expand your thinking” activity for this section:
Multiplying by Powers of 10 Explains the Cycling of Decimal Representations of
Fractions.
Understanding why multiplication by 10 works the way it does will be an important component in understanding and explaining the common multiplication algorithm in section 4.6.
Class Activity 4D: Multiplying by 10
1. (a) No. 10 × 1.2 6= 1.20.
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(b) Although the statement is true, you would not want to describe multiplication by 10 this way with students who are just studying whole number
multiplication and who are not working with decimals.
(c) This statement is correct and is appropriate for both whole numbers and
decimals.
2. Bundling vertically, each of the 3 ones becomes a bundle of 10 and each of the
2 tens becomes a bundle of 100. Thus, when we multiply 23 by 10, the 3 ones
become 3 tens, and the 2 tens become 2 hundreds, so each of the digits in 23
shift one place to the left.
4.3
The Commutative and Associative Properties
of Multiplication, Areas of Rectangles, and Volumes of Boxes
More on Explaining and Illustrating the Commutative Property of Multiplication,
Applying the Associative and Commutative Properties of Multiplication,
Different Ways to Calculate the Total Number of Objects, and
More on Using Multiplication to Estimate How Many.
You might want to begin this section by having students think back to the commutative and associative property of addition and how important those properties
were for mental math and for learning single-digit additions. The situation is similar
for multiplication. Note that the CCSS ask third, fourth, and fifth grade students
to apply the commutative and associative properties, but don’t require students to
explain why these properties make sense. Thus explaining why the properties make
sense could be considered mathematical knowledge for teaching that goes beyond
what is expected of students.
Class Activity 4E: Explaining and Applying the Commutative
Property of Multiplication
1. A 4 × 3 problem: ( Boris runs 3 miles 4 times a week. How many miles does
Boris run each week?) We could change this problem into a 3 × 4 problem by
replacing the fist sentence with ( Boris runs 4 miles 3 times a week.). It would
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not be obvious to many children that the answers to both questions are the
same.
The point of this is to help students see that the commutative property of
multiplication is not obvious and shouldn’t be taken for granted.
As an extension, you could discuss that other operations are not commutative.
For example, subtraction and division are not commutative. For another example, consider the operations of “rotating clockwise by 90 degrees” and “flip
across the horizontal.” These operations don’t commute. In other words, we
get different results depending on which of the two operations we do first and
which we do second. To demonstrate why, you can take two small paper squares
and poke a small hole at the top right corner of each square. Do the operations
in a different order for each square. Note that the small hole comes out in a
different location.
2. Organizing the array by rows, we can think of the array as consisting of 3
groups, where each group has 5 stars, so there are 3 × 5 stars total. On the
other hand, organizing by columns, we can think of the array as consisting of
5 groups, where each group has 3 stars, so there are 5 × 3 tiles total. It’s the
same either way, so 5 × 3 = 3 × 5.
Some students may explain that the two products are equal because both are
15. Although this is true, this line of reasoning won’t allow us to explain why
the commutative property is valid in other cases. See the next part.
Notice that writing an equation in the form 5 × 3 = 3 × 5 requires being able
to view the equal sign as standing for “is the same amount as” and not just
as “becomes” or “evaluates to.” You may wish to review this perspective on
equality, which was discussed in the section on the commutative and associative
properties of addition.
3. Students should come to see that by referring to the underlying array, rather
than the specific numerical products (such as 15), they can give a general,
conceptual explanation. They should explain that the total number of items in
the array is the same no matter how the items are counted.
In my experience, many students find it difficult to give this kind of general
explanation.
4. Ben calculates 21 × 2. Kaia calculates 2 × 21. The answers are the same by the
commutative property of multiplication. Another way to se why it’s the same
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amount is to think about putting the first marble in each bag in one pile, and
the second marble in a second pile. In the end there will be 2 piles with 21
marbles in each pile. This kind of “dealing out” strategy is in fact a way that
young children often do divide up quantities.
Class Activity 4F: Multiplication, Areas of Rectangles, and
the Commutative Property
1. We can visualize the rectangle as 5 groups (rows) of nine 1 cm by 1 cm squares.
We could also visualize this rectangle as 9 groups (columns) of five 1 cm by 1
cm squares.
2. If we group by rows the area = 5 × 9. If we group by columns the area = 9 × 5.
Since the area of the rectangle is the same no matter how we group 5×9 = 9×5.
3. Just repeat the previous argument for an A by B rectangle.
Note that this activity provides another opportunity for students to give a
general explanation for why the commutative property is valid.
Class Activity 4G: Ways to Describe the Volume of a Box with
Multiplication
I usually go straight from this activity into the first part of the next activity.
3. First page: Top left: 2 groups of 3 × 4 blocks, so 2 × (3 × 4) blocks total.
Top right: 3 groups of 2 × 4 blocks, so 3 × (2 × 4) blocks total.
Bottom left: 4 groups of 2 × 3 blocks, so 4 × (2 × 3) blocks total.
Bottom right: 2 × 3 groups of 4 blocks, so (2 × 3) × 4 blocks total.
Second page:
Left: 4 × 3 groups of 2 blocks, so (4 × 3) × 2 blocks total.
Right: 4 × 2 groups of 3 blocks, so (4 × 2) × 3 blocks total.
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Class Activity 4H: Explaining the Associative Property
Parts 2 and 3 provide you with other ways to explain the associative property if you
choose not to do the first part. I think one of these parts will be enough to do in
class.
In light of parts 2 and 3, it might be interesting to look for designs on fabric,
wallpaper, wrapping paper, or elsewhere that are relevant to the associative property
of multiplication.
1. The grouping on the first page bottom left consists of 4 groups of 2 × 3 blocks
while the grouping on the second page right consists of 4 × 2 groups of 3 blocks.
Since both groupings came from the same box it follows that (4 × 2) × 3 =
4 × (2 × 3).
2. There are 6 × (4 × 2) dots total. We can also describe the total as (6 × 4) × 2
dots. Therefore 6 × (4 × 2) = (6 × 4) × 2.
3. There are 5 × 3 groups (because there are 5 columns with 3 sets in each column)
and there are 4 wedges in each group, so there are (5 × 3) × 4 wedges total. If
students have difficulty describing the 5 × 3 groups you could ask them if they
see groups of 4. Then ask how many groups of 4 there are. If they say 15, ask if
these 15 groups have a structure to them. Can they see these groups themselves
as grouped into 5 groups of 3?
On the other hand, we can also say that there are 5 groups with 3 × 4 wedges
in each group, so there are 5 × (3 × 4) wedges total. It’s the same total number
either way, hence the equation.
Class Activity 4I: Using the Associative and Commutative
Properties of Multiplication
IMAP video clip 18 has a segment that is especially relevant to part 3. However,
a lot of the clip is about seeing an amount as grouped in different ways, which is
relevant overall. Also relevant is video clip 19, in which 192,000 is viewed as 19200
tens, 1920 hundreds, and 192 thousands and there is a discussion about how to go
from viewing 1920 hundreds as 19200 tens, which can be viewed as an application of
the associative property.
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1. The equations show how the 6 in 60 starts out associated with ten (to make 60)
and is then re-associated with the 4 to make 24.
We can view the math drawing as showing 4 groups of 60 or as 4 × 6 groups of
10. To help students see the latter structure, you may need to prompt them by
asking them if they see groups of 10, and if so, how many of those groups of 10
are there? If they say there are 24 groups of 10 ask if they see any structure to
the 24 groups. Are those 24 groups arranged in a way that shows multiplicative
structure?
2. Students may find it helpful to think in terms of money because 4 quarters make
a dollar. By viewing 28 as 7 groups of 4, we see that 28 ×0.25 = (7 ×4) ×0.25 =
7 × (4 × 0.25) = 7 × 1 = 7.
3. You may want to watch video clip 18 in conjunction with this part, which is
about determining the total in 25 groups of 10. About 1:50 into that clip a
second grader seems to have recognized that 20 groups of 10 is the same as 2
groups of 100.
Class Activity 4J: Using Multiplication to Estimate How Many
Elementary schools sometimes have contests to guess how many small pictures there
are on a long piece of wrapping paper. You could mention these contests or even bring
in a roll of wrapping paper with repeated small pictures and have students estimate
out how many pictures there are on it. Otherwise, you may wish to do the additional
Class Activity on estimating how many that is online in My Math Lab first.
Another fun extension of this activity is to bring in a large see-through container,
in the shape of a rectangular prism, containing small wrapped candies. Count the
candies before you bring them in and put the number on a hidden slip of paper. Hold
up the container and allow students to ask questions that they can answer by coming
up and making counts of candies they can see without opening the container. After
various data about the candies has been obtained and shared with the entire class,
have students do calculations to estimate how many candies there are. They are then
allowed to adjust this estimates to try to make them better. Each student should put
their calculation and estimate on a piece of paper. Announce the actual number and
give a small prize to the person who comes closest. Finally, share the candies with
the class.
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1. The swatch has 25 stars on it and is about 4 inches wide and 2 inches tall.
Because the wrapping paper is 40 inches wide, the swatch will fit about 10
times across the width of the wrapping paper. Because there are 12 inches in a
foot, the swatch will fit 6 times down each foot of length and therefore 50 × 6
times down the 50-foot length. Therefore the total number of stars should be
about
50 × 6 × 10 × 25 = 75000
2. In case your students are eager to know (as they surely will be): it turns out
that there were 330 candies in the container.
Students might use these methods to estimate the number of candies:
(a) Each bottom edge has about 6 or 7 candies along it, making about 40 candies
in a horizontal layer. There are about 8 candies along each vertical edge, so
that means there are about 8 horizontal layers. This makes a total of about
8 × 40 = 320 candies, which is pretty close!
(b) Another way of estimating is to determine that there are about 50 candies
up against the front face of the container (facing us). Since there are about 6 or
7 candies along a bottom edge, there should also be 6 or 7 candies or so along
the other edges, making 6 or 7 vertical slabs of candies, each with about 50,
making about 6.5 × 50 = 325 candies in all.
4.4
The Distributive Property
Order of Operations,
The Distributive Property and FOIL, and
Squares and Products Near Squares.
Class Activity 4K: Explaining the Distributive Property
1. (a) This expression 6 × 3 stands for the total number of eraser tops. The
expression 6 × 4 stands for the total number of stickers. When we add
these together, we get the total number of items in the goodie bags.
(b) Each goodie bag contains 3 + 4 items. There are 6 goodie bags. So according to the meaning of multiplication, the total number of items in the
goodie bags is 6 × (3 + 4).
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(c) We have two ways of describing the total number of items in the goodie
bags. Since the number must be the same no matter which way we count
them, the two expressions for the total must be equal.
(d) See text. Part (c) is a general explanation for the distributive property
because we can change the numbers 6, 3, 4 to any other counting numbers
and the explanation will remain the same.
2. On the one hand, there are 3 × (2 + 4) small squares in all because there are 3
rows with 2 + 4 squares in each row. On the other hand, there are 3 × 2 + 3 × 4
small squares in all because there are 3 rows of 2 shaded squares and 3 rows of
4 unshaded squares. The total number of small squares is the same either way
you count them, hence the equation.
3. Show a rectangle that is 8 squares by 15 squares broken into an 8 by 10 rectangle
and another 8 by 5 rectangle (use different shading or different colors). The
individual squares don’t need to be shown explicitly. The drawing can just
indicate how many squares there are in each direction.
Class Activity 4L: Applying the Distributive Property
1.
17 × 15 = 15 × 15 + 2 × 15 = 225 + 30 = 255
as we see from the array of dots by breaking it into 15 rows of 15 and another
2 rows of 15.
2.
6 × 9 = 6 × 10 − 6 × 1 = 10 − 6 = 54
3. 20 × 15 = 300 so 19 × 15 = 20 × 15 − 1 × 15 = 300 − 15 = 285. You can
suggest to students that they draw a rectangle and label one side 20 and the
other side 15 to stand for a 20 by 15 array. They should then show that one
row (or column) of 15 will need to be removed to calculate 19 × 15.
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Class Activity 4M: Relating Multiplication Problems
1. No, Kylie also needs to add 20 × 3 and 3 × 20 as we can see if we subdivide the
rectangle into 20 columns and 3 columns and 20 rows and 3 rows.
2. No, this is not correct, but it is close. If we consider 19 of the 20 rows of 21 dots
and imagine moving the last column of dots down to the 20th row of dots, then
we will almost get 20 rows of 20 dots except we will be 1 dot short. Therefore
19 × 21 = 20 × 20 − 1 = 400 − 1 = 399.
3. Using Mary’s reasoning we could write
19 × 21 = 20 × 21 − 1 × 21 = 20 × 20 + 20 × 1 − 1 × 21 = 400 − 1 = 399
4.5
Properties of Arithmetic, Mental Math, and
Single-Digit Multiplication Facts
Which Properties of Arithmetic do These Calculations Use? and
More on Writing Equations that Correspond to a Method of Calculation.
Class Activity 4N: Using Properties of Arithmetic to Aid the
Learning of Basic Multiplication Facts
Be sure that prospective teachers don’t misunderstand this activity and think the
message is that children don’t need to learn their basic multiplication facts. Of course
children need to be fluent with all the basic multiplication facts! However, learning
the facts is not a matter of rote memorization, but rather of learning relationships as
a process toward becoming fluent.
1. We can use the commutative property of multiplication to obtain the unshaded
facts from the shaded facts.
2. Follow the examples shown. Notice that the 6× facts can be obtained from a
5× fact or as a double of a 3× fact. Also, 4× facts are doubles of 2× facts.
3. The 5 × facts arise from the first 5 rows of each array.
1
1
5 × 7 = ( × 10) × 7 = × (10 × 7)
2
2
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using the associative property of multiplication.
4. This uses the distributive property. For example,
9 × 7 = (10 − 1) × 7 = 10 × 7 − 1 × 7 = 70 − 7 = 63
5. The digits add to 9 and the tens digit is 1 less than the number you are multiplying by 9.
Students love to learn about the “hand trick” for the 9× facts. To use this hand
trick to multiply 9 × 3, for example, you hold out both hands, then lower your
3rd finger from the left. You then see 2 fingers to the left of the lowered finger
and 7 fingers to the right, so 9 × 3 is 27. Similarly, for 9 × 4 lower your fourth
finger from the left. You’ll see 3 fingers to the left of your lowered finger and 6
to the right, so 9 × 4 is 36! Note that this hand trick fits with the two patterns
in the 9s table mentioned above.
The fact that for multiples of 9 the sum of the digits is also always a multiple
of 9 is discussed in section 8.5 on divisibility tests.
Class Activity 4O: Solving Arithmetic Problems Mentally
I like to do this activity as follows: after students have described their mental methods
for (some or most of) these problems, I show them how to write equations that
correspond to the methods. In effect, I write many of the equations that are in the
activity Which Properties of Arithmetic do These Calculations Use?, which is in the
online activity bank. You might prefer to give that banked activity to your students
instead of writing the equations out on the board. Students will need to see and
think about examples of how to write such equations before they are ready for the
last “core” activity in the section.
1. Multiply 4 by 100 and then take away 4 to get 396.
2. View 16 as 4 × 4 and re-associate one of the 4s with 25 to make 100, so the
answer is 4×100 which is 400. Or view the problem as 4×4×5×5 and combine
each 4 with a 5 to turn the problem into 20 × 20. You can also break 25 apart
into 20 + 5 or break 16 apart into 10 + 6.
3. 50% of 680 is half of 680 which is 340. 5% is
answer is 340 − 34 = 306.
1
10
of that which is 34. So the
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4. Add 10 times 125 and 2 times 125 to get 1250 + 250 = 1500. Or think of 12 as
3 sets of 4. Four times 125 is 500 (4 groups of 100 and another 4 groups of 25).
Then 3 times 500 is 1500.
5. 100% is 120. Another 25% is
compare to part 2.)
1
4
of 120, which is 30, so the answer is 150. (Also,
Class Activity 4P: Writing Equations that Correspond to a
Method of Calculation
See the online activity bank in My Math Lab for additional parts to this activity
(most are easier than the ones here).
1.
55% × 120 = (50% + 5%) × 120
= 50% × 120 + 5% × 120
1
1
× 120 + ( × 10%) × 120
=
2
2
1
= 60 + × (10% × 120)
2
1
= 60 + × 12
2
= 60 + 6 = 66
2.
35% · 80 = (25% + 10%) · 80
= 25% · 80 + 10% · 80
1
=
· 80 + 10% · 80
4
= 20 + 8 = 28
3.
90% · 350 =
=
=
=
(100% − 10%) · 350
100% · 350 − 10% · 350
350 − 35
315
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4.
1 1 1
12.5% · 1800 = ( · · ) · 1800
2 2 2
1 1
1
= ( · ) · ( · 1800)
2 2
2
1 1
= ( · ) · 900
2 2
1 1
· ( · 900)
=
2 2
1
=
· 450 = 225
2
Class Activity 4Q: Showing the Algebra in Mental Math
Be sure that students also describe their mental method in words and that the
method really can be carried out mentally.
Although the previous two activities, and the activity Which Properties of Arithmetic do These Calculations Use? in the online activity bank are not “core,” I’ve
found that students will need such examples before they are ready to write the kinds
of equations they are asked to write here.
1. I recommend having students first think about different ways they might solve
this problem. Since they may already know what the answer is, tell them: “put
your kid eyes on” and think about how a child in elementary school might solve
the problem mentally. You could have the students write their equations at this
point or wait until after you show the video.
Next, you could show IMAP video clip 6 and observe Javier’s method for calculating 6 × 12. Before watching, tell the students to observe closely since their
method of calculation might be different from Javier’s. Now have students write
equations that go along with Javier’s method.
Javier is also asked about 12 × 12 (or rather, about 12 dozen eggs). You could
pause the clip when he’s asked the question and have students come up with
several ways to calculate 12 × 12, then finish watching and see what Javier did.
2.
24 × 25 = (6 × 4) × 25
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= 6 × (4 × 25)
= 600
3.
1
5% × 48 = ( × 10%) × 48
2
1
× (10% × 48)
=
2
1
=
× 4.8
2
= 2.4
4.
15% × $44 = (10% + 5%) × $44
= 10% × $44 + 5% × $44
= $4.40 + $2.20 = $6.60
5.
26% × 840 = (25% + 1%) × 840
= 25% × 840 + 1% × 840
= 210 + 8.4 = 218.4
6.
9 × 99 = 9 × (100 − 1)
= 9 × 100 − 9 × 1
= 900 − 9 = 891
4.6
Why Algorithms for Multiplying Whole Numbers Work
Go online to My Math Lab for an additional activity for this section:
More on Why the Multiplication Algorithms Work.
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Before getting to the activities, I think it’s important to make a big deal about the
distinction between the meaning of multiplication and the multiplication procedure.
It’s important for teachers to know that the procedure can be explained in terms of
place value and the properties of arithmetic and is not merely a “given.”
Class Activity 4R: The Standard Versus the Partial Products
Multiplication Algorithm
1.
495
×7
3465
495
×7
35
630
2800
3465
84
×69
756
5040
5796
84
×69
36
720
240
4800
5796
2. The standard algorithm is essentially a condensed version of the partial products
algorithm. Several lines of the partial products algorithm are collapsed into one
line in the standard algorithm. Students may bring up other things they notice.
3. Using the partial products algorithm, one could just as well start with the places
of highest value. This just reverses the order in which the lines are produced.
Also, one could multiply from the top down. Again, this just puts lines in a
different order.
Class Activity 4S: Why the Multiplication Algorithms Work,
Part I
Internet video opportunity
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Of direct relevance to his class activity is a wonderful Lesson Study video on developing the multiplication algorithm, which is available at
http://hrd.apec.org/index.php/Classroom_Videos_from_Lesson_Study
and more specifically at
http://hrd.apec.org/index.php/Multiplication_Algorithm_Grade_3_%28Japan%29
The third grade students in this lesson have previously studied single-digit multiplication and multiplication of tens with single-digit numbers (e.g., 20 × 3). The students
then develop and discuss how to multiply 23 × 3 with the aid of an array. About a
third of the way through the lesson, a student explains how he sees 23 as 20 + 3 and
describes how to calculate based on what they have already learned. The website has
a video of the full lesson as well as video clips of highlights of the lesson.
1.
38
×6
48
180
228
2. 6 × 38 stands for the total number of small squares in the array because there
are 6 rows of 38 small squares. The array can be broken into 6 rows of 30 and 6
rows of 8, so the total number of squares is the combined amount. The partial
products algorithm breaks 6 × 38 into these same components.
3.
6 × 38 = 6 × (30 + 8)
= 6 × 30 + 6 × 8
= 180 + 48 = 228
The calculations 6 × 30 and 6 × 8 were exactly the multiplications performed
in the partial products algorithm. This is essentially the same explanation as
before, just that one is pictorial and the other involves equations.
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Class Activity 4T: Why the Multiplication Algorithms Work,
Part II
1.
45
×23
15
120
100
800
1035
2. According to the definition of multiplication, the total number of small squares
in the array is 23 × 45 because there are 23 rows with 45 squares in each row.
The four products produced by the partial products algorithm are exactly the
numbers of squares in each of the 4 sections shown in the subdivided array.
Therefore the partial products algorithm is just a way to break the multiplication problem into several other easier multiplication problems and add up those
pieces to get the full amount.
3. Using the standard algorithm the array is divided into 3 rows of 45 squares which
corresponds to the product 3 × 45 and 20 rows of 45 squares which corresponds
to the product 20 × 45. The 3 rows of 45 squares correspond to 3 rows of 5
squares plus 3 rows of 40 squares (3 × 5 + 3 × 40). The 20 rows of 45 squares
correspond to 20 rows of 5 squares plus 20 rows of 40 squares (20 × 5 + 20 × 40).
4.
23 × 45 =
=
=
=
(20 + 3) × (40 + 5)
20 × (40 + 5) + 3 × (40 + 5)
20 × 40 + 20 × 5 + 3 × 40 + 3 × 5
800 + 100 + 120 + 15 = 1035
The four multiplication problems produced by applying the distributive property are the same multiplication problems we do when we use the partial products algorithm. So the partial products algorithm is just a way to use the
distributive property to calculate a product.
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CHAPTER 4
This part and part 2 are essentially the same explanation, but one is pictorial
and the other is with equations. In both cases, we are seeing why it makes
sense to break the original multiplication problem into several smaller and easier
multiplication problems and then determine the full amount by adding up the
pieces.
5.
23 × 45 = (20 + 3) × 45
= 20 × 45 + 3 × 45
= 900 + 135 = 1035
Class Activity 4U: The Standard Multiplication Algorithm
Right Side Up and Upside Down
1.
347
×26
2082
6940
9022
2. See Figure 4.1.
347
20 × 347
26
6 × 347
Figure 4.1: 26 × 347
83
3.
26
×347
182
1040
7800
9022
The steps are different but the answer is the same.
4. Subdivide the rectangle vertically instead of horizontally, as in Figure 4.2
347
26
26 × 300
26 × 40
26 × 7
Figure 4.2: 26 × 347
5. The commutative property of arithmetic tells us that the answers to Problems
1 and 3 must be the same.
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Copyright Chapter 5
Multiplication of Fractions,
Decimals, and Negative Numbers
85
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5.1
Multiplying Fractions
Misconceptions With Fraction Multiplication,
More on Explaining Why the Procedure for Multiplying Fractions Is Valid,
More on Multiplying Mixed Numbers (which you might prefer to use in the next
section anyway, to connect with decimal and whole number multiplication).
Class Activity 5A: Extending the Definition of Multiplication
to Fractions
.
1. Either during this activity, or later, right before the next Class Activity, you
may wish to use the annotated equations to discuss that in a multiplication
equation,
A·B =C
B and C refer to the same underlying whole (or unit amount, or unit), whereas
A refers to a different underlying whole (or unit amount or unit).
Even though multiplication is commutative, ask students to pay attention to
the order in which the numbers occur and to write scenarios that fit with the
given definition of multiplication. The issue probably won’t come up until Part
b.
(a) Scenario: If you need 13 of a bag of fertilizer for 1 acre of grass, then for 2
acres of grass you will need 2 × 13 bags of fertilizer.
Table: 1 group (1 acre) → 13 bag of fertilizer; 2 groups (2 acres) → 2 × 13
bags of fertilizer.
Double number line: Show two coordinated number lines: one for bags of
fertilizer and one for groups (acres). Align 1 acre (groups) and 13 of a bag
of fertilizer. Align 2 acres (groups) and 2 × 13 bags of fertilizer.
(b) Annotated equation: Discuss with students that writing “amount in 1
group” or “amount in 1 whole” group is better than saying “each” because
we are considering less than a whole group.
Scenario: You need 13 of a bag of fertilizer for 1 acre of grass. So for 12 acre
of grass you will need 21 × 31 bags of fertilizer.
87
Some students will probably write scenarios such that the 12 refers to the
number of groups and other students will write problems such that 12 refers
to the amount in a whole group. Then you can ask which of those fit with
multiplication as it has been defined. (Here, the 12 should refer to the
number of groups.)
Table: 1 group (1 acre) → 31 bag of fertilizer; 12 groups ( 21 acres) → 12 × 31
bags of fertilizer.
fertilizer and one for groups (acres). Align 1 acre (groups) and 13 of a bag
of fertilizer. Align 12 groups ( 12 acres) and 12 × 13 bags of fertilizer.
(c) Annotated equation: Discuss with students that writing “amount in 1
group” or “amount in 1 whole” group is better than saying “each” because
we are considering less than a whole group.
Scenario: If you need 2 bags of fertilizer for 1 acre of grass, then you will
need for 13 × 2 bags of fertilizer for 13 acre of grass.
Table: 1 group (1 acre) → 2 bags of fertilizer; 13 groups ( 31 acres) → 13 × 2
bags of fertilizer.
fertilizer and one for groups (acres). Align 1 group (1 acre) and 2 bags of
fertilizer. Align 13 groups ( 31 acres) and 13 × 2 bags of fertilizer.
(d) Scenario: You need 54 of a bag of fertilizer for 1 acre of grass. So for 3 acres
of grass, you need 3 × 54 bags of fertilizer.
Table: 1 group (1 acre) → 45 bags of fertilizer; 3 groups (3 acres) → 3 × 54
bags of fertilizer.
fertilizer and one for groups (acres). Align 1 group (1 acre) and 45 bags of
fertilizer. Align 3 groups (3 acres) and 3 × 54 bags of fertilizer.
2. Top left: 14 of 13 , or 41 × 13 is shaded. Top right: 53 of 14 , or 53 × 14 is shaded.
Bottom left: 41 of 32 , or 41 × 23 is shaded. Bottom right: 43 of 35 , or 34 × 53 is shaded.
Class Activity 5B: Explaining Why the Procedure for Multiplying Fractions Is Valid
See text for further details.
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CHAPTER 5
In part 1, students should see that 14 of 31 of the square is darkly colored, which is
therefore 14 × 31 of the square according to the definition of multiplication. The final
square shows how breaking each of the thirds into 4 parts creates 3 groups of 4 (or 4
1
groups of 3 small parts). Therefore each such part is 4×3
of the square and so 41 of 31
1
.
is 4×4
In part 2, note that although we read the 23 first in 23 · 85 , since we interpret the
fraction multiplication as 23 of 85 we should show the 58 first and then take 23 of that
amount. This can be a source of confusion for students.
The original 8 equal parts that the square was divided into are each broken into
3 parts. This makes 8 × 3 or 3 × 8 equal parts. This is why the denominators are
multiplied.The 5 parts that were originally shaded are each broken into 2 parts. This
makes 5 × 2 or 2 × 5 equal parts. This is why the numerators are multiplied.
Students will sometimes explain fraction multiplication by considering overlapping strips (for example 23 vertically overlapping with 85 horizontally). Textbooks for
children also show such pictures. My response is: what is the logic if we have defined
2 5
· as 32 of 58 ? How does such a picture show 32 of 58 ? I think this deserves careful
3 8
explanation.
Class Activity 5C: When Do We Multiply Fractions?
Make clear to students that this activity is not intended for their students but rather
it is for them to sort out the subtleties in the wording of problems. Be warned that
some students may feel frustrated by how difficult and subtle it is to word fraction
problems correctly and may feel that therefore, they should stay away from word
problems altogether. Remind students that it’s important to persevere in making
sense of mathematics. You may need to find some way to reduce the stress and keep
the students “with you.”
1. (a) It is neither of these. It’s actually a problem for
3
4
− 13 · 43 .
(b) For 13 · 43 : Originally, there was 34 of a cubic yard of mulch in a mulch pile.
Then 31 of the mulch in the mulch pile was removed. How much mulch was
removed?
For 34 · 31 : Originally, there was 34 of a cubic yard of mulch in a mulch pile.
Then 13 of a cubic yard of mulch was removed from the mulch pile. How
much mulch was left in the mulch pile?
2. (a) This is a story problem for
2
3
· 41 .
89
(b) We can’t solve this problem because we don’t know how many students
are in Mrs. Watson’s class.
(c) This is a story problem for
2
3
· 41 .
(d) We can’t solve this problem because we don’t know how many students
are in Mrs. Watson’s class.
(e) We can’t solve this problem. Even if we knew that Carla had used all her
jelly worms in the snack bags, we would only be able to say that 32 of her
jelly worms had been bought, not 32 · 14 .
Class Activity 5D: What Fraction Is Shaded?
In figure 1 the larger shaded triangle is 21 of 53 of the whole rectangle. The smaller
shaded triangle is 12 of 15 of the whole rectangle. Overall the shaded region is 21 · 35 +
1 1
· = 52 of the whole rectangle.
2 5
In figure 2 the large shaded rectangle is 13 of the whole rectangle. The smaller
shaded rectangle is 31 · 83 of the whole rectangle. Overall the shaded region is 13 + 13 · 38 =
11
of the whole rectangle.
24
In figure 3 the lower left shaded triangle is 12 · 41 of the whole rectangle. The upper
right shaded rectangle is 41 · 41 · 13 of the whole rectangle. Overall the shaded region is
1 1
7
· + 14 · 41 · 13 = 48
of the whole rectangle.
2 4
For part 2, see Figure 5.1.
Class Activity 5E: Multiplying Mixed Numbers
Consider delaying this activity and doing it together with Decimal Multiplication and
Areas of Rectangles.
1. The rectangle decomposes into 2 whole unit squares, 3 rectangles, each of which
is 21 of a unit square, and another 14 of a unit square. All together, that makes
3 43 square units of area.
2. See Figure 5.2 for decompositions of the rectangle that correspond with the two
calculation methods.
(a)
1
1
1
1
= (1 + ) × (2 + )
1 ×2
2
2
2
2
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CHAPTER 5
Figure 5.1: Shading for an expression
1
1
1
= 1 × (2 + ) + × (2 + )
2
2
2
1 1
1 1
= 1×2+1× + ×2+ ×
2 2
2 2
3
= 3
4
(b)
1
1
3 5
1 ×2
=
×
2
2
2 2
15
=
4
3
= 3
4
5.2
Multiplying Decimals
More on Explaining Why We Place the Decimal Point Where We Do When We
Multiply Decimals
More on Decimal Multiplication And Areas of Rectangles.
91
1×2
1/2 × 2
1 × 1/2
1/2 × 1/2
The four sections correspond
with the 4 terms produced
by the distributive property.
3 × 5 squares, each of which
has area 1/2 × 1/2 = 1/4
square units.
Figure 5.2: Decomposing the rectangle according to different area calculation methods
Class Activity 5F: Multiplying Decimals
1. How much does 2.7 pounds of sugar cost if the price is $1.35 per pound?
2. Since 3.46 is between 3 and 4 and 1.8 is between 1 and 2, the product must be
between 3 and 8, so the decimal point must go after the 6 for the product to be
the right size.
1
3. 3.46 is between 3 and 4 and 0.18 is between 10
and 1, so the product must be
between 0.3 and 4. Therefore only 0.6228 makes sense.
We can see that this method of estimation is easier to use when both factors
are between two non-zero whole numbers.
4. The rule does work but 15 × 12 ends in a 0 and 1.8 = 1.80.
Class Activity 5G: Explaining Why We Place the Decimal
Point Where We Do When We Multiply Decimals
1. To get from 136 × 27 = 3672 back to 1.36 × 2.7 we must divide by 10 a total
of 3 times (2 times plus 1 more time). Therefore we move the decimal point in
3672 3 places to the left to get 3.672.
2. See text and use the ideas above.
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Class Activity 5H: Decimal Multiplication And Areas of Rectangles
Consider doing the Class Activity on Multiplying Mixed Numbers, right after (or
before) this one and comparing the situations. If so, you might also want to ask
the students how the distributive property is relevant in the decimal multiplication
situation.
If students don’t bring it up on their own, ask them if the rectangle reminds them
of other math drawings they have used. They will hopefully bring up the case of
whole number multiplication.
1. Since 1 large square represents 1 square unit, each long strip represents 0.1
square units, and each small square represents 0.01 square units. We see that
the total in the figure on the bottom includes small squares representing 0.01
square units but no smaller amounts. By rearanging these pieces one can form
4 large squares, 1 long strip, and 4 small squares. Hence the total rea is 4.14
square units.
2. The 2.3-unit-by-1.8-unit rectangle can be viewed as a 23-by-18 array of 0.1-by0.1 small squares. Since each of these small squares has area 0.01 square units,
the total area is 23 × 18 × 0.01 square units.
In the case of decimal, whole number, and mixed number multiplication, we
can apply the distributive property to calculate answers, which we can view as
decomposing rectangles.
5.3
Multiplying Negative Numbers
Using Checks and Bills to Interpret Multiplication with Negative Numbers
Does Multiplication Always Make Larger?.
Class Activity 5I: Patterns With Multiplication and Negative
Numbers
1. In the first list, the answers decrease by 6 going down the list. In the second
list, the answers decrease by 7 going down the list. In the third list, the answers
increase by 8 going down the list.
93
2. (a) (positive number) × (negative number) = (negative number)
(b) (negative number) × (positive number) = (negative number)
(c) (negative number) × (negative number) = (positive number)
Class Activity 5J: Explaining Multiplication with Negative
Numbers (and 0)
1. You use your credit card to purchase 3 $5.00 items at BigMart. How much
more do you owe the credit card company after this transaction? Three −$5.00
dollar charges yields a total charge of −$15.00.
2. Since 0 groups of any number of objects gives 0 objects, 0× (any number)= 0.
Since any number of groups of 0 objects gives 0 objects, (any number)×0 = 0.
3.
0 =
=
=
=
0×5
((−3) + 3) × 5
(−3) × 5 + 3 × 5.
(−3) × 5 + 15
So whatever number (−3)×5 is, when you add 15 to it, you get 0. Since the only
such number is −15, it must be the case that (−3) × 5 = 15. Or: subtracting 15
from both sides of the equation 0 = (−3) × 5 + 15 we see that −15 = (−3) × 5.
4.
0 =
=
=
=
(−3) × 0
(−3) × ((−5) + 5)
(−3) × (−5) + (−3) × 5.
(−3) × (−5) − 15
So whatever number (−3) × (−5) is, when you subtract 15 from it you get 0.
Since the only such number is 15, it must be the case that (−3) × (−5) = 15.
Or: adding 15 to both sides of the equation 0 = (−3) × (−5) − 15 we see that
15 = (−3) × (−5).
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5.4
Powers and Scientific Notation
Go online to My Math Lab for additional extension activities for this section:
How Many Digits Are in a Product of CountingNumbers?,
Explaining the Pattern in the Number of Digits in Products.
Class Activity 5K: Multiplying Powers of 10
1. (a) 103 × 104 = (10 × 10 × 10) × (10 × 10 × 10 × 10) = 107
(b) 102 × 105 = (10 × 10) × (10 × 10 × 10 × 10 × 10) = 107
(c) 103 × 103 = (10 × 10 × 10) × (10 × 10 × 10) = 106
2. The exponent in the product is the sum of the exponents in the two factors.
3. 10A × 10B stands for the product of A 10s multiplied by the product of B 10s.
All together, that’s A + B 10s multiplied together, so 10A × 10B = 10A+B .
4. If the equation is still true when B is 0, then 10A × 100 = 10A+0 so 10A × 100
is equal to 10A . This means 100 must be 1.
5. In general, 10N × 10−N = 10N +(−N ) = 100 = 1. In other words, when we
multiply 10−N by 10N we get 1. Therefore 10−N must be equal to 101N .
Class Activity 5L: Scientific Notation Versus Ordinary Decimal Notation
My calculator’s display reads 1.21932631113E17, which stands for
1.21932631113 × 1017
1. No, the calculator only has room to display so many digits.
2. The digits shown, except for possibly the last 3 (which might have been rounded),
are correct. We don’t know what the other digits are. In other words, even
though
1.21932631113 × 1017 = 121, 932, 631, 113, 000, 000
those last 6 zeros are not the correct digits for the answer to the multiplication
problem.
95
3. Yes, there are 18 digits, but we don’t know the last 6 (and the 3 might actually
be a 2 that was rounded).
4.
123456789 × 987654321 =
(12345 × 104 + 6789) × (9876 × 105 + 54321)
= 12345 × 9876 × 104 × 105 + 12345 × 54321 × 104
+6789 × 9876 × 105 + 6789 × 54321
=
121, 919, 220, 000, 000, 000
+6, 705, 927, 450, 000
+6, 704, 816, 400, 000
+368, 785, 269
121, 932, 631, 112, 635, 269
Use the calculator to multiply the numbers without the powers of 10, then put
on extra zeros for the powers of 10.
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Copyright Chapter 6
Division
97
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6.1
Interpretations of Division
Class Activity 6A: What Does Division Mean?
IMAP Video Opportunity
I recommend asking students to do Part 1 of this activity before you discuss the
two types of division. Then have a few people share their story problem for Part 1.
You will probably get mostly (or only) “how many in each group?” story problems.
Comment on the similar structure of these problems. Then lead students through
Part 2 of the activity, noting that most (or all) of the problems fit with the equation
2×? = 10
Even if some people share “how many groups?” story problems with the class, it will
probably be productive for students to do Part 3 for themselves in order to think
through this other type of problem.
You could also show IMAP Video clip 2, in which Dillon solves the two types of
division problems, starting at around 5:50 in the clip.
Students will need time and opportunity to think about and make sense of the
two definitions of division.
Class Activity 6B: Division Word Problems
1. 8 cups ÷2 cups in each batch = ? batches;
? batches × 2 cups in each batch = 8.
Bill can make 4 batches of muffins. “How many groups?” because we can view
each batch as a group.
2. 96 inches ÷12 inches in each foot =? feet;
? feet × 12 inches in each foot = 96 inches.
The rope is 8 feet long. “How many groups?” because we can think of each
foot as a group.
3. 32 yards ÷8 pieces =? yards in each piece;
8 pieces × ? yards in each piece = 32 yards.
Each piece is 4 yards long. “How many in each group?”
99
4. 400 pounds ÷8 pounds in each gallon = ? gallons;
? gallons × 8 pounds in each gallon = 400 pounds.
It’s 50 gallons. “How many groups?” because each gallon is a group of 8 pounds
and we want to know how many groups of 8 pounds are in 400 pounds.
5. 220 miles ÷4 hours = ? miles in each hour (or miles per hour);
4 hours × ? miles in each hour = 220 miles.
You drove 55 miles per hour. “How many in each group?” because the 220
miles are distributed equally among the 4 hours, so each hour is like a group
that has 55 miles in it.
6. $3.00 ÷ 6 limes = ? dollars for each lime (or dollars per lime).
6 limes × ? dollars for each lime = 3 dollars.
Each lime costs $0.50. “How many in each group?” because we can think of
each lime as a group. The money is distributed equally among the limes.
7. 6 limes ÷$3.00 = ? limes for each dollar (or limes per dollar).
3 dollars × ? limes for each dollar = 6 limes.
You can buy 2 limes for $. “How many in each group?” because we can think of
each dollar as a group. The limes are distributed equally among the 3 dollars.
Since a ”How many groups” division problem has the form number of objects ÷
number of objects per group = number of groups, the units pattern is units of object
÷ units of objects = units of groups.
Since a ”How many in each group” division problem has the form number of
objects ÷ number of groups = number of objects per group, the units pattern is units
of object ÷ units of groups = units of objects.
Class Activity 6C: Why Can’t We Divide by Zero?
1. If 2 ÷ 0 = C, then C × 0 = 2. But C × 0 = 0 no matter what C is, therefore
there is no number C such that C × 0 = 2.
2. How many in each group?: You have 2 gallons of juice to distribute equally
among 0 bottles. How much juice does each bottle get? Notice that you can’t
actually pour out the 2 gallons into the bottles in this case, i.e., you can’t
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CHAPTER 6
actually divide up the 2 gallons among the bottles because you don’t have any
bottles. So you can’t actually use up your 2 gallons. Note the link to 0×? = 2.
How many groups?: You have 2 gallons of juice to pour into “bottles” that each
hold 0 gallons. How many bottles will you need? Again, notice that you can’t
actually distribute the 2 gallons among the bottles. Note the link to ? × 0 = 2.
I don’t think it’s necessary to emphasize the word problem approach to seeing
why division by 0 is not defined, but I think it’s worth having prospective
teachers think about these word problems, if only for the opportunity to expand
their thinking.
3. How many in each group?: You have 0 gallons of juice to pour equally into 2
bottles. How much juice will you pour into each bottle? You will pour 0 gallons
into each bottle. Note the link to
How many groups?: You have 0 gallons of juice that you want to pour into jugs
that hold 2 gallons. How many jugs will you need? You will need 0 jugs.
Notice that in both cases, you can do the dividing and you can give an answer
to the problem. In contrast, in the 2 ÷ 0 case, when you try to pour 2 gallons
into “bottles” that hold 0 gallons, or into 0 bottles, you can’t actually pour out
the 2 gallons, i.e., you can’t actually divide it up.
4. 0 ÷ 0 should be the number C such that C × 0 = 0. But for every number C,
C × 0 = 0. Therefore 0 ÷ 0 can’t be defined as just a single number.
6.2
Division and Fractions and Division with Remainder
Division with Remainder Notation.
Class Activity 6D: Relating Fractions and Division
You might like to show IMAP Video clip 14, in which Felisha, a 4th grader, solves
and explains the solution to 2 ÷ 5 almost perfectly except there is some confusion
about what the appropriate whole is. This video was also recommended in the fraction
chapter.
101
1. Divide each of the 3 pizza into 4 equal slices and then give each of the 4 people
3 slices. Therefore each person gets 3 pieces, each of which is 14 of a pizza that
is, each person 34 of a pizza. It follows that 3 ÷ 4 = 34 .
I think it’s valuable to discuss the different ways to describe the amount that
each person gets: it is 3 pieces; it is 3 pieces, each of which is 14 of a pizza; it is
1
of all the pizza (all three pizzas).
4
2. 3 ÷ 5 = 35 .
3. A ÷ B =
A
.
B
Class Activity 6E: What to Do With the Remainder?
Note that decimal answers to division problems are discussed in the next section,
along with the common error of putting the remainder behind the decimal point.
1. Cookie baking problem: you can make 4 batches and you will have 2 cups of
flour left over. Brownie problem: you can put 4 brownies in each bag and will
have 2 brownies left over.
Notice that the units/wholes associated with the 4 and the 2 are different for
the cookie baking problem but are the same for the brownie problem. This
is because the cookie baking problem in a “how many groups?” problem, for
which the quotient has a different unit than the dividend (batches and cups),
whereas the brownie problem is a “how many in each group?” problem, for
which the quotient has the same unit as the dividend (brownies).
Cookie baking problem: you can make 4 23 batches, which means 4 full batches
and 32 of another batch (or: your last batch is a smaller batch, namely 23 the size
of the other batches). This is assuming that it’s possible to make 23 of a batch
(there might be issues with using 32 of an egg, for example). The 2 cups of flour
that are left over after making 4 full batches can be used to make another 23
of a batch. Notice that these 2 cups of flour are 2 out of the 3 cups needed to
make a full batch (in this case, unlike in the brownie case, there is no need to
break the 2 cups up into parts).
Brownie problem: you can put 4 32 cookies in each bag, which means 4 whole
brownies and 23 of another brownie, which could consist of 2 pieces, each of
which is 13 of a brownie. The 2 brownies that remain after putting 4 brownies in
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CHAPTER 6
each bag can be subdivided and given out, resulting in each bag getting another
2
of a brownie.
3
2. Some people might say that the cookie baking problem from part 1 is suitable
for that. Better might be some context such as putting people or things into
containers that can only hold up to 3 people/things (e.g. rides at an amusement
park, tennis ball containers). You’ll need a fifth container to hold the remaining
2 people/things.
3. Each week has 7 days and every seventh day is the same day of the week. Since
31 ÷ 7 = 4, remainder 3, there are 4 full weeks plus another 3 days in 31 days.
After 4 full weeks, you are back to the same day of the week as today. After
the additional 3 days, you are at the day of the week that is 3 days from today
(e.g., if today is a Monday, then you will be at a Thursday).
4. You might suggest that students use double number lines (with miles on one
line and hours on the other) as an aid with these problems.
You might also ask the students what kind of division problems these are. They
are “how many groups?” problems because we want to see how many groups
of 55, of 60, or of 30 are in the total number of miles.
15
.
(a) 180 ÷ 55 = 3, remainder 15 or 3 55
15
195 ÷ 60 = 3, remainder 15 or 3 60 = 3 14 .
15
105 ÷ 30 = 3, remainder 15 or 3 30
= 3 12 .
(b) In each case, we can answer 3 full hours with another 15 miles to go for
the whole number with remainder answer. The mixed numbers tell you
how many hours it will take in all, which in each case is 3 full hours and
another fraction of an hour.
(c) The remainder of 15 is really miles, not minutes. But when you are traveling 60 miles per hour, then you go 15 miles in 15 minutes. If you are
traveling close to 60 mph, as in the case of 55 mph, then it takes close to
15 minutes to travel 15 miles.
6.3
Why Division Algorithms Work
Can We Use Properties of Arithmetic to Divide?
103
More on Dividing Without Using a Calculator or Long Division
Using the Scaffold Method Flexibly
More on Interpreting the Standard Division Algorithm as Dividing Bundled Toothpicks
Using Division to Calculate Decimal Representations of Fractions
Rounding to Estimate Solutions to Division Problems.
As with the multiplication algorithms, I think it’s important to emphasize the distinction between the meaning of division and the division algorithms. It’s important
for teachers to know that the algorithms can be explained in terms of place value and
the properties of arithmetic and are not merely “given.”
Internet Video opportunity
As an introduction to this section or in conjunction with one of the activities on
the standard division algorithm, you and your students might enjoy watching and
analyzing the error in the “Ma and Pa Kettle” video at
http://www.youtube.com/watch?v=y1g-b3SUjik
in which 25 ÷ 5 is calculated as 14 in several different ways. It’s always the same
place value error that occurs.
Class Activity 6F: Dividing Without Using a Calculator or
Long Division
1. The second equation collects the packages together.
2. Ashley finds that 10 groups of 6 is 60. She then repeatedly doubles to find that
20 groups of 6 is 120 and 40 groups of 6 is 240. She then adds on 2 more groups
of 6 and finally one last group of 6 for a total of 43 groups of 6, which make
258.
2 × 2 × 10 × 6 + 2 × 6 + 1 × 6
40 × 6 + 2 × 6 + 1 × 6
(40 + 2 + 1) × 6
43 × 6
=
=
=
=
258
258
258
258
3. By first multiplying 15 by 2 and then multiplying the product by 2 and then 4,
Zane has multiplied 15 by a total of 4 × 2 × 2 = 8.
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4 × (2 × (2 × 15)) + 5 = 245
(4 × 2 × 2) × 15 + 5 = 245
16 × 15 + 5 = 245
4. Think about dividing 1000 bottle caps equally among 6 groups (or into groups
of 6) in chunks. First you could put 100 in each of the 6 groups. That leaves
400 bottle caps left. Then you could put another 50 bottle caps in each group,
which uses up 300 more bottle caps. Now there are 100 left so you could put
10 more in each group, leaving 40 left, then another 6 in each group, leaving
4 bottle caps remaining. Altogether, each group has 100 + 50 + 10 + 6 = 166
bottle caps and there are 4 left. Of course other ways are possible too.
Class Activity 6G: Why the Scaffold Method of Division Works
A modification (or extension) to consider here is to change the word problems to
“how many in each group?” problems. It would also be good for the items that are
to be divided to be in base ten bundles, e.g., 3 bundles of a thousand, 4 bundles of a
hundred, 7 bundles of ten, and 4 ones. In this case the discussion will be very similar
to the discussion for the standard algorithm.
1. First make 400 bags of marbles. That uses 3200 marbles and leaves 275 marbles
to be put in bags. Then make 30 bags of marbles. That uses another 240 marbles
and leave 35 marbles left. Finally, make 4 bags of marbles. That uses 32 more
marbles and leave 3 marbles left. All together, 400 + 30 + 4 = 434 bags of
marbles were made and 3 marbles are left, so 3274 ÷ 8 = 434, remainder 3.
Starting with 3475 marbles, you use up 400 × 8 marbles, then another 30 × 8
marbles, then another 4 × 8 marbles, leaving 3 marbles at the end. The second
equation shows the 400, then 30, then 4 bags being collected. The third equation
shows that after 434 bags of 8 marbles each have been made, there are still 3
marbles left.
105
2.
6
80
300
1000
6)8321
−6000
2321
−1800
521
−480
41
−36
5
Therefore 8321 ÷ 6 has whole number quotient 1386, remainder 5.
3.
8321 − 1000 × 6 − 300 × 6 − 80 × 6 − 6 × 6 = 5
8321 − (1000 + 300 + 80 + 6) × 6 = 5
8321 − 1386 × 6 = 5
Class Activity 6H: Interpreting the Standard Division Algorithm as Dividing Bundled Toothpicks
If you have enough bundled toothpicks available, you might have students use them to
illustrate the steps of the standard division algorithm. I recommend using a smaller
dividend, e.g., 157 ÷ 3. The bundle of 100 should be bundled into 10 bundles of ten so
that the physical dividing will parallel the steps of the standard division algorithm.
I recommend discouraging students from using the common “goes into” language,
which I’m not sure is understood in a meaningful way.
In the standard long division algorithm we first see if we can distribute 1 thousand
among 3 groups. In order to do so, we must first unbundle the 1 thousand into 10
groups of 100 and combine them with the 3 groups of 100 to make 13 hundreds. We
can now divide the 13 hundred bundles of toothpicks into 3 groups of 4 hundred with
Copyright 106
CHAPTER 6
a remainder of 1 hundred. When we now “bring down” the 7 to make a 17. This is
like unbundling the remaining 1 hundred into 10 tens and combining these 10 tens
with the 7 tens we had to make 17 tens. Now we can give each group 5 tens and 2
tens will remain. When we “bring down” the 2 next to the 2 we have, this is like
unbundling the 2 tens to make 20 ones and combining them with the 2 ones we have
to make 22 ones. We now distribute 7 toothpicks to each group and we are left with
a remainder of 1 toothpick. All these steps can be shown with a math drawing by
showing how bundles are divided among the three groups and showing the unbundling
of remaining bundles.
The additional part of this activity that can be found online has a step that
involves a 0, which often leads to errors. You could use that part to lead into the
activity on errors.
Class Activity 6I: Interpreting the Standard Division Algorithm in Terms of Area
You might want to point out that of course the rectangle in the example is not to
scale (and of course it would not be reasonable to try to draw it to scale).
1. There are 948 square units of area inside the rectangle, which can be chunked
as 9 hundreds, 4 tens, and 8 ones (units of area). At each step you are finding a
portion of the length of the top side of the rectangle. Ask students to attend to
and explain the “bringing down” steps in the algorithm in terms of unbundling
hundreds into tens and tens into ones (see the text for a discussion of this).
2. Write the area inside the rectangle as 1 Th, 5 H, 4 T, 3. The first step consists
of unbundling the 1 Th as 10 H and joining it with the 5 H to make 15 H. Then
proceed as in the previous example. Note that there will be 0 T in the side
length, which can be a difficult step for students in elementary school.
Class Activity 6J: Student Errors in Using the Division Algorithm
1. First case: There should be a 0 between the 3 and the 9 in the quotient because
12 goes into 11 zero times.
Second case: The student didn’t “bring down” the 9. The student may think
that whenever a 0 appears, it’s time to stop.
107
Third case: The student didn’t “bring down” the 0. Again, it may be that the
student thinks a 0 anywhere means it’s time to stop.
2.
9
300
12)3711
−3600
111
−108
3
2
80
100
4)729
−400
329
−320
9
−8
1
4
30
100
7)940
−700
240
−210
30
−28
2
Class Activity 6K: Interpreting the Calculation of Decimal
Answers to Whole Number Division Problems in Terms of
Money
891.33
3) 2674
−24
27
−27
04
−3
10
−9
10
−9
1
← each person gets 8 hundreds
← 2 hundreds and 7 tens = 27 tens remain
← each person gets another 9 tens
← 0 tens and 4 ones = 4 ones remain
← each person gets another 1 one
← 1 one and 0 tenths = 10 tenths remain
← each person gets another 3 tenths
← 1 tenth and 0 hundredths remain
← each person gets another 3 tenths
← 1 hundredth remains
Class Activity 6L: Errors in Decimal Answers to Division
Problems
1. No, you need to know the divisor to be able to give the decimal or mixed number
version of the answer.
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CHAPTER 6
2. Ben is simply putting the remainder behind the decimal point.
6.4
Fraction Division from the How Many Groups?
Perspective
Class Activity 6M: “How Many Groups?” Fraction Division
Problems
If available, you might like to use pattern tiles (yellow hexagons, red trapezoids,
blue rhombuses, and green triangles) for the fraction division problems. The division
word problems could all be of the form “how many of these pattern tiles are in these
other pattern tiles?” For example, you could let 2 yellow hexagons (combined) stand
for 1. Then 3 trapezoids would stand for 34 and for 3 ÷ 34 you’d ask “how many sets
of 3 trapezoids are in 6 hexagons?” You could have students pose and solve various
fraction division problems using the pattern tiles in this way.
As a warmup, you could ask students what 1 ÷ 13 means from the “how many
groups?” perspective, have them think about it for a bit, and then show the first
part of IMAP Video clip 16, in which Elliot solves that division problem. Save the
second problem in the video clip for discussion in conjunction with part 3b, c of this
activity.
1. You have 3 bags of bird seed. Each of your bird feeders take
them. How many bird feeders can you fill?
3
4
of a bag to fill
Note that a strip diagram, table, and double number line will be nearly identical
to the example. The only difference is that 3, 6, 9, and 12 are replaced with 3
fourths, 6 fourths, 9 fourths, and 12 fourths and the strips that represent 1 now
represent 1 fourth.
2. Tonya is correct in saying that there is 1 serving of rice and 31 of a cup of rice
left over, but this does not mean that 1 13 servings of rice were eaten. That 31
cup that is left over is 12 of a serving, so 1 12 servings of rice were eaten.
This part of the activity will prepare students for thinking about the problems
coming up next.
In IMAP Video clip 16, Elliot makes an “incorrect whole” error of the type
being discussed here on the second problem that he works (1 12 ÷ 13 , which is in
part 3b, c of this activity).
109
3. A punch recipe calls for 31 of a cup of grape juice. You have 1 12 cups of grape
juice. How many batches of the punch recipe can you make (you have plenty
of all the other ingredients)?
Note that in the second part of IMAP Video clip 16, Elliot works on 1 21 ÷ 13
and has a lot of good ideas but doesn’t get the correct answer. You might show
the clip first (without commenting on whether Elliot gets the correct answer or
not), then have students work on this part of the class activity, and then show
the clip again and discuss it in light of their own work.
If you draw a strip diagram showing a strip and another 12 of the same size strip
and divide the large strip into 6 equal parts and the 21 strip into 3 equal parts,
then each batch of the recipe requires 13 = 26 of the recipe which uses 2 of the
small parts. So 4 batches will use up 8 of those parts, leaving 1 part left. This
1 part can make 21 of a batch because 2 of those parts make a whole batch. So
you can make 4 12 batches.
Looking ahead to the next activity, you may wish to point out that it was
helpful to give the fractions a common denominator and that once they have a
common denominator, you are essentially just working with the numerators. In
this case, the problem became “how many groups of 2 sixths are in 9 sixths?”
which is solved by 9 ÷ 2.
4. If 43 of a cup of juice provides your full daily value of vitamin C, then what
fraction of your daily value of vitamin C is in 13 cup of juice?
9
If you draw a picture showing 43 = 12
of a cup of juice that you need for your
1
4
daily value and 3 = 12 of the cup of juice that you have, then you see that you
get 4 out of the 9 parts you need, so you get 49 of your daily value of vitamin C.
Looking ahead to the next activity, you may wish to point out that it was
helpful to give the fractions a common denominator and that once they have a
common denominator, you are essentially just working with the numerators.
Class Activity 6N: Dividing Fractions by Dividing the Numerators and Dividing the Denominators
Internet Video opportunity
See the Lesson Study video clips under “Let’s think about how to multiply and
divide fractions by fractions!,” a grade 6 lesson in Japan available at
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CHAPTER 6
Clips 2 and 3 are especially relevant to this activity.
1. “How many 52 are in 56 ?” is what 65 ÷ 52 means from the “how many groups?”
perspective. Ignoring that each piece is labeled 15 , we really just want to know
how many sets of 2 blocks are in 6 blocks, which is what 6 ÷ 2 means from the
“how many groups?” perspective.
2. To solve the multiplication formulation of the problem, we need to find out
what times 3 is 6 (i.e., find 6 ÷ 2) and what times 4 is 20 (i.e., find 20 ÷ 4),
given that we already know how fraction multiplication works.
3. Any example where the numerator and denominator of the first fraction are
evenly divisible by the numerator and denominator, respectively, of the second
fraction, will work. For example,
12 4
12 ÷ 4
3
÷ =
=
25 5
25 ÷ 5
5
4. The idea is that you can’t initially use the method of dividing the numerators
and dividing the denominators, so you turn 57 into an equivalent fraction where
you can use that strategy. By leaving the products “unmultiplied”, you can see
that the net effect of dividing by 34 is in fact to multiply by its reciprocal, 34 .
The strategy is general, and explains why the “invert and multiply” method of
fraction division is valid.
Note that this explanation is yet another explanation for why “invert and multiply” is valid that is a little different from the two that are in this section of
the text (and quite different from the one given in the next section).
6.5
Fraction Division from the How Many in One
Group? Perspective
Go online to My Math Lab for an additional activities for this section:
“How Many in One Group?” Fraction Division Problems, Alternate
Using “Double Number Lines” to solve “How Many in One Group?” Division
Problems.
111
Class Activity 6O: “How Many in One Group?”
Division Problems
Fraction
1. Problem: If 41 of a meter of wire weighs 3 grams, then how much does 1 meter
of the wire weigh?
Strip diagram: Show a strip in 4 parts, with 1 part shaded to show the 14 meter
of wire. That amount weighs 3 grams, so the full meter, which consists of 4
such parts weighs 3 · 4 = 12 grams.
Table:
1
4
2
4
3
4
4
4
meter −→ 3 grams
= 1 meter −→ 12 grams
Double number line: As in the example, show 0 grams and 0 meters aligned.
Then show 3 grams and 14 meter aligned, show 6 grams and 24 meters aligned,
9 grams and 43 meters aligned, and show 12 grams and 44 = 1 meter aligned.
Regardless of method, we see that since 4 fourths is 4 times as much as 1 fourth,
the weight of 1 meter is 4 times as much as the weight of 41 meter. Therefore
the weight of 1 meter is 3 · 4 = 12 grams.
2. The problem is a 6 ÷ 34 “how many in one group?” division problem because 6
miles makes 43 of a group (mile) and we want to know how much makes 1 full
group (mile).
In solving the problem with a strip diagram, table, or double number line,
proceed as in the previous part, and be sure to find how far Anna runs in 14 of
an hour first. In 14 of an hour she runs 2 miles because the 6 miles are divided
equally among each of the 3 fourths. Because 1 hour is 4 of those fourths, she
runs 2 · 4 miles. Therefore to find how far she ran, we first divided the 6 miles
by 3 and then we multiplied that result by 4. The net effect is to multiply 6 by
4
.
3
3. The example wire problem or the running problem from part 2 could be adapted
for these numbers. For example: Grandpa walks 21 mile in 23 of an hour. At
that pace, how far will Grandpa walk in an hour?
In solving the problem, proceed as in the previous parts. Be sure to find out
how far Grandpa walks in 13 of an hour first. Because 32 is 2 thirds, he walks
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CHAPTER 6
1
2
÷ 2 miles in each 1 third of an hour. Then in 1 hour—which is 3 thirds of an
hour—he walks 3 times as far, namely 12 ÷ 2 · 3 miles. Dividing by 2 and then
multiplying by 3 is the same as multiplying by 32 , so Grandpa walks 12 · 32 miles
in one hour.
Class Activity 6P: Are These Division Problems?
Remind students that if they change the numbers in the problems to whole numbers
it will often become easy to see if a problem is a division problem or not.
Use this activity to discuss the difference between dividing in half and dividing
by 21 (see the text discussion).
1. Yes, this is a problem for
2. No, this is a problem for
3
4
1
2
3
4
÷
1
2
of the “how many groups?” type.
× 34 .
÷
1
2
of the “how many in one group?” type.
4. This is a problem for 21 ÷ 34 of the “how many in one group?” type ( 21 of a
mile “fills” 34 of the road and we want to know how many miles make the whole
road).
3
4
÷
1
2
of the “how many in one group?” type.
3
4
÷
1
2
of the “how many groups?” type.
7. No, this is a problem for
6.6
1
2
× 34 .
Dividing Decimals
Class Activity 6Q: Quick Tricks and Estimation with Decimal
Division
1. Since 0.5 = 12 , 32.5 ÷ 0.5 = 32.5 × 12 = 65. Or you can think: how many 50
cents are in $32.50? There are 2 for each of the 32 dollars and one more.
2. 1.2 ÷ 0.25 = 1.2 ÷
1
4
= 1.2 × 4 = 4.8.
3. 7.2 ÷ 0.333 is approximately equal to 7.2 ÷
1
3
= 7.2 × 3 = 21.6.
113
4. Since 2.45 is between 2 and 3 and 1.5 is between 1 and 2, the quotient must
be between 1 and 3, so the only place where it makes sense to put the decimal
point is after the 1. So the answer must be 1.633.
Class Activity 6R: Decimal Division
If you have time, you might like to point out that the reasoning used in problems 2
and 3 of this activity (and on page 346 of the text), which allows us to shift decimal
points, is the very same reasoning used on pages 331, 332 of the text in which fractions
are first given a common denominator so that one then just divide the numerators.
Problem 5 on page 350 of the text examines this connection.
1. ”How Many Groups?”: It takes 2.7 gallons of gasoline to fill your lawnmower’s
gas tank. How many times can you fill this tank with 23.45 gallons?
”How Many in One Group?”: 23.45 gallons of gas will fill your car’s tank 2.7
times. How much gas does your car’s tank hold?
2. The problem “how many $0.25 are in $12.37?” is equivalent to “how many 25
cents are in 1237 cents?”. The first problem can be solved by
0.25)12.37
and the second can be solved by
25)1237
so the two problems are equivalent, i.e., have the same solution.
3. If each box is worth a penny, or $.01, the the number of groups of 2 boxes in
6 boxes is the same as the number of $.02 in $.06. If the boxes are worth a
dime, then we have the interpretation 0.6 ÷ 0.2. If each box is worth $1.00, the
interpretation is 6 ÷ 2. If each box is worth 1, 000, 000 then it’s 6, 000, 000 ÷
2, 000, 000. If each box is worth $10.00, then it’s 60 ÷ 20, etc.
If students don’t bring it up on their own, ask if the boxes could even have
a fractional value. If each box is worth 17 , then the picture represents 67 ÷ 72 .
Notice that this taps into a line of reasoning that was used in section 6.4 to
divide fractions by giving them a common denominator first and then dividing
the numerators.
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CHAPTER 6
The moral is that all of these division problems are equivalent to the easy
whole number division problem 6 ÷ 2. Therefore, to solve any of the decimal
or large number division problems, we can just replace it with 6 ÷ 2. Notice
that replacing those other division problems with 6 ÷ 2 is the same as shifting
the decimal points in both the dividend and the divisor by the same number of
places in the same direction.
4. Students could represent 0.15 with drawings showing 5 individual items and 1
bundle of 10 items and they could represent 1.2 with 2 bundles of ten and 1
bundle of one hundred. These “bundles” could be base ten blocks, as in the
text. Phrased in terms of the drawings of bundled objects, the question “How
many (1 bundle of ten and 5 individuals) are in (1 bundle of a hundred and 2
bundles of ten)?” can stand for the original division problem 1.2 ÷ 0.15 or for
the whole number division problem 120 ÷ 15 (as well as many other division
problems) by reinterpreting the value of 1 individual. That the quotient is 8
could also be seen from the drawings.
Copyright Chapter 7
Ratio and Proportional
Relationships
115
116
7.1
CHAPTER 7
Motivating and Defining Ratio and Proportional Relationships
Comparing Paint Mixtures,
Using Strip Diagrams and Double Number Lines to Represent Ratios
Using Ratio Tables to Compare Two Ratios.
Class Activity 7A: Comparing Mixtures
I like to ask students to: “put your kid eyes on,” and ask them to think about using
these mixtures as a way to introduce and motivate the concept of ratio.
If possible, make the first mixture twice, in two containers. Then, before you
add the additional juices to the second container to make the second mixture, ask
students to predict if there will be a difference. This is fun and interesting!
Another good variation is to compare three mixtures: the first two are the ones
given in the activity and the third mixture is a double batch of the first one. If you
do this, then make the first mixture 3 times in 3 containers. Then tell students what
you will add to the second two containers and ask them to predict what will happen.
Then add the juices and see.
If you use different kinds of juices to make the mixtures, test them out first to see
if their colors and flavors will be different. I have tried mixing other pairs of juices,
such as purple and white grape juices, but often didn’t get a good color separation.
You might also try water colored with food coloring instead (for the color, not the
flavor).
You can use parts 2 and 3 to discuss the definition of ratio from the first perspective.
See the online Activity Bank in My Math Lab for a paint mixture version of the
first part of this activity.
1. If the punch mixtures were to taste the same it would have to be true that that
the fraction, or %, of each mixture that is red punch is the same. For the first
mixture 14 , or 25%, is red punch, but with the second mixture 38 , or 37.5% is
red punch.
Another way to compare the mixtures is to consider what would happen if you
made 2 batches of the first mixture. In that case, you’d only use 2 cups of red
punch and 6 cups lemon-lime soda as compare to 3 cups red punch and 5 cups
lemon-lime in the other mixture.
117
Some children will think that the mixtures should taste the same because you
are adding the same amount of each drink or because the difference between the
amounts of the two juices is the same in both cases. These children are making
additive, not multiplicative comparisons.
2. The combined contents of the containers will always taste the same because it’s
the combination of things that taste the same.
#
#
#
#
batches
1
cups red punch
1
cups lemon-lime soda 3
cups mixture
4
2 3 4
2 3 4
6 9 12
8 12 16
5 6
5 6
15 18
20 24
Use this table and the next table to talk about how each column is obtained
from the first column by repeating a “batch,” which is the combined unit of 1
cup red punch and 3 cups lemon-lime soda (which makes 4 cups of the mixture).
This motivates the definition of ratio from the “composed unit” perspective that
is discussed in the text.
Note that it’s also possible to put other entries in the table, such as for 100
batches, or even fractional entries, such as for 12 of a batch or 31 of a batch.
3. As in part 2, the combined contents of the containers will always taste the same
because it’s the combination of things that taste the same.
#
#
#
#
batches
1
cups red punch
3
cups lemon-lime soda 5
cups mixture
8
2 3 4 5 6
6 9 12 15 18
10 15 20 25 30
16 24 32 40 48
4. Each column within a table makes a mixture that tastes the same as all the
other mixtures, so any two columns from the two tables can be compared.
If we make 8 cups of each mixture (second column of table 1 and first column
of table 2), then the first mixture uses 2 cups of red punch whereas the second
mixture uses 3 cups of red punch, so the second mixture tastes more like red
punch. You can also look at mixtures that use the same amount of red punch,
or mixtures that use the same amount of lemon-lime soda.
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CHAPTER 7
Class Activity 7B: Walking Times and Distances
1. The double number lines should show the same entries as the tables.
Devonte: Every 2 seconds, he walked another 3 meters.
Devonte: Every 3 meters took him 2 seconds.
Kellie: Every 3 seconds, she walked another 4 meters.
Kellie: Every 4 meters took her 3 seconds.
2. There is no consistent number of meters that correspond with each fixed number
of seconds. At first Heather walks 1 meter in 2 seconds, but then she walks 2, 3,
4, and 5 meters in 2 seconds. Her walking can’t be shown on a double number
line (in the usual way) because equal time intervals don’t correspond with equal
distance intervals. Therefore Heather’s table is not a ratio table.
Devonte’s and Kellie’s tables are ratio tables because of the sentences in Part
1.
3. The graphs for Kellie and Devonte are straight lines that go to the origin. The
graph for Heather is not a straight line. Students may make some observations
that are prompted by Parts 4 and 5. See below.
It makes sense to connect the points because at times in between, the students
are still moving and will have locations in between. The origin fits on the graphs
because after 0 seconds, the students have moved 0 meters.
4. Those sentences are reflected in “slope triangles” like the ones shown in the
text. For Kelly: every 3 units to the right correspond to 4 units up on the
graph. For Devonte: every 2 units to the right correspond with 3 units up on
the graph. This consistent way of changing shows that that Kellie and Devonte
are each walking at a constant speed. We can tell that Heather is not walking
at a constant speed because the same amount of increase in time (horizontal
change) results in different amounts of increase in distance (vertical change).
5. At the vertical line through 6 seconds, Devonte’s location is at 9 meters, which is
above Kellie’s 8 meters, so Devonte walked farther in that amount of time than
Kellie. Therefore Devonte walked faster. Similarly, a horizontal line through 12
meters shows Devonte at 8 seconds and Kellie at 9 seconds. So Devonte took
less time than Kellie for the same distance and therefore Devonte walked faster.
Also, Devonte’s graph is steeper and is above Kellie’s, which shows that for each
given time, he has walked farther than Kellie and therefore walked faster.
119
Class Activity 7C: Using strip diagrams to represent ratios as
“fixed numbers of parts”
If:
Then:
1. Then:
Then:
1 part is
3 parts red
juice is
5 parts lemonlime is
8 parts total
punch is
3 liters
9 liters
1
2
1 21
5 cups
15 cups
6 cups
18 cups
gallon
gallons
25 cups
30 cups 15 liters
2 21 gallons
40 cups
48 cups 24 liters
4 gallons
Students should see the table as “filling the parts” rather than repeating batches.
2. Similar to Part 1.
7.2
Solving Proportion Problems by Reasoning with
Multiplication and Division
The CCSS expect students to reason about strip diagrams, ratio tables, and double
number lines to solve problems about ratios and proportional relationships.
Class Activity 7D: Using Strip Diagrams to Solve Ratio Problems
1. (a) The 2 parts of blue paint contain 40 pails, so each part contains 40÷2 = 20
pails of paint. The three yellow parts then stand for 60 pails of paint.
(b) The 3 parts of yellow paint contain 48 pail, so each part contains 48÷3 = 16
pails of paint. The two blue paint rectangles then stand for 2 · 16 = 32
pails of paint.
(c) The 5 total parts contain a total of 150 pails of paint, so each part contains
150 ÷ 5 = 30 pails of paint. The three yellow parts then stand for 90 pails
of paint and the two blue paint rectangles stand for 60 pails of paint.
2. The 4 total parts in Figure 7.1 represent 160 students. Therefore each part
represents 160 ÷ 4 = 40 students. Therefore 40 students ate a hot dog and
3 · 40 = 120 students ate pizza.
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CHAPTER 7
Hot Dog
Pizza
Figure 7.1: Hot Dog or Pizza?
3. Represent fruit juice with a strip of 3 parts and bubbly water with a strip of 5
parts. Since you want 24 liters of punch, and there are 8 parts total, each part
represent 24 ÷ 8 = 3 liters of liquid. Hence you need 3 × 3 = 9 liters of fruit
juice and 5 × 3 = 15 liters of bubbly water.
To make 10 liters of punch, each part now consists of 10 ÷8 = 1 41 liters of liquid.
You need 3 × 1 41 = 3 34 liters of fruit juice and 5 × 1 41 = 6 14 liters of bubbly water.
Class Activity 7E: Ratio Problem Solving with Strip Diagrams
1. Before Shauntay gave 15 cards to Jessica we have the strip diagram given in
Figure 7.2.
Shauntay:
Jessica:
Figure 7.2: Before
After Shauntay gave Jessica 15 cards we have the strip diagram give in Figure 7.3.
Shauntay:
Jessica:
Figure 7.3: After
121
One rectangle holds 15 cards and hence both Jessica and Shauntay now have
60 cards.
2. Using before and after strip diagrams, as in the previous problem, we see that
1 21 rectangles holds 12 cards which implies that one rectangle holds 12 ÷ 1 12 = 8
cards. Hence both Jessica and Shauntay now have 3 21 × 8 = 28 cards.
3. The ratio of Shauntay’s cards to Jessica’s cards is 8 to 3. After Shauntay gives
Jessica 15 cards, both girls have the same number of cards. How many cards
do Shauntay and Jessica each have now?
Both have 33 cards now. Note that Jessica got 2 21 rectangles when the before
strip diagram was transformed into the after. Hence the number cards she gets
from Shauntay divided by 2 12 must equal an integer which happens only when
the number of cards is a multiple of 5.
Class Activity 7F: Multiplicative Relationships in Ratio Tables
1. The entries in the pails of blue paint row are 2, 4, 6, 8, 10, 12, 14, 16, 18, and
20.
The entries in the pails of yellow paint row are 3, 6, 9, 12, 15, 18, 21, 24, 27,and
30.
The entries in the pails of green paint produced row are 5, 10, 15, 20, 25, 30,
35, 40, 45,and 50
2. The entries in the second row increase by 2 each time you step one place to the
right, by 3 each time you step one place to the right in the third row, and by 5
in the fourth row. This pattern is a consequence of the fact that each time the
number of batches increases by 1, you must use 2 more pails of blue paint and
3 more pails of yellow paint in which case 5 more pails of green paint will be
produced.
Students will hopefully bring up that they have seen rows like these in multiplication tables. They might also think of strings of equivalent fractions.
3. Each entry in the eighth column of the table is 8 times the corresponding entry
in the first column. For the sixth column the multiplication factor is 6. In
general, every entry in the nth column is n times the corresponding entry in the
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first column because the nth column is n batches of the first column, so every
quantity is n times as much.
4. There is always 1 21 times as much yellow paint as blue paint, 2 21 times as much
green paint produced as blue paint used, and 1 23 times as much green paint
produced as yellow paint used (among other relationships). A strip diagram
will make these relationships more visually evident.
You can use this activity to discuss that when you make horizontal comparisons,
you are comparing like quantities (blue paint to blue paint, yellow paint to
yellow paint), but when you make vertical comparisons within a column you
are comparing blue paint to yellow paint (or blue paint to green paint, etc.).
Class Activity 7G: Solving Proportions by Reasoning About
How Quantities Compare
I highly recommend the Lesson Study video clips under “Let’s think about multiplying and dividing decimal numbers, Grade 5 (Japan)” at
See especially video clip 3, which fits with Parts 3 and 4 here and which ends with a
segue into the next topic, which is “going through 1.”
# pails blue paint
1. # pails red paint
# pails purple paint produced
2.
# ounces of cheese
$
12
3
2 100 30 400 34
3 150 45 600 51
5 250 75 1000 85
8 20 28 18
1
2 5 7 4.50 0.25
3. The video clips referred to above are good to watch after the students have
solved this problem and the next one. I’ve had students do these problems and
watch the video for homework. Students expressed delight at seeing kids solving
the problem in more ways than they thought of!
6 meters is 3 times as much as 2 meters, so 6 meters weigh 3 · 24.8 = 74.4 grams.
Another way to reason is that 1 meter weighs 24.8÷2 = 12.4 grams, so 6 meters
weighs 6 times as much.
123
4. See the video clips referred to above. This part can be used as a segue to the
next topic of “going through 1” because the second method of finding the weight
of 1 meter first makes the problem easier. 15 meters will weigh 15 · 12.4 = 186
grams.
5. In 8 minutes, 1 21 miles; in 12 minutes 2 14 miles; in 2 minutes,
3
minute, 16
of a mile.
3
8
of a mile, in 1
Class Activity 7H: Going Through 1
An interesting connection to make would be to link “going through 1” to “how many
in one group” (fraction) division. We can think of going through 1 as solving such a
division problem. So we can view solving a proportion as consisting of solving a “how
many in one group” division problem first, followed by a multiplication problem.
# liters juice
3 38 3 43
1. (a) # liters bubbly water 5 58 6 41
# liters punch
8 1 10
If 3 liters of juice is needed for 8 liters of punch, then dividing by 8 (using
the ”How Many in Each Group?” interpretation of division) we see that
3 ÷ 8 = 38 liters of juice is needed for 1 liter of punch. Using the same
reasoning, 5 ÷ 8 = 85 liters of bubbly water is needed for 1 liter of punch.
For 10 liters of punch we need 10 groups of 38 liters of juice and 10 groups
of 58 liters of bubbly water.
(b) The strategy was to first use division to find the number of liters of juice
and bubbly water needed for 1 liter of punch, and to then use multiplication
to find the number of liters of each ingredient needed for 10 liters of punch.
2. (a) For 34 cups of red paint use 23 cups of yellow paint to make 17
cups of orange
12
paint.
Multiplying by 12: use 9 cups red paint and 8 cups yellow paint to make
17 cups orange paint.
8
9
cups red paint and 17
cups yellow paint to make 1
Dividing by 17: use 17
cup orange paint.
1
cups red paint and 7 17
to make 15 cups orange
Multiplying by 15: use 7 16
17
paint.
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A different approach would be not to “clear denominators” first but instead
1
of a cup of orange paint
to find how much red and yellow are needed for 12
and then get to 1 before getting to 15.
(b) First, clear denominators, then “go through 1”.
Class Activity 7I: More Ratio Problem Solving
1. (a) First we clear denominators and see that Chandra can mix 2 cups of ice
cream with 3 of milk to get 5 cups of milkshake. Then 1 cup of ice cream
with 32 cups of milk yield 2 21 cups of milkshake. Finally 3 cups of ice cream
with 92 = 4 12 cups of milk yield 7 21 cups of milkshake.
(b) She would use 1 51 cups of ice cream with 1 45 cups milk to make 3 cups of
the milkshake.
7.3
cups of water so as to use 2 31 cups of water in all.
7
12
2. He should add
Unit Rates and the Values of a Ratio
You may wish to review the definition of fraction at the beginning of this section.
Class Activity 7J: Unit Rates and Multiplicative Comparisons
Associated with a Ratio
1. There are 3 cups red paint for every 5 cups yellow paint in the mixture.
First column: 8 cups orange paint.
3
8
Second column:
3
5
Third column:
Fourth column:
cups red,
cups red,
5
3
8
5
5
8
cups yellow.
cups orange.
cups yellow,
8
3
cups orange.
2. The paint is 3 parts red and 5 parts yellow (where all parts are the same size
but could be any size). Draw a strip diagram showing 3 rectangles for the red
paint and 5 rectangles of the same size for the yellow paint.
3. (a) There are 83 cups of red paint for every 1 cup of orange paint.
3
of the mixture is red paint. Or: The mixture contains 83 times as much
8
red paint as orange paint.
125
(b) There are 85 cups of yellow paint for every 1 cup of orange paint.
5
of the mixture is yellow paint. Or: The mixture contains 58 times as
8
much yellow paint as orange paint.
(c) There are 35 cups of red paint for every 1 cup of yellow paint.
The mixture contains 35 times as much red paint as yellow paint. Or: The
amount of yellow paint times 35 gives the amount of red paint.
(d) There are 35 cups of yellow paint for every 1 cup of red paint.
The mixture contains 53 times as much yellow paint as red paint. Or: The
amount of red paint times 53 gives the amount of yellow paint.
Class Activity 7K: Solving Proportions by Cross-Multiplying
Fractions
21
x
and 62 are two ways to express the same unit rate, namely the number
1. The 10
of cups of flour needed per person. Therefore these two fractions must be equal.
2. Cross-multiplying is a shortcut for giving the two fractions a common denominator—
we simply ignore the common denominator and work only with the resulting
numerators.
tells us how many times as many people 10 is as 6. Similarly,
3. The fraction 10
6
the fraction 2x1 tells us how many times as many cups of flour x is as 2 21 . These
2
two fractions tell us how the like quantities compare. (The unit rate told us
how the unlike quantities compare.) In general, for N times as many people,
we should use N times as much flour, so the two fractions must be equal.
7.4
Proportional Relationships Versus Inversely Proportional Relationships
Class Activity 7L: Can You Use a Proportion or Not?
1. No, the area of the patio is 4 times as large, not 2 times as large, so that
proportion doesn’t work. So Ken will need 4 · 3 = 12 loads of stone pavers.
2. No, more lines will not require more time but rather less time to fill a truck.
Using twice as many assembly lines it should take half as long, so 5 hours.
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3. Watch out for the reasoning shown in part 4. To reason correctly, if 4 lines take
10 hours, then 1 line should take 4 times as long, namely 40 hours. So 6 lines
should take 16 as long, namely 6 32 hours, which is 6 hours, 40 minutes.
4. 1 line takes 40 hours, 2 lines take 20 hours, 8 lines take 5 hours, 16 lines take
2 12 hours, and 32 lines take 1 14 hours. 16 lines is about half way between 1 and
32 but 2 12 is not half way between 40 and 1 41 .
Class Activity 7M: A Proportional Relationship Versus an
Inversely Proportional Relationship
1. (A) Left: 2 times as many people frost 2 times as many cupcakes. 21 as many
people frost 21 as many cupcakes. N times as many people frost N times as
many cupcakes.
(A) Right: 2 times as many people take 12 as long. 21 as many people take 2
times as long. N times as many people take N1 times as long.
(B) Left: From left to right, the number of cupcakes is 25, 50, 75, 100, 125, 150,
175, 200.
(B) Right: From left to right, the number of minutes is 24, 12, 8, 6, 24÷5 = 4.8,
4, 24 ÷ 7 = 3 37 , 3.
(C) Left: # cupcakes ÷ # people = 25. This expresses the constant unit rate
of cupcakes per person.
(C) Right: # minutes · # people = 24. The expresses that the number of
person-hours is constant.
(D) Left: Proportional relationship, because every 2 people can frost 50 cupcakes.
(D) Right: Inversely proportional relationship because of part (A).
2. The graph for the relationship on the left is a straight line through the origin.
The graph on the right is curved, like the one in the text. (It is a hyperbola.)
127
Class Activity 7N: Who Says You Can’t Do Rocket Science?
1. You can reason that it will take between 15 and 20 minutes to fill the tank.
In 15 minutes, hose A fills the tank 21 full and hose B fills the tank 31 full, so
together they fill the tank 21 + 13 = 65 full, which is not quite full. In 20 minutes
hose A fills the tank 23 full and hose B fills the tank a little less than 21 full, so
together the hoses fill the tank more than full.
2. If students are stuck, give them the hint of figuring out how much of the tank
will be filled in some amount of time.
1
A “going through 1” way of reasoning: hose A fill the tank 30
full in 1 minute
1
and hose B fills the tank 45 full in one minute, so together they fill the tank
1
1
1
+
=
30 45
18
full in one minute. Therefore the two hoses together will take 18 minutes to fill
the tank.
7.5
Percent Revisited: Percent Increase and Decrease
Percent Of Versus Percent Increase or Decrease.
Class Activity 7O: How Should We Describe the Change?
Use this activity as a starting point for discussing percent change.
Class Activity 7P: Calculating Percent Increase and Decrease
You can use parts 1 and 2 of this activity to discuss the two methods for calculating
percent increase and decrease presented in the text.
1. (a) There was a 3 ounce increase in the new package of Brand A. Since the
15
3
= 100
= 15%.
old package was 20 ounces, the % increase is 20
23
(b) The new weight is 115% of the old weight because 20
= 115
. When we
100
subtract 100%, we get the 15% we got in part (a).
The next part asks students to show why this is not a coincidence.
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(c) Use a strip diagram/double number line and the distributive property, as
shown in the text. Students can also explain that the 115% of part (b)
is 100% + 15% and that the 100% represents the original weight, so the
remaining 15% represents the increased weight.
2. (a) The number of gallons decreased by 5, which is 25% of 20.
(b) Fifteen is 75% of 20 and 100% − 75% = 25%, which is the same answer
obtained in part (a).
(c) Use a strip diagram/double number line and the distributive property, as
shown in the text.
Class Activity 7Q: Calculating Amounts from a Percent Increase or Decrease
You can use parts 1 and 2 of this activity to discuss the two methods presented in
the text.
1. 20% → $80.00
120% → $400 + $80 = $480
Use a strip diagram/double number line like the one in the text. Also, we can
view 120% as 100% + 20%. The 100% represents the old price and the 20% is
the increase, so added, they give the new price.
You may also wish to discuss with students that they can use the definition to
find the new price: the new price equals the old price ($400) plus 20% of old
price (0.20 × $400 = $80), so $480.
2. 10% → $6.00
5% → $3.00
85% → $60 − $6 − $3 = $51
Use a strip diagram like the one in the text. Also, when we take 15% away from
100% we are left with 85%, so this 85% represents the new price.
You may also wish to discuss with students that they can use the definition to
find the new price: 15% of $60 is $9, so the new price is $60 − $9 = $51.00.
129
3. 20% → $30
100% → 5 · $30 = $150
The 120% was the result of adding 20% to 100%.
To reach $180 the price of the suit went up by 20% of a smaller amount. You
cannot get back to that smaller amount by subtracting 20% of a larger amount.
4. 20% → $100
100% → 5 · $20 = $500
The 80% was the result of subtracting 20% from 100%.
To decrease to $400 the price of the sofa decreased by 20% of a larger amount.
You cannot get back to that larger amount by adding 20% of a smaller amount.
Class Activity 7R: Can We Solve it This Way?
1. Matt’s method is not correct because he is finding 15% of $230, not 15% of
the original price. If P is the original price, then 115% · P = 2300, so P =
2300 ÷ 1.15 = 2000, and the original price was $2000.
2. The new amount is 25% more than the old amount (but to go back to the
original amount would require a 20% decrease).
Use this problem to make the point that we must pay close attention to the
“whole” or reference amount when working with percentages and fractions.
3. Forty-five is 300% of 15, but from 15 to 45 is a 200% increase, so Kaia is correct.
It’s common for people to confuse “increase” and “of” when a percent increase
is over 100%.
Class Activity 7S: Percent Problem Solving
1. When Prarie’s money went from 10% more to 15% less it decresed by 25% of
the cost of the game. Since this decrease wa $7.50 we know that 25% of the
cost equals $7.50 which implies that the cost equal 4 × $7.50 = $30.00.
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2. Together, the two mice weigh 220% of the lighter mouse. Since there are eleven
20% pieces in 220%, 20% of the wreight of the lighter mouse equal 66 ÷ 11 = 6
grams. Hence the lighter mouse weighs 5 × 6 = 30 grams. The heavier mouse
weighs 6 × 6 = 36 grams.
3. After 10% of the first vat is poured into the second, the first vat has 90% of
what it had before which is three times what the second vat has now. Hence
30% of the original content of the first vat must equal the contents of the second
vat. Since 10% of the original contents is now in the second vat, the original
contents of the second vat must equal 20% of the original contents of the first
vat. Since the second vat went from 20% to 30% the contents of the second vat
increased by 50% (because 30 is 50% more than 20).
Class Activity 7T: Percent Change and the Commutative Property of Multiplication
In general: let P be the original price of the pair of pants.
First way: the final price is 0.8 · 1.1 · P .
Second way: the final price is 1.1 · 0.8 · P .
According to the commutative property of multiplication, 0.8 · 1.1 = 1.1 · 0.8.
Therefore the final price is the same either way.
Copyright Chapter 8
Number Theory
131
132
8.1
CHAPTER 8
Factors and Multiples
Students tend to confuse the terms “factors” and “multiples” and some of the related
terms, so be sure to emphasize the distinct meanings of the terms. Remind students
that they should take care to use terms correctly when they teach.
Problems about Factors and Multiples.
Class Activity 8A: Factors and Rectangles
1. From the abstract perspective, there are 4 such rectangles: 1 by 24, 2 by 12, 3 by
8, and 4 by 6. But if you think of the rectangles as having an orientation (e.g.,
on the page), then there are also the 90 degree rotations of the four, namely 24
by 1, 12 by 2, 8 by 3, and 4 by 6. In any case, the side lengths produce all the
factors of 24.
2. No, if she has 25 tiles she’ll only be able to make 2 different rectangles: a 25 by
1 one and a 5 by 5 one. Therefore larger numbers don’t necessarily have more
factors.
Class Activity 8B: Finding All Factors
1. Tyrese can stop dividing at this point because, as we can see, if we divide 156
by a number greater than 13 the resulting quotient will be less than 12 and so
if this quotient is a whole number it would already have been found previously.
The factors of 156 are therefore 1, 2, 3, 4, 6, 12, and 13.
2. We can first find the following pairs by dividing by 2, 3, etc.:
2, 99; 3, 66; 6, 33; 9, 22; 11, 18
Once the quotient gets smaller than the divisor, we can stop. So the factors are
2, 3, 6, 9, 11, 18.
Class Activity 8C: Do Factors Always Come in Pairs?
No, factors do not always come in distinct pairs. If a number is a perfect square then
it will have an odd number of factors. For example, 25 has the factors 1, 5, and 25.
The classic “locker problem,” which is Problem 9 in the text, is a good follow-up
to this activity.
8.2
133
Even and Odd
I like to start this section by asking students to think for a moment about characterizations of even (and odd). Then I collect their responses on the board. Next I group
them to point out that there are different ways to characterize the same thing, as
described in the text. As is usual in mathematics, we pick one description as a definition. But then we need to explain why other ways characterize the same numbers,
which leads to the Class Activity.
Class Activity 8D: Why Can We Check the Ones Digit to
Determine if a Number is Even or Odd?
For part 1: Starting at 2, if we count by twos the ones digits repeatedly cycle through
2, 4, 6, 8, 0. So thinking of the even numbers as the numbers we get by counting
by twos from 2 (which we can think of as gathering pairs of toothpicks), we see that
they must end in 2, 4, 6, 8, or 0. A similar discussion can be made for counting by
twos starting from 1.
See text for parts 2 and 3. This provides a “base-ten bundles” way of explaining
why even numbers have those specific ones digits.
Class Activity 8E: Questions About Even and Odd Numbers
At some point, perhaps after students have worked on the problems but before you
discuss them as a class, it would be good to discuss the following with your students:
What if a student in your class says that he checked a lot of examples and
odd plus odd was always even, so it must always be true. Is that good
enough? Is it even possible to know for sure that odd plus odd is always
even? Because after all, you can’t check every single such sum.
Then have students think through whether their arguments in problems 1 and 2 really
are general. Even if their arguments use specific numbers and specific drawings, are
they done in such a way that we can see why the result will still be true in general?
1. The sum of two odd numbers is always even. For one explanation, consider
combining an odd number of small objects with another odd number of small
objects. Both collections of objects can be put in groups of 2 with 1 left. If we
combine the leftovers from the two collections, we will have put the combined
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number of objects into groups of 2 with none left, proving that the combined
amount is even.
For another explanation, consider 2 odd numbers. Then they can be put in the
form 2N + 1 and 2M + 1 where N and M are whole numbers. Note that N
and M might be different numbers – we must be careful not to use the same
letter twice here. The sum of the numbers is 2N + 1 + 2M + 1 = 2(N + M) + 1,
showing that the sum is odd.
2. The product of an even number with an odd number is even. For one explanation, consider an even number of groups with the same odd number of small
objects in each group. Then in each group, the objects can be put into pairs
with 1 left over. Since there are an even number of groups, we can pair all
the left over objects so that no leftovers remain unpaired. Therefore the total
number of objects can be put into groups of 2 with none left over and hence is
even.
For another explanation, consider an even number and an odd number. Then
we can put the even number in the form 2N for some whole number N and we
can put the odd number in the form 2M + 1 for some whole number M (note
that we should use a different letter than before because the numbers need not
be consecutive). Then the product is 2N(2M + 1) = 2[N(M + 1)], showing that
the product is even.
Class Activity 8F: Extending the Definitions of Even and Odd
1. 0 is even because 0 = 2 · 0. −5 is odd because −5 = 2(−3) + 1. Or you can
think in terms of the alternating pattern between even and odd.
I think it’s a good idea to let students think about what the answers should be
before jumping into the definitions in part 2. You can then come back to this
part and verify these using the definitions.
2. An integer is even if it is 2 times some integer. An integer is odd if it is not
even, or if it can be written in the form 2N + 1 for some integer N.
You might want to return to part 1 at this point.
3. No because every number would have to be both even and odd which would not
make the concepts useful.
8.3
135
Divisibility Tests
Class Activity 8G: The Divisibility Test for 3
My students usually ask me if they should explain why the test works to kids. I
tell them that they will have to use their professional judgment in deciding this.
If the children are curious and seem ready for it, they might find the explanation
interesting and worthwhile. If I were teaching this test to kids, I would try to present
it as something a bit mysterious, surprising, and neat. Maybe then the kids would
want to know what makes it work. But I certainly wouldn’t start by discussing what
makes it work.
A tricky point in explaining this and other divisibility tests is that we want to
explain why the test is a valid way to check if a number is divisible by 3. Students
sometimes get confused about this and think that we are just checking to see if a
particular number is divisible by 3 rather than proving a general statement. I think
it’s worthwhile to repeatedly emphasize the distinction.
1. No.
3. & 4. See text. Part 4 gives students the opportunity to connect physical and pictorial
representations to an algebraic representation. I think such connections are
worth emphasizing.
8.4
Prime Numbers
Class Activity 8H: The Sieve of Eratosthenes
1. The prime numbers between 2 and 120 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,
37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113.
2. See text.
Class Activity 8I: The Trial Division Method for Determining
whether a Number Is Prime
You will need the list of prime numbers from Class Activity 8H on the Sieve of
Eratosthenes for this activity.
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I like to turn students loose on this activity without first discussing when they
should stop dividing. After they’ve worked on the first one for a few minutes I ask if
they have any questions, at which point someone usually brings up this issue, which
is the topic of part 2 of this activity.
1. 239 is prime; 323 is not prime, it is 17 · 19; 4001 is prime.
2. See text. As an extension, you could ask students if they can characterize
the “break even point,” when they know they can stop with the trial division
method. This point is where the divisor and the quotient are equal, or in other
words, it’s at the square root of the number.
3. See text.
Class Activity 8J: Factoring into Products of Primes
1. 240 = 2 × 2 × 2 × 2 × 3 × 5
2. All factor trees must produce the same collection of prime numbers in the end
because of the Fundamental Theorem of Arithmetic, which is not obvious (and
unfortunately too advanced to prove in this book). I like to put this question
in “kid terms” because it seems to me that there should be some kids who are
surprised that you can make different factor trees but the primes at the end
come out the same, although perhaps in a different order.
3. Lindsay could simply say that 6372 = 7 × 7 × 13 × 7 × 7 × 13 = 74 × 132 .
Although she could re-order the 7s and 13s, the factorization can’t come out in
any different way because of the Fundamental Theorem of Arithmetic.
4. Since 77 = 7 × 11, and since neither 7 nor 11 are in the prime factorization of
527, we can say right away, without dividing, that 77 cannot be a factor of 527.
8.5
Greatest Common Factor and Least Common
Multiple
Go online to My Math Lab for an additional activities for this section:
Finding Commonality,
More Problems Involving GCFs and LCMs,
137
Relationships between the GCF and the LCM and Explaining the Flower Designs,
Using GCFs and LCMs With Fractions.
I recommend introducing the concept of LCM with the strips in Figure A.3 on
page 280 in the appendix of this manual. Copy the strips onto transparencies and
cut the transparencies into individual strips. Have students use markers to color the
multiples of 4 on one strip, the multiples of 6 on another strip, the multiples of 10
on a third strip, and perhaps even multiples of 8 on a fourth strip. Make sure the
colors are not too dark so that overlaps will be visible. Students can take two strips
and place one strip over another to see overlapping colors. You could have different
students shade different multiples and then ask the students to pair up in different
ways to see how the different strips overlap. The first overlap is the LCM of the two
numbers. It turns out that all the overlaps are exactly the multiples of the LCM.
Interestingly, the GCF is also visible: it turns out to be the smallest gap that occurs
between colors (but it’s not obvious why this is the case—courses on number theory
prove that this is true).
The Euclidean algorithm is another excellent method for determining LCMs and
GCFs. Most books on elementary number theory discuss this method and its implications.
Class Activity 8K: The “Slide Method”
1. On each row the lefthand entry is a common factor, if there is one, of the center
and righthand entry. Once you have found a common factor, the center and
righthand entries of the next row are found by dividing the center and righthand
entries of the present row by this common factor. Once you have found center
and righthand entries with no common factor, the GCF = the product of the
entries in the first column and the LCM = the product of the GCF with the
entries in the last row. Another slide for 900 and 360 would be:
5 900, 360
4 180, 72
3 45, 18
3
15, 6
5, 2
2. For 1080 and 1200 one slide would be:
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10 1080, 1200
4
108, 120
3
27, 30
9, 10
Hence GCF(1080,1200)= 10×4×3 = 120 and LCM(1080,1200)= 120×9×10 =
10800.
For 675 and 1125 one slide would be:
5 675, 1125
5 135, 225
9
27, 45
3, 5
Hence GCF(675,1125)= 5×5×9 = 225 and LCM(675,1125)= 225×3×5 = 3375.
3. I don’t think it’s worth aiming for a detailed, fully precise answer here. The
goal here is just to have a good feel for what makes this method work. See the
text for a “good enough for now” explanation. You can get more precise by
thinking in terms of products of primes, as in the next Class Activity.
Class Activity 8L: GCF and LCM with Products of Powers of
Primes
1. GCF = 25 · 34 · 73 , LCM = 26 · 310 · 75 .
The GCF is the product of the greatest powers of primes that the numbers have
in common. This means selecting the smaller power from each number. For
example, both numbers are divisible by 25 . But the first number is not divisible
by 26 , so the highest power of 2 that divides both numbers is 25 . Similarly for
the other prime factors.
The LCM is the product of the least powers of primes that both of the numbers
divide. This means selecting the greater power from each number. For example,
the LCM must be a multiple of both 25 and 26 , which means the LCM must
have 26 in its prime factorization, otherwise the first number would not divide
it. Similarly for the other prime factors.
2. Knowshon is right (see explanations for part 1). In each case, those prime
powers should be multiplied to give the GCF, LCM.
139
The reason we don’t take the greatest power for the GCF is so that the GCF
will divide both numbers. If we took the greatest power, it might not.
Knowshon’s method works for Matt’s numbers too because the first number can
be written as 27 ·38 ·53 ·70 and the second number can be written as 29 ·34 ·50 ·71 .
Class Activity 8M: Problems Involving GCFs and LCMs
1. The LCM of 18 and 24 is 72, which is 4 · 18 and 3 · 24, so you should buy 4
packages of pencils and 3 packages of erasers.
2. Half the class is clapping on multiples of 3, the other half is clapping on multiples
of 2. The class will clap together for the first time on the beat that falls on the
least common multiple of 3 and 2, which is 6. (Thereafter, the common snaps
occur at all multiples of the least common multiple.)
3. See Figure 8.1. Since the greatest common factor of 12 and 8 is 4, the largest
squares that Mary can use are 4 by 4.
Figure 8.1: Subdivided Rectangles
4. For the stars to be aligned again, each gear must have made a whole number
of revolutions. Therefore the number of teeth that must have meshed must
be a multiple of both 15 and 36. Therefore the first time the stars come back
together again is after the LCM of 15 and 36 teeth have meshed. The LCM of
15 and 36 is 180, which is 12 · 15 and 5 · 36. So the small gear will make 12
revolutions and the big gear will make 5 revolutions for the stars to be aligned
again.
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Class Activity 8N: Spirograph Flower Designs
The coolest math toy of all time is a spirograph (a drawing toy, which is available
at toy stores). If you can get your hands on several of these to show your class,
they will enjoy learning that this neat drawing toy is very mathematical and can be
explored in the same way as the flower designs of this Class Activity. Spirographs
have numbers on the holes and wheels that tell you how many teeth they have (mine
do, anyway). Figure A.4 on page 281 of this manual is a scan of some designs I drew
with a spirograph. You could copy this page to show the class. Under each design, I
wrote the number of teeth in the hole and in the gear I used to create the design, as
well as the number of “petals” in the design. It’s the same relationship that we get
with the flower designs. I ask students to predict how many “petals” they will make
with various holes and gears and then let them try it out.
1. The number of petals is the number you must multiply the “jump number” by
to get the LCM of the number of dots and the “jump number.” You can also
describe the number of petals as the number of dots divided by the GCF of
the number of dots and the “jump number.” You can use the additional online
Class Activity for thinking about why these are true.
2. Design 5: 15 petals
Design 6: 5 petals.
8.6
Rational and Irrational Numbers
Class Activity 8O: Decimal Representations of Fractions
The argument students develop here is fairly involved and intricate, so it’s a good
idea to summarize and discuss it several times. To help students think about the big
picture, at the end of the activity, ask students what some natural next questions
might be.
1. The decimal representations on the left are repeating, on the right they are
terminating. We can also say that terminating decimal representations are
repeating because they have a repeating 0.
2.
4
7
= 0.571428571428 . . . and the remainders are 4, 5, 1, 3, 2, 6, and the repeating
4, 5, 1, 3, 2, 6.
3
8
141
= 0.375 and the remainders are 3, 6, 4, 0, and 0 thereafter.
When you get a remainder you got before, the decimal representation repeats.
When you get a remainder of 0, the decimal representation terminates, i.e.,
repeats with zeros.
3. (a) The only possible remainders are the whole numbers from 0 to 30 because
the remainder must be less than the divisor.
(b) If you got a remainder of 0 (which doesn’t actually happen in this case),
the decimal representation would terminate.
(c) If you got a remainder you’d gotten before, the decimal representation
would repeat from there on.
(d) After at most 30 places, if you haven’t gotten a 0, you will have to have
gotten the same non-zero remainder twice, in which case the decimal representation must have begun to repeat.
4. See text.
5. No, this could not be the decimal representation of a fraction because there is
not a fixed block of numbers that repeats forever in the decimal.
Class Activity 8P: Writing Terminating and Repeating Decimals as Fractions
1. 0.137 =
0.25567
13.89 =
329.2 =
2.
137
1000
25567
= 100,000
1389
100
3292
10
1
= 0.1111111 . . .
9
1
= 0.0101010101 . . .
99
1
= 0.001001001001 . . .
999
1
= 0.000100010001 . . .
9,999
1
99,999
= 0.000010000100001 . . .
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3. 0.2 = 0.222222 . . . =
2
9
0.08 = 0.080808 . . . =
8
99
0.52 = 0.525252 . . . =
52
99
0.1234 =
49
9900
0.00049 =
49
99000
0.000049 =
73
10
0.1249 =
1.249 =
123456
999999
49
990
0.0049 =
5. 7.349 =
3
999
1234
9999
0.123456 =
4. 0.049 =
0.003 = 0.003003 . . . =
49
990000
+
12
100
49
990
+
=
49
9900
99·73+49
990
=
=
7276
990
1237
9900
1237
990
0.11149 =
111
1000
+
49
99000
=
11038
99000
Class Activity 8Q: What is 0.9999. . . ?
1.
2
9
= 0.2
3
9
= 0.3
4
9
= 0.4
5
9
= 0.5
6
9
= 0.6
7
9
= 0.7
8
9
= 0.8
9
9
= 0.9
Therefore 0.9 = 1.
2.
143
0.9999999999. . .
+0.1111111111. . .
1.1111111111. . . Therefore 0.9 = 1.
−.1111111111 . . .
1.0000000000. . .
3. See text. Again we conclude that 0.9 = 1.
4. 17 = 16.9
23.42 = 23.419
139.8 = 139.79
Class Activity 8R: The Square Root of 2
At √
the end of this activity, you might ask students if the proof of √
the irrationality
4 – 8 could be adapted to show that 3 is irrational.
of 2 developed in problems
√
Then ask: what about 6? Even though 6 has an even number of prime factors, we
can just focus on how many times the prime factor 2 (or 3) occurs in 6 · B 2 and in
A2 .
Note that part 1 assumes students know the Pythagorean theorem, which has not
yet been discussed in the text, but which should be familiar to students.
√
1. According to the Pythagorean theorem, the diagonal has length 2.
√
2. 2 = 1.414213562 . . . It looks like the decimal representation is not repeating
(and in fact it isn’t) but we can never tell for sure by looking at a finite portion
of the decimal representation.
3.
1414213562
999999999
= 1.41213562141213562 which is repeating. This is a rational number.
√
The decimal representation begins the same as 2.
4. Square both sides, then multiply both sides by B 2 .
5. A2 must have an even number of prime factors.
6. 2 × B 2 must have an odd number of prime factors.
7. A number with an odd number of prime factors cannot be equal to a number
with an even number of prime factors because of the uniqueness of factorization
into products of prime numbers (the fundamental theorem of arithmetic).
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8.
CHAPTER 8
√
2 must be irrational because we arrive at a contradiction if we assume it is
rational.
Class Activity 8S: Pattern Tiles and the Irrationality of the
Square Root of 3
To lead into the activity, you might allow students some time to make designs with
pattern tiles and to challenge each other to make the same overall shape in a different
way using different pattern tiles.
Notice that part 4 of the activity requires knowledge of the Pythagorean theorem
and the area formula for triangles, which have not yet been discussed in the text, but
which most students will probably know how to use.
1. Relationships include: Y = 6G, R = 3G, B = 2G.
2. O = 2T which we can see by determining that O +G = 2T +G. If students have
a hard time relating T and O ask them to use the green triangle in addition to
these shapes to find such a relationship.
3. The next several parts indicate why the orange square and thin tan rhombus
can’t be related to the other shapes.
4. The height of the triangle is
5. G =
B=
R=
Y =
√
3
2
inches and so its area is
√
3
4
square inches.
√
3
square inches.
4
√
3
square inches.
2
√
3 3
square inches.
4
√
3 3
square inch.
2
6. O = 1 square inch.
T =
1
2
square inches.
7. You may want to explain this rather informally with your students. To give a
more thorough proof that you cannot relate G, Y, R and B with O and T by
putting together tiles, you can argue as follows. If there were such a relationship,
there would be whole numbers a, b, c, d, e, and f such that
aG + bY + cR + dB = eO + f T
145
gathering terms appropriately we would then have an equation of the form
√
3
= another whole number
(whole number) ·
4
and so we would have
√
3 = rational number
√
But we saw in the text that 3 is irrational. Therefore there can be no such
relationship.
8.7
Looking Back at the Number Systems
There are no Class Activities for this section. You may wish to combine a lecture
with a class discussion of this material.
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Copyright Chapter 9
Algebra
147
148
CHAPTER 9
Go online to My Math Lab for another section for this chapter:
Series.
9.1
Numerical Expressions
Class Activity 9A: Writing Expressions for Dot, Star, and
Stick Designs
1. Design A: 1 + 6 · 5 + 4 · 3
Design B: 4 + 6 · 3 + 4 · 6
Design C: 2 + 6 · 4 + 4 · 5
All the expressions can be written in the form number + 6· number + 4· number.
2. Several different expressions are possible depending on how one decomposes the
design into parts. For example, we have the expressions 6 · 6 + 5 · 5, 1 + 3 + 5 +
7 + 9 + 11 + 9 + 7 + 5 + 3 + 1 and 2 · (1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1) + 11.
3. The expression 10·8−2·32 comes from viewing the design as a 10 by 8 array with
two 3 by 3 square arrays missing from the corners. The expression 2·(7·5)−4·2
comes from viewing the design as created by overlapping two 7 by 5 rectangles.
One copy of the “overlapped” 4 by 2 rectangle must be removed so that it isn’t
duplicated.
4. Be sure students find the expression 102 − 4 · 2, which comes from viewing the
design as a 10 by 10 square with a 4 by 2 rectangle removed.
5. Several different expressions are possible depending on how one decomposes the
design into parts.
6. Many expressions from part 5 should generalize to the case of an 8-stick-wide
design.
149
Class Activity 9B: How Many High-Fives?
1. On the one hand there are
19 + 18 + 17 + . . . + 2 + 1
hi-fives, which you can see by thinking of the 1st person as hi-fiving with 19
people, then standing off to the side, the 2nd person hi-fiving with 18 people,
then standing off to the side, the 3rd person hi-fiving with 17 people, then
standing off to the side, and so on.
Another way to express the number of hi-fives is as
(20 × 19) ÷ 2
because 20 people each hi-five with 19 others, but this way you have counted
every hi-five twice, so you have to divide by 2 to get the correct count.
2. The expression
is easier to evaluate; it is 190.
(20 × 19) ÷ 2
3. Two expressions are:
49 + 48 + 47 + . . . + 2 + 1 and
(50 × 49) ÷ 2
The second one is easier to evaluate; its value is 1225.
Class Activity 9C: Sums of Counting Numbers
Note that this situation is basically the same as the high-five problem of the previous
activity, except that the sums you deal with are slightly different in the two activities.
1. The number of darkly colored small squares can be expressed as the sum or as
1
(5 · 6) or (5 · 6) ÷ 2
2
because they are half of the small squares in the 5 by 6 rectangle.
2. A 200 by 201 unit rectangle could be shaded like the previous one to explain
why
1
1 + 2 + 3 + . . . + 198 + 199 + 200 and
(200 · 201)
2
both stand for the total number of darkly colored small squares. The expression
on the right is easy to evaluate as 20100.
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Class Activity 9D: Sums of Odd Numbers
1. 1 + 3 = 2 · 2
1+3+5= 3·3
1+3+5+7= 4·4
1+3+5+7+9= 5·5
1 + 3 + 5 + 7 + 9 + 11 = 6 · 6
1 + 3 + 5 + 7 + 9 + 11 + 13 = 7 · 7
2. The solutions are all perfect squares.
3. Based on Problem 1, the sum should be 100 · 100 = 10, 000.
4. 50 · 50 = 2500. Why 50? If you look at the pattern above, to get the number
to multiply by itself on the right hand side of the equation we add 1 to the last
number in the sum on the left side and divide by 2. (99 + 1) ÷ 2 = 50.
5. The coloring of the squares shows why the perfect squares can be expressed as
sums of odd numbers as in part 1.
Class Activity 9E: Expressions with Fractions
1. Rectangle 1:
Rectangle 2:
1
2
3
7
· 13 + 25 ·
· 25 + 13 ·
1
3
1
5
2. See Figure 9.1.
Class Activity 9F: Evaluating Expressions With Fractions Efficiently and Correctly
This class activity brings up some common errors in evaluating expressions with
fractions.
1. We use fraction multiplication at the first equals sign, the commutative property of multiplication at the second equals sign (in the denominator), fraction
35
= 1 is
multiplication is used again at the third equals sign, and the fact that 35
used at the last equal sign (or we could invoke what we know about equivalent
fractions instead of the work done at the last two equals signs).
151
Figure 9.1:
2
3
· 54 + 14 ·
1
10
2.
2·9
7
2·9·7
2·7·9
2·7 9
2·7
14
18 7
·
=
·
=
=
=
· =
=
5 99
5
11 · 9
5 · 11 · 9
5 · 11 · 9
5 · 11 9
5 · 11
55
3. (a) Incorrect
(b) Correct
(c) Correct
(d) Incorrect
9.2
Expressions with Variables
More Identities Arising from Rectangular Designs.
Class Activity 9G: Describing the Structure of an Expression
1. The expression is a sum of two terms. The first term is the number 20. The
second term is the coefficient 2 multiplied by the square, x − 3 times x − 3.
2. The expression is a difference of two terms. The first term has coefficient 5,
which multiplies the product of A with A + 9. The second term is the coefficient
4 times A2 .
3. The expression has only one term, which has coefficient 6. It is 6 times B − 4
times B + 7.
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Class Activity 9H: Defining Variables and Writing Expressions
Consider having students use a spreadsheet to create the tables shown here. A spreadsheet might also help students think about other examples.
Pounds ordered
1
2
1.
3
4
5
Expression for the total cost
10 + 7
2 · 10 + 7
3 · 10 + 7
4 · 10 + 7
5 · 10 + 7
Let C be the number of pounds of beads a customer purchases. Then the total
cost for the customer is 10C + 7 dollars.
Pounds sold
in a week
100
2.
200
300
400
Expression for the net weekly income
100 · 4 − 500
200 · 4 − 500
300 · 4 − 500
400 · 4 − 500
Let P be the number of pounds the company sold in a week. Then the net
weekly income that week is 4P − 500 dollars.
3. Both students’ work is correct, except that student 1 wrote an equation, not
just an expression. The first student’s method makes it easier to find the next
entry in the table. However, it won’t allow for finding an entry far down the
table without finding all the intervening entries. The second student’s method
makes it easier to formulate an expression which will allow one to calculate the
number of pounds left after any number of days.
4. Answers will vary. Encourage students to think in terms of making tables if
they have difficulty coming up with examples. Encourage them also to think
of expressions as calculation recipes. They should think about examples where
one would repeatedly do the same calculation but with different numbers.
153
Class Activity 9I: Equivalent Expressions
You might have students draw strip diagrams for parts 1 and 2.
1. (b), (c), and (d)
2. (c), (d), and (e)
3. Yes. P − 25 P = 1 · P − 25 · P = (1 − 25 ) · P = 53 P Highlight the hidden coefficient
of 1 in P .
4. No. We can see this by evaluating at P = 0, for example.
5. No. We can see this by evaluating at x = 0, for example.
7. No. We can see this by evaluating at x = 1, y = 1 for example.
8. Yes, by the distributive property.
Class Activity 9J: Expressions for Quantities
1. (a) After 41 of the sand in the pile is removed, 34 is left, so then there are 34 T
tons of sand in the pile. When another 23 of a ton of sand is dumped onto
the pile, there are then
2
3
T+
4
3
tons of sand in the pile.
(b)
3
(T
4
+ 23 ) tons. This expression is not equivalent to the one in part a.
(c) Part (a):
3
4
· 1 23 +
2
3
= 1 11
tons. Part (b): 43 (1 23 + 32 ) = 1 43 tons.
12
A
B
)(1 − 100
)P dollars.
100
A
B
Second scenario: (1 − 100 )(1 − 100
)P dollars.
Third scenario: (1 − A+B
)P dollars.
100
2. First scenario: (1 −
The final prices in the first and second scenarios are the same because of the
commutative property of multiplication, but these are not the same as the third
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scenario (which results in a lower price). Other equivalent expressions are also
possible. For example, students might write P − A% · P − B% · (P − A% · P ) for
the first scenario. But notice that P − A% · P − B% · P would not be correct
for the first scenario because the B% discount is on the new, lowered price of
P − A% · P , not just on P .
Class Activity 9K: Encoding and Decoding Secret Numbers
Let S be a secret number. Encoding can be described with this expression:
(5S + 3) · 6 − 11
which is equivalent to
30S + 7
Decoding does this:
[(30S + 7) − 7] ÷ 10 ÷ 3
which is equivalent to S.
As an extension, you could have students come up with their own encoding and
decoding rules.
Class Activity 9L: Rocket Science Reasoning
1. (a) The expression is a product of 16 with 2 + t and 22 − t.
(b) For that product to be 0, at least one of the factors must be 0. To make
2 + t be 0, t would have to be −2, but that doesn’t make sense in this
context. To make 22 − t be 0, t must be 22. When t is 22, the expression
has value 0.
(c) Part b tells the rocket scientists that the rocket hits the ground after 22
seconds.
2. (a) The expression is a difference of two terms. The first term is the number
2304. The second term is 16 multiplied with t − 10 and t − 10.
(b) The expression 16(t − 10)2 can’t be negative because it’s 16 times a square
and squares are always greater than or equal to 0 because even a negative
number times a negative number is positive. Since 16(t−10)2 is subtracted
from 2304, the largest that the expression 2304−16(t−10)2 can be is 2304,
and this occurs when 16(t − 10)2 is 0, which occurs when t is 10.
155
(c) The rocket scientists will want to know how high the rocket goes, which is
2304 feet, 10 seconds after launch.
9.3
Equations for Different Purposes
Class Activity 9M: Different Ways We Use Equations
You can use this activity as a preview and overview of this section.
The first equation expresses the result of a calculation.
The second equation is an identity: it is true for all values of the variable. We
often apply identities in the process of carrying out calculations.
The third equation relates two quantities that vary together.
The fourth equation must be solved to solve the problem.
Class Activity 9N: Identities Arising from Rectangular Designs
1. The shading breaks the 6 by 7 rectangle into 4 by 4, 4 by 3, 2 by 4, and 2 by 3
rectangles.
2. The shading breaks the N + 2 by N + 3 rectangle into N by N, N by 3, 2 by
N, and 2 by 3 rectangles. Hence we have the equation (N + 2) · (N + 3) =
N · N + N · 3 + 2 · N + 2 · 3.
3. We could have used the distributive property to get the equation.
4. See Figure 9.2.
Class Activity 9O: Equations About Quantities That Vary Together
You might consider having students work with spreadsheets for parts 2 – 4 or for
similar problems.
1. Variables need to stand for numbers, so Aaron should let S stand for some
number of units of sand (e.g., buckets of sand, cubic yards of sand, etc.) and
C stand for the number of units of cement. The correct equation is S = 3C.
Drawing a strip diagram can help show this.
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N×N
1×N
N×1
1×1
Figure 9.2: (N + 1)2
# T-shirts ordered
(one design)
20
2.
25
30
35
40
Expression for the total cost
(in dollars)
75 + 20 · 12
75 + 25 · 12
75 + 30 · 12
75 + 35 · 12
75 + 40 · 12
Let T be the number of T-shirts ordered and let C be the total cost of the order
in dollars. Then C = 75 + 12T .
3. Both students’ work is correct. The first student’s method makes it easier to
find the next entry in the table. However, it won’t allow for finding an entry far
down the table without finding all the intervening entries. The second student’s
method makes it easier to formulate an expression for the total cost and to find
an equation relating the total cost to the number of T-shirts.
4. You may wish to encourage students to make a table first. Even if the prospective teachers know how to define variables and write an equation right away,
they should recognize that tables can be helpful for students who are struggling
with these tasks.
Let S be the number of the T-shirts the company still has in stock. Let D be
the number of days since today. Then S = 500 − 15D.
157
Class Activity 9P: Formulating Equations for Word Problems
Instead of discussing these types of equations in this section, you could delay them
until Section 9.5, when students formulate and solve equations using strip diagrams.
1. (a)
(b)
(c)
4
M + 200 = 800
5
4
(K − 200) = 800
5
3 1
( L + 2 21 ) = 4
4 3
2. (a) You had x dollars in your bank account Monday morning. You withdraw
1
of the money in the account on Monday afternoon and the next day
4
you deposit 30 dollars. After these two transactions you have $150 in your
account. How much money was in your account Monday morning?
(b) Use the same problem as above with the Monday afternoon withdrawal
changed to 14 of a dollar instead 14 of the money in the account.
(c) There are 30 students in Mr. Lai’s class and x students in the rest of
the school. After 41 of the school’s students leave on a field trip there are
150 students left in the school. How many students are in the rest of the
school?
(d) There are x students enrolled in your school with 60 of these students in
kindergarten. The 60 kindregarten students are too young to go on the field
trip but 23 of the rest of the students go. These students are accompanied
by 20 parents and teachers. What is the school’s total enrollment if 80
people go on the field trip?
9.4
Solving Equations
Class Activity 9Q: Solving Equations by Reasoning About Expressions
1. Since 382 = 380 + 2 we can think of taking the 2 and combining it with 49, so
x must be 51.
2. Because the expressions on each side of the = sign are a product with 19, the
other factors, x − 5 and 7 must be equal. When 5 is subtracted from x it must
equal 7, so x must equal 12.
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3. By the distributive property, we can see that A must equal 10 · 84 = 840.
5
4. Because the expressions on both sides of the equal sign have a term of 16
, the
2
other terms must be equal to each other. So 2x must be equal to 3 . Since 23
means 2 pieces, each 31 , this means x = 31 .
5. 7 · 48 = 14 · 24, because we can think of breaking 48 into 2 · 24 and putting the
2 with the 7. So Z = 24.
6. Because of the common factor of 17 on both sides of the equal sign, x − 12 must
equal 1. So x must be 13.
7. Since 8 is 4 times 2, this means x − 19 must equal 2, so x must be 21.
8. x + 2 must equal 87, so x must be 85.
Class Activity 9R: Two Ways of Reasoning to Solve “One
Step” Equations
From a more advanced perspective, we can think of the “undo the expression” method
as an “inverse function” method. In other words, viewing the expression x + 0.34 as
representing the function x → x + 0.34, we are applying the inverse function to find
the input that corresponds to the output of 1.8.
Looking ahead, the “do the same operation to both sides of the equation” method
is the one we use more generally because it can be extended to solve all linear equations
in one variable.
1. If it doesn’t come up in the discussion, ask students why the left side becomes
just x when 0.34 is subtracted in the “do the same operation to both sides”
method. Explaining this basically involves the perspective of “undoing the
expression.” That is, the expression is x with 0.34 added on, so subtracting
0.34 just leaves x. Bring this point up again in discussing parts (b) and (c) in
part 2.
2. (a) “Undo the expression:” . . . to go backwards, add 0.83 to 6.4 to get x. So
x = 7.23.
“Do the same to both sides:” . . . add 0.83 to both sides. The left side will
just be x . . . .
159
(b) “Undo the expression:” The expression says that when you multiply x by
1
you get 27 . To go backwards, divide 72 by 13 . So x = 76 .
3
“Do the same to both sides:” . . . divide both sides by 13 OR multiply both
sides by 3 because 3 · 13 = 1, therefore leaving just x on the left . . . .
Note that with this multiplication method, we are using the fact that 3
and 13 are multiplicative inverses.
4
(c) “Undo the expression:” . . . To go backwards, divide 11
by 53 . So x = 20
.
33
3
“Do the same to both sides:” . . . divide both sides by 5 OR multiply both
sides by 53 because 53 · 53 = 1, therefore leaving just x on the left.
Note that with this multiplication method, we are using the fact that 53
and 35 are multiplicative inverses.
Class Activity 9S: Solving Equations Algebraically and With
a Pan Balance
Remove 2 big squares from both sides of the pan balance, correspondingly, subtract
2x from both sides of the equation, obtaining
3x + 1 = 7
Remove 1 small square from both sides of the pan balance, correspondingly, subtract
1 from both sides of the equation, obtaining
3x = 6
Take 31 of both sides of the pan balance, i.e., divide by 3 and take one part. Correspondingly, divide both sides of the equation by 3, obtaining
x=2
So the solution is 2.
Class Activity 9T: What Are the Solutions of These Equations?
1. The solution is x = −2. Note that even though this equation has only positive
coefficients, it nevertheless has a negative solution (and using a pan balance
would be problematic for this equation).
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2. Students may arrive at the equation 7 = 5 or 2 = 0 or 0 = −2 when attempting
to solve this equation. Because these equations are false no matter what x is,
they don’t have a solution, so neither does the original equation. This is because
whenever we add or subtract the same amount from both sides of the equation
(or multiply or divide both sides by the same nonzero amount), the resulting
equation has the same solutions as the original equation.
3. The solution is x =
5
3
or 1 23 .
4. Students may arrive at the equation 5x − 8 = 5x − 8 or 0 = 0, both of which
are true for every value of x. Every number is a solution; the equation is an
identity.
5. Any equation that leads to a = b during the process of solving the equation has
no solutions if a and b are different numbers.
6. Any equation that leads to a = a during the process of solving the equations is
an identity and has infinitely many solutions.
9.5
Solving Algebra Word Problems with Strip Diagrams and with Algebra
Modifying Problems.
Class Activity 9U: Solving Word Problems with Strip Diagrams and with Equations
1. Since the 4 boxes have a total value of $35.00, each box is worth $35.00 ÷ 4 =
$8.75. Hence the t-shirt costs $8.75. If we let x = the cost of the t-shirt then
3x = the cost of the hat and we then solve the equation 4x = 35 which has the
solution x = 8.75.
2. Since the two longer boxes must contain 180 − 30 = 150 blankets, each longer
box must contain 150 ÷ 2 = 75 blankets. Hence the second group consists of 75
blankets. Algebraically we would let x = the number of blankets in the second
group and we would then solve the equation 2x + 30 = 180.
161
3. Since 4 boxes hold 36 (white) sheep, one box holds 36÷4 = 9 sheep. Since there
are 7 boxes, there are a total of 7 · 9 = 63 sheep. Algebraically, we would let
x = the total number of sheep and we would then solve the equation 74 · x = 36.
4. Since one of the boxes in the second strip holds $240, the second strip holds
$480 in total. Hence the three unshaded boxes in the first strip hold $480 which
implies that each box holds $480 ÷ 3 = $160, and this implies that the first
strip holds 4 × $160 = $640 which is the amount of money that Mrs. Jones had
at first. Algebraically we would let x = the amount of money that Mrs. Jones
had at first and we would then solve the equation 21 · 34 · x = $240.
5. Draw a single strip with two boxes. If both boxes are full the total weight is
1.1 kilograms. If one box is empty then the total weight is .7 kilograms which
implies that 1.1−0.7 = 0.4 kilograms was lost emptying that box. Hence if both
boxes are emptied 2 × 0.4 = 0.8 kilograms are lost and the remaining weight
equals 1.1 − 0.8 = 0.3 kilograms. Hence the empty chocolate box must weigh
0.3 kilograms. Algebraically we let x = the weight of the empty chocolate box
and y = the weight of the chocolates. We must then solve the pair of equations
x + y = 1.1 and x + 21 y = 0.7.
6. We immediately see that 3 of the before boxes hold 18 problems and hence
one box must hold 6 problems which iplies that Quint’s 4 before boxes hold
24 problems. Algebraically we must solve the pair of equations Q = 4A and
Q − 20 = A − 2.
7. Draw a single strip of 6 boxes to represent the money that Carmen started with.
Shade in one of these boxes to represent the 61 that Carmen spent on a CD.
The 5 unshaded boxes must then hold the $45 that remained. Each box must
then hold $45 ÷ 5 = $9. Hence since Carmen started with 6 boxes, Carmen
started with 6 × $9 = $54. Algebraically if we let x = the amount of money
that Carmen had at the start then we must solve the equation 65 · x = 45.
8. Draw two stips. The first strip holds all the girls at the party and consists
of a long rectangle and a short rectangle. The short rectangle represents the
extra girls at the party and holds 25. The second strip holds all the boys at
the party and consists of 1 long rectangle which is the exact same length as the
long rectangle in the first strip. Since the total capacity of the 2 strips is 105,
the two long strips hold 105 − 25 = 80 which iplies that each long strip holds
40. Hence there wer 65 girls at the party and 40 boys. Algebraically we must
solve the pair of equations G = B + 25 and G + B = 105.
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9. Draw a strip for the boys and divide it into 10 equal pieces to show 10% of
the boys in each piece. The strip for the girls should therefore be made of 11
of those small pieces, to show 10% more girls than boys. All together, the 21
small strips make 168, so each piece represents 168 ÷ 21 = 8 children. There are
therefore 80 boys and 88 girls. Algebraically, if x stands for the number of boys,
then 1.10x is the number of girls and x + 1.10x = 2.10x is the total number
of children. Since the total is 168, it follows that 2.10x = 168 and therefore
x = 168 ÷ 2.10 = 80. Notice that the arithmetic is almost the same as when
solving using the strip diagram, except that we used x for 10 pieces instead
of for 1 piece (which would be 10% of the boys). It would also be perfectly
legitimate to let x stand for 10% of the boys and solve algebraically that way.
10. Draw 2 strips. The first strip represents the muffins that the bakery started
with and consists of 5 boxes. Shade in the last 3 of these boxes to represent
the muffins that the bakery sold. The second strip consists of 3 shorter boxes
and has overall length equal to the length of the 2 unshaded boxes in the first
strip. Since the secopnd strip represents the muffins taken by the 3 employees,
the second strip holds 3 · 16 = 48 muffins. Hence the 2 longer unshaded boxes
in the first strip hold 48 muffins which implies that each box in the first strip
holds 24 muffins. It follows that the bakery had 5 · 24 = 120 muffins at first.
Algebraically if we let x = the number of muffins that the bakery had at first
then we must solve the equation 25 x ÷ 3 = 16.
Class Activity 9V: Solving Word Problems in Multiple Ways
and Modifying Problems
1. One approach would be to translate each statement about the fraction gone into
the corresponding statement about the fraction left. This leads to the equation
3 2 1
· · · x = 15
4 3 2
which has the solution
x = 60.
This equation could also be found by a process of “guess an amount and write
an expression” without defining a variable. For example, one could guess there
were 100 pencils. Then check if that works and use the checking process to
163
formulate the expression 34 ·
could also be formulated.
2
3
·
1
2
· 100. A more complicated looking expression
Another approach is to “work backwards,” from the information at the end.
And another way to solve the problem is with the aid of a strip diagram in
which a strip stands for the total number of pencils. Some students might
draw a picture that looks different from a strip diagram but allows for the same
reasoning.
2. The number of pencils left at the end could be changed to any multiple of 3.
This is true because the whole number at the end is 34 of another whole number
and this true if and only if 34 of the whole number at the end is a whole number
which will happen only if the whole number at the end is a multiple of 3. Note
also that this second whole number must be a multiple of 4 which implies that
3
of this whole number is also a whole number. Clearly 21 of this last number
2
will be a whole number.
3. Suppose that you change the fraction 14 to 51 . Then the number, N, of pencils
left at the end must be a multiple of 8. This is true because each of the products
3
2
2·
5
N
4
· 54 N
3 5
· N
2 4
must be a whole number and this will happen if and only if N is a multiple of
8.
Class Activity 9W: Solving Word Problems
1. Markus had $750 to start with.
2. Keisha had $1200 to start with.
3. There were 8 12 liters of liquid in the container to start with.
9.6
Sequences
More Arithmetic Sequences of Numbers Corresponding to Sequences of Figures,
More Explaining Formulas for Arithmetic Sequences,
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Geometric Sequences,
The Fibonacci Sequence in Nature and Art,
An Infinite Geometric Series,
Making Payments Into an Account.
Class Activity 9X: Repeating Patterns
Have students do part 3 before you discuss part 2 in detail.
1. The 100th shape will be the 3rd circle in the pattern because the pattern repeats
in sets of 6. There are 16 full sets of 6 in 100 and 4 remaining. The 4th shape
in the sequence is a circle.
2. The pattern repeats in sets of 6. Because 100 ÷ 6 = 16, remainder 4, i.e.,
100 = 16 × 6 + 4, there are 16 full repeats, plus another 4 shapes in the first
100 shapes. By the end of the 16th repeat, there are 16 × 3 = 48 circles. The
next 4 shapes contribute another 3 circles. So all together, that’s 48 + 3 = 51
circles among the first 100 shapes.
3. Amanda’s and Robert’s reasoning is not correct. They are not dealing with full
repeats of the pattern. In fact it turns out that different sets of 10 or 20 will
have different numbers of circles in them. Kayla’s reasoning also doesn’t apply
because of this variation. For example, even among the first 10 shapes, there
are 6 circles, not 5.
To reason correctly, one must attend to the unit of repeat and also to how much
of a partial unit is used at the end.
Class Activity 9Y: Solving Problems Using Repeating Patterns
1. In conjunction with this part of the activity, you might like to watch Lesson
Study video clips on the 5th grade lesson, “Do I Have a Window Seat or an
Aisle Seat?” available at
http://hrd.apecwiki.org/index.php/Do_I_Have_a_Window_Seat_or_an_
Aisle_Seat%3F_Grade_5_%28Japan%29
165
It takes a fair amount of time to watch all the videos, so consider assigning
these for students to watch outside of class. I’ve had students just watch the
last minute of the first video clip and then watch the full 4th and 5th clips.
If it doesn’t come up in the discussion, ask the students if there are any arithmetic sequences related to this problem. They should see that all the seats in
the same location relative to the windows and aisle form an arithmetic sequence.
A nice connection to make is that all the entries in these arithmetic sequences
have the same remainder when divided by 4.
The seats that are multiples of 4 are at the window on the right hand side of
the train. So since seat 43 is 3 seats after such a seat, it is an aisle seat on the
right. Since 137 is 1 after a multiple of 4, it is a window seat on the left. Since
294 is 2 after a multiple of 4, it is an aisle seat on the left.
2. 100 days from today will be 2 days after the day of the week it is today. Here’s
why: 7 days from today is the same day of the week, and so is every additional
7 days after that. Since 14 · 7 = 98, 98 days from today will be the same day
of the week as today. So 100 days from today will be 2 days of the week later
than today. Just like with patterns of shapes, there 7 days of the week repeat
in a pattern.
3. (a) If Antrice starts with Benton, goes counterclockwise, and points to people
22 times, then Carmina will be it. We can predict this because 5 divides
into 22 4 times with a remainder of 2. At every multiple of 5, Antrice will
be pointing to herself. So 2 after a multiple of 5, Antrice will be pointing
to Carmina.
(b) Ellie will be it. Now at every multiple of 6 Antrice will be pointing to
herself. Since 6 leaves a remainder of 4 when divided into 22, Antrice will
be pointing to the 4th person after herself at the end.
(c) If it’s just the original 5 children, then Ellie will be it (4 after a multiple
of 5). If it’s the 6 children then Antrice will be it (since 24 is a multiple of
6).
4. The ones digit is 6. The digits in the ones places of the numbers in the sequence
2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64,
27 = 128, 28 = 256, . . . , 2N , . . .
repeat in the pattern 2, 4, 8, 6. So if N is a multiple of 4, 2N ends in 6.
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Class Activity 9Z: Arithmetic Sequences of Numbers Corresponding to Sequences of Figures
1. 4, 7, 10, 13, 16, . . . , 3N + 1.
2. There are 3 “arms”. Each arm has N circles in it, so the arms contribute 3N
circles (we could also say that the arms of the Nth figure are made of N layers,
each layer having 3 circles). There is also one dark circle in the middle. Hence
there are 3N + 1 circles altogether.
3. 3 · 38 + 1 = 115 circles.
4. Yes, the 33rd. We can find this by solving 3N + 1 = 100 or by thinking that
after taking away the dark circle in the middle, there must be 99 circles to
distribute equally among the 3 arms.
5. No. There is no whole number such that 3N + 1 = 125. Or, taking away the
dark circle in the middle, we would have to distribute 124 circles equally among
the 3 arms. By 124 is not divisible by 3.
Class Activity 9AA: How Are Expressions for Arithmetic Sequences Related to the Way Sequences Start and Grow?
1. 4N + 1, 4N + 3, 5N + 2.
2. The coefficient of N is the amount by which the sequence increases going from
one entry to the next.
3. The first entry is the sum of the coefficient of the N term and the constant
term. In other words, if the formula is aN + b the first entry is a + b. Said
another way, the constant term is the first entry minus the increase amount.
In the next activity, the constant term, b, will be described as the 0th entry of
the sequence. In the section on linear functions, this constant term, b, will be
seen as the y-intercept of the graph, and the formula for arithmetic sequences
will be connected to the y = mx + b equation for lines.
4. 3N − 2, −4N + 11
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Class Activity 9BB: Explaining Expressions for Arithmetic
Sequences
Use the previous activity first to pique students’ interest in why arithmetic sequences
must have the kinds of formulas that they do.
Following this activity, you might like to move quickly to connecting the formula
you develop here with the y = mx + b equation for a line. To do so, you could follow
this activity by a quick introduction to functions (section 9.8) and you could then go
directly to the Class Activity on arithmetic sequences as functions.
1. (a) The 0th entry would be 2.
This 0th entry will be seen as the y-intercept of the graph, and the formula
for arithmetic sequences will be connected to the y = mx + b equation for
lines.
(b) To
To
To
To
To
To
find
find
find
find
find
find
the
the
the
the
the
the
1st entry: Start at 2 and add 3 1 time.
2nd entry: Start at 2 and add 3 2 times.
3rd entry: Start at 2 and add 3 3 times.
4th entry: Start at 2 and add 3 4 times.
5th entry: Start at 2 and add 3 5 times.
Nth entry: Start at 2 and add 3 N times.
(c) 1st entry = 2 + 1 · 3
2nd entry = 2 + 2 · 3
3rd entry = 2 + 3 · 3
4th entry = 2 + 4 · 3
5th entry = 2 + 5 · 3
Nth entry = 2 + N · 3
(d) This part of the activity previews ideas that will be discussed in the section
on linear functions.
The points lie on a line. Students should see that to get from one point
to the next, one goes over 1 and up 3, in the same way that one adds
3 in going from one entry in the sequence to the next. This 3 (or really
3/1) is the coefficient 3 in the 3N term of 3N + 2. The 2 comes from the
“additional” (0, 2) point.
For a real thrill, have students substitute x for N and set the expression
equal to y to see what they get: y = 3x + 2, which is an equation of a line
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with slope 3 and y-intercept 2! The students will be excited to see this
connection.
2. An expression for the Nth entry is 3N − 1. Use an argument like the one
outlined in part 1 a, b, c. Students might also like to make a graph and use
that as a support as well.
Another approach is for students to compare the given sequence to the sequence
3, 6, 9, 12, . . . which is also increasing by 3 and whose Nth term is 3N. The
given sequence is shifted down one, so is 3N − 1.
Class Activity 9CC: Comparing and Contrasting Sequences
1. 5, 7, 9, 11, 13, 15, 17, . . .
1, 4, 9, 16, 25, 36, 49, . . .
2, 4, 8, 16, 32, 64, 128, 256, 512, . . .
2. The first sequence grows by adding a fixed number, 2, each time. The second
sequence grows by increasing amounts: by 3, by 5, by 7, etc., increasing by
odd numbers. Interestingly, the increases form an arithmetic sequence. (In
a more advanced setting this is related to the fact that the derivative of a
quadratic function is a linear function.) The entries in the third sequence are
multiplied by 2 going from one to the next. Interestingly, the increases (obtained
by subtracting consecutive entries) form the same sequence again. (In a more
advanced setting, this is essentially the fact that the derivative of an exponential
function is a constant times the function.)
3. See Figure 9.3.
4. Students should notice the difference in growing by 2 each time versus growing
by larger and larger amounts. The increase of 2 in the first sequence is caused by
the 2 in 2N. The square shapes in the second sequence fits with “N squared.”
169
2N
N2
Figure 9.3: Sequences
Class Activity 9DD: What’s the Next Entry?
1. 2, 4, 8, 16, 32, 64, . . . Multiply previous entry by 2, or the nth entry = 2n .
2, 4, 8, 14, 22, 32, . . . The nth entry = n2 − n + 2.
2, 4, 8, 32, 256, 8192, . . . The 1st 2 entries are 2 and 4. Each subsequent entry
is the product of the previous 2 entries.
2. 2, 5, 11, 20, 32, 47, . . . The nth entry = 2 + 23 n(n − 1).
2, 5, 11, 20, 35, 59, . . . The 1st 2 entries are 2 and 5. Each subsequent entry is
the sum of the previous 2 entries plus 4.
2, 5, 11, 23, 47, 95, . . . The nth entry = 3 · 2n−1 − 1.
3. 2, 3, 6, 10, 17, 28, . . . The 1st 2 entries are 2 and 3. Each subsequent entry is
the sum of the previous 2 entries plus 1.
2, 3, 6, 11, 18, 27, . . . The nth entry = 3 − 2 · n + n2 .
2, 3,6, 13, 28, 59, . . . The nth entry = 2n − n + 1.
9.7
Functions
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Class Activity 9EE: Is It a Function?
1. (a) This is not a function because there isn’t a single output for each input.
Even if you just buy 1 item, the cost can be different depending on what
it is. The rule can be modified to be the cost of N specific items, such as
notebooks of a specific type.
(b) This is not a function because there are always two numbers associated
with an input. The rule could be modified to associate a number that is x
units to the right of 5 on the number line.
(c) This is not a function because a speed such as 60 miles per hour might
be associated with many different distances. The rule could be modified
to go the other way around: to a distance, D, associate Melissa’s speed
when she has driven D miles. Notice that if one said “when she is D miles
from Indianapolis” this might not be a function because there might be
places in the road that are the same distance from Indianapolis (if the
road curves, for example), but Melissa might be driving different speeds
at those places.
2. Example 1 is a function, even though it is a pretty boring one. Its rule is to
send any input to the output 1. Students sometimes think a constant function
such as this one is not a function because it doesn’t have a rule that depends
on the input or because it sends many numbers to the same number (i.e., is not
one-to-one, in more advanced language).
Example 2 is a function. A rule for it is ”round to the nearest whole number.”
Example 3 is a function. A rule for it is “round down to the nearest whole
number that is less than the given number.”
Example 4 is not a function because it sends an input such as 1 to more than
one output.
Class Activity 9FF: What Does the Shape of a Graph Tell Us
About a Function?
1. (a) – graph 2 and Table Z; (b) – graph 3 and Table X; (c) – graph 1 and Table
Y.
Have students focus on the increase amounts in the tables. In Table Y, there is
a constant increase of 3000 fish per year. In Table X, the increase is greater at
171
first, then less later on. Table Z has the reverse pattern of increase: smaller at
first, then greater later on.
Discuss that we can see the pattern of change at a glance by observing the shape
of the graph.
2. The second graph is the correct graph of water temperature versus time. The
temperature of the water drops quickly at first and then drops more and more
slowly. You can see that the temperature of the water gets closer and closer to
room temperature as time goes by.
The first graph shows the water temperature dropping at a steady rate.
With the third graph the temperature of the water drops faster the closer the
temperature is to room temperature which is totally unreasonable.
Class Activity 9GG: Graphs and Stories
1. In the story, the manatee should swim quickly at first, then slow down and stop
for a little while, then turn around and head back toward the dock quickly, slow
down again, then head back away from the dock again quickly, stop again and
head even more rapidly toward the dock than he swam away from it initially.
2. One problem with the graph in the book is that it is not the graph of a function.
For a correct graph see Figure 9.4.
3. The graph could be the graph of Jenny’s speed against time but not distance
against time. In that case, the y-axis should be labeled “speed,” not “distance.”
Another problem is that the units on the y-axis are not equally spaced because
1 mile is not 5000 feet. And the units are also not labeled as feet.
4. The graph of a distance function is in Figure 9.5. Note that as the graph
becomes steeper during the first 4 minutes Jenny’s speed must be increasing.
Also that she stops after 4 minutes and then resumes running at a much slower
speed. The graph in the manual could be relabeled to be the speed function.
However, the shape of the various parts could also be different than what is
shown.
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CHAPTER 9
50
40
distance
in
30
miles
20
10
1
2
time in hours
3
Figure 9.4: A Graph of a Distance Function
Distance (in feet)
6000
4000
2000
0
0
5
Time (in minutes)
10
Figure 9.5: Jenny’s Run
173
Class Activity 9HH: How Does Braking Distance Depend on
Speed?
1. & 2. The third graph is correct. The first cannot possibly be correct because it shows
the braking distance becoming less with greater speeds. The second graph shows
that the braking distance is the same no matter what the speed, so also can’t
be correct. It also shows a non-zero braking distance when going 0 miles per
hour, which doesn’t make sense.
1
1
(2 · 37)2 can also be expressed as 2 · 2 · 20
(37)2 , and
3. Students should see that 20
similarly for the other cases. Because of the 2 · 2 factor, the braking distance is
always multiplied by 4 when the speed is doubled.
Class Activity 9II: Reasoning About Equations for Functions
1. The profit is 0 when the expression in x on the right side of the equation is 0.
Since that expression is a product, it is 0 when at least one of the factors is 0.
This occurs when either x − 10 is 0 or when 80 − x is 0, i.e., when x = 10 or
when x = 80.
2. The expression in x is a difference of 2450 and 2(x − 45)2 . This difference will
always be less than or equal to 2450 because 2(x − 45)2 is a positive number
times a square, and squares are always greater than or equal to 0. So no matter
what the value of x is, you are always subtracting a positive or 0 amount from
2450. The greatest value therefore occurs when 2(x − 45)2 is 0, which occurs
when x − 45 = 0, i.e., when x = 45. So the company makes a maximum profit
when they sell gizmos for $45 each and that profit is $2.45 million.
3. In that case, x = 0 and so their “profit” will be −1600, meaning the company
will have a loss of $1.6 million.
4. A = (0, −1600), B = (10, 0), C = (45, 2450), D = (0, 80).
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9.8
CHAPTER 9
Linear Functions
Linear functions involve an important connection between geometry and algebra.
Class Activity 9JJ: Proportional Relationships as Functions
1. The pink paint function can be described by the table:
R
0
1
2
3
4
5
6
W
0
1 21
3
4 21
6
7 21
9
The graph is a line through the origin.
A strip diagram that shows 2 strips for red paint and 3 strips for white paint
represents the relationship. Using it we see that if we multiply R by 23 we get
W . Therefore an equation is
3
W = ·R
2
2. Whenever the input increases by 2 the output increases by 3, and whenever the
input increases by 1 the output increases by 23 . On the graph, whenever one
goes 2 units to the right, one goes up 3 units. Or whenever one goes 1 unit to
the right, one goes up 32 units. The 32 is a unit rate, which tells how much white
per 1 cup of red and it is a value of the ratio which tells us how many times as
much white there is as red.
You may wish to bring up the notion of slope and describe it in terms of the
unit rate: it is the increase in output per 1 unit of increase in the input.
3. We would switch the columns in the table and flip the graph across the diagonal.
We would use the equation
2
R= ·W
3
2
The 3 is the unit rate of cups red per 1 cup white and it is the value of the
ratio, which tells us how many times as much red there is as white.
175
Class Activity 9KK: Arithmetic Sequences as Functions
1. Viewed as a function, the arithmetc sequence can be described by the table:
y
x
input output
1
3
2
5
3
7
4
9
5
11
When we graph this function, it is a collection of isolated points that lie on a
line.
2. Whenever the input increases by 1 the output increases by 2. If we add a “0th
entry” it will have value 1. The y-value associated with x is x jumps of 2 from
1, so is 2 · x + 1. Therefore y = 2x + 1 is an equation for the function. The
coefficient 2 is the rate of change of y with respect to x; it tells us how much y
increases for every increase of 1 in x. The constant term 1 is the y-coordinate
where the line crosses the y-axis. We can think of it as the “0th entry” of the
sequence.
At this point, you might like to draw in the full line (by connecting the isolated
points that form the graph of the sequence) and discuss the reasoning presented
in the text for why a line with slope m and y-intercept b must have the equation
y = mx + b.
Class Activity 9LL: Different Equations for Linear Functions
1. Niles’s equation is valid if T is defined to be the number of minutes of phone
calls and if T is restricted to be greater than or equal to 5. Then T −5 stands for
the number of minutes of phone calls in excess of 5 minutes and so 0.06(T −5) is
the additional cost for those extra minutes. If T is less than 5 then the equation
doesn’t give the correct cost.
Anja’s equation is valid if T is defined to be the number of minutes of phone
calls in excess of 5 minutes.
2. Mariah might have been thinking of the equation
2B + 3T = 150
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DeShun’s idea could help Chris because it tells us that whenever B increases
by 3, T decreases by 2, therefore whenever B increases by 1, T decreases by 23 .
This unit rate must be the coefficient of B, so
2
T = 50 − B
3
Class Activity 9MM: Linear Versus Inversely Proportional Relationships
1. In both cases, as x increases, y decreases. Some students may automatically
think that such a relationship is not linear, but in fact, linear relationships with
negative slopes behave this way, as they will see in part 2.
2. Relationship 1 is an inversely proportional relationship with equation x · y = 24.
The graph is a collection of isolated points which lie on a hyperbola.
Relationship 2 is linear and has equation y = 100 − 5x. Its graph is a line with
slope −5.
Class Activity 9NN: Is it Linear or Not?
1. (a) Yes, it’s linear because the rate of change is a constant 20.
(b) The organizers will lose money for those cases. This makes sense because
they have to pay the band and rent space for the concert.
(c) Tickets are being sold for $20. You can tell because this is the rate of
change of dollars per ticket.
(d) Where the graph crosses the x-axis the y value is 0, so this is where the
organizers break even.
2. (a) No, because the rate of change is not constant.
(b) The building is 256 feet tall and the marble hits the ground after 4 seconds.
(c) The marble falls faster and faster because for each second that elapses, the
marble falls a greater distance.
(d) The changes from one y-value to the next form an arithmetic sequence:
−16, −48, −80, −112.
177
(e) The changes in the changes are always −32. You can use this to lead
into a discussion about how quadratic functions have “constant second
differences” and that falling objects give rise to quadratic functions because
the acceleration due to gravity is constant.
3. (a) No, because the rate of change is not constant.
(b) $100
(c) Every 5 years, the amount of money doubles. You could use this to lead
into a discussion about exponential functions.
Class Activity 9OO: What Kind of Relationship Is It?
Table A: The first differences are an arithmetic sequence, 5, 7, 9, 11 and the second
differences are constant, namely 2, so this pattern tells us the function is quadratic.
In fact it has equation y = x(x + 2) or y = x2 + 2x.
Table B: This might look like a quadratic function at first, but it is actually linear
because it has constant rate of change of 32 .
Table C: This might look like an exponential function but it is linear because it
has a constant rate of change of 3. In fact, it is y = 3x.
Table D: This is an exponential function because of the repeated doubling as x
goes up by 1.
Table E: This is a linear function because it has constant rate of change −5.
Table F: This is an inversely proportional relationship because whenever x is
multiplied by a number, y is divided by that number. It has equation xy = 60.
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Copyright Chapter 10
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10.1
Visualization
Elementary teachers often teach children about time zones and about the phases of
the moon. My students have told me that even though the activities on the earth and
moon were difficult, they were glad we did them. On the other hand, you can safely
omit this section if you do not have time for it or prefer to focus on other material.
Class Activity 10A: Visualizing Lines and Planes
1. Two lines in a plane divide the plane into 3 disjoint pieces if the lines are parallel
and 4 disjoint pieces if the lines are not parallel. If nobody comes up with it,
draw lines on the board which are not parallel but don’t meet on the board and
ask students how many pieces those lines will divide the plane into (4 because
the plane of the board extends beyond the board itself).
2. The possibilities are 4, 6, and 7 pieces. Three lines in a plane divide the plane
into 4 pieces if all 3 lines are parallel, they divide the plane into 6 pieces if
exactly 2 of the lines are parallel, they divide the plane into 6 pieces if the 3
lines meet at a single point, and they divide the plane into 7 pieces if none of
the lines are parallel and the lines do not meet at a point. As before, consider
drawing on the board a case where lines intersect off the board.
An interesting extension of this activity is to consider increasing numbers of
lines in the plane, but with the restriction that no lines are parallel and that
no 3 lines meet at a point. You could have students make a table showing how
many disjoint pieces the plane is divided into for increasing numbers of lines.
The students should observe that you get the nth entry in the table by adding
n to the previous entry. Students might be able to determine (or you could
prompt students to see) that when another line is added, the line will cross
each existing line once. If you think of traveling along the new line, every time
you cross an old line you will divide the region you enter into into 2 regions. So
adding an nth line will add n regions to the ones that are already there, thus
explaining the pattern in the table. If the students know that
1+2+·+n=
n(n + 1)
2
they might then be able to determine that for n lines there will be
1 + (1 + 2 + · + n) = 1 +
n(n + 1)
n2 + n + 2
=
2
2
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regions.
3. Two parallel planes in space divide space into 3 disjoint regions. Two nonparallel planes in space divide space into 4 disjoint regions.
4. The possibilities are 4, 6, and 8. If all three planes are parallel, then there are
4 regions. If exactly 2 of the 3 planes are parallel, then there are 6 regions. If
none of the planes are parallel and if the planes do not meet along a line, then
there are 8 regions. If none of the planes are parallel but the planes meet along a
line, then there are 6 regions. You know you’ve found all possible cases because
either all 3 planes are parallel, exactly 2 are parallel, or none are parallel.
Class Activity 10B: The Rotation of the Earth
I do parts 1 and of this activity with the whole class, using an inflatable globe and
pretending that the sun rays are coming from the side of the room. If you do this, be
sure to have the sun rays coming from the students’ right side (your left side if you
are facing them) so as to agree with the figure in the activity. To start, I show the
two ways that the earth could be rotating and ask: how do we know which one it is?
This is what we will try to figure out.
I ask students to show where on the globe it is about noon, where it is about
midnight, and where the sun is low on the horizon. We then sort out which is sunrise
and which is sunset by remembering that the sun rises in the east and sets in the west.
With this information, we can deduce that the earth must rotate counterclockwise,
when viewed from looking down on the North Pole. It’s fun to show a rotating globe
pointed so that students are looking down on the North Pole, and then continuing to
rotate the globe while turning around so that the students are now looking down on
the South Pole. They will see that from this perspective, the rotation is clockwise!
At the end of the activity, you could discuss why we have time zones and how the
times on the east and west coasts of the US differ.
Class Activity 10C: Explaining the Phases of the Moon
Note that this activity is for deepening prospective teachers’ own thinking and may
not be suitable for students below about eighth or seventh grade.
You may wish to show simulations, such as the one at
http://www.astro.wisc.edu/~dolan/java/MoonPhase.html
which is also linked to the textbook’s webpage.
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2. The left part of the moon is the part that is illuminated. The bottom part of
the moon is the part that faces the earth. So the brightly colored portion shows
the part of the moon that is illuminated and faces the earth, and hence is the
part that is visible from earth.
The moon will appear like half of a disk (circle) to people on earth.
3. Over the course of a day, as the earth rotates on its axis, the moon will rise and
set from the perspective of the earth. Because the moon takes a full month to
rotate around the earth its location relative to the earth will not change much
over the course of an earth day. People who can’t see the moon at a given time
of day because they are not on the part of the earth facing the moon at that
time will be able to see the moon at some other time in the day (even a new
moon is somewhat visible despite not being brightly illuminated).
Students might be interested in which way the moon orbits around the earth
and how we know. Viewed from above the north pole, the moon orbits counterclockwise. We can tell this by observing the moon at the same time every
night and seeing that it moves to the east.
4. In (A), the moon is not visible from earth; it is a new moon. The part of
the moon that is illuminated faces away from the earth. In (B) the moon
appears as half a disk (as in part (2), but this time the other side of the moon
is illuminated). In (C) the full disk of the moon is visible; it is a full moon.
Most of the time the earth does not block the sun’s rays (except when there
is a lunar eclipse) because the earth, moon, and sun are rarely in aligned in a
straight line. The sentence at the top of the picture indicates this non-alignment
(see the paragraph at the end of the activity). In (D) about three-quarters of
the disk of the moon is visible. In part (E), half of the disk of the moon is
visible.
In (A), (B), (C), and (D), the moon is new, waxing, full, and waning respectively.
If (E) is interpreted as like (B), then the moon is waxing.
10.2
Angles
Angle Explorers,
More on Seeing That the Angles in a Triangle Add to 180◦,
Describing Routes Using Distances and Angles.
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In this section I did not introduce all of the standard angle terms such as complementary angles and supplementary angles. It is too easy for geometry instruction
to focus on terminology without much substance! In my opinion, it is best to introduce terminology only when there is a real need for it, namely, when the term will
be used repeatedly and is needed to facilitate clear communication about substantial
mathematics.
Class Activity 10D: Angles Formed by Two Lines
I like to demonstrate the phenomenon (which is sometimes called Thales’s Theorem)
by holding two straws together at their middles with a rubber band and moving the
straws to show different angles. This is a good place to talk about the distinction
between observing that something appears to be true and proving that it must be
true.
1. Observing the lines, it appears that a = c and b = d.
3. See the answer to practice problem 4. Students may come up with slightly
different proofs. For example, someone might say that since
a + b = 180◦
and
b + c = 180◦
it follows that
a+b =b+c
Subtracting b from both sides shows that a = c. Another student might give
the same proof in words but without equations, in which case you might ask for
equations that go along with the words.
Class Activity 10E: Angles Formed when a Line Crosses Two
Parallel Lines
1. By the Parallel Postulate, the angle that is vertically opposite from f is equal
to e. By what was proved previously about angles that are vertically opposite
each other, it follows that e = f .
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2. By either applying the Parallel Postulate directly, or by applying the previous
part, either angle that is next to g and together with g forms a straight angle
is equal to h. Therefore g + h = 180◦.
Class Activity 10F: Seeing That the Angles in a Triangle Add
to 180◦
Students love this way of seeing how the angles in a triangle add to 180◦ . I think it’s
important for students to recognize that this method is not an actual proof, but is
nevertheless a very cool way to show why the theorem is plausible.
A variation on this demonstration is to take three copies of a triangle, label the
angles a, b, and c on each, and put them together so that a, b, and c are adjacent.
Together, they will form a straight angle.
3. When the angles are put together they appear to form a straight line, which
makes 180◦ , but how do we know it isn’t 179◦ or 181◦ ? Also, even if everyone
in the class made a different triangle, that would still only be a finite number of
triangles that were checked. How do we know it will work for all the infinitely
many other triangles that weren’t checked?
Class Activity 10G: Using the Parallel Postulate to Prove that
the Angles in a Triangle Add to 180◦
One thing that seems to be difficult for many students is understanding that for a
proof such as this one, we must show that the theorem is true for all triangles. Even
though we work with a specific triangle, the argument we develop must apply to any
triangle. So it’s not enough just to verify for one specific triangle that the angles
add to 180◦, because how would you know that the theorem would be true for other
triangles?
You may wish to point out that proofs in geometry often involve adding in an
extra line.
1. According to Part 1 of the Class Activity on Angles Formed when a Line Crosses
Two Parallel Lines, the angle to the left of a is congruent to c and the angle to
the right of a is congruent to b. All three angles together form a straight line,
so a + b + c = 180◦ .
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Class Activity 10H: Explaining Why the Angles in a Triangle
Add to 180◦ by Walking and Turning
Most students find this activity really interesting.
One thing that seems to be difficult for many students is understanding that for a
proof such as this one, we must show that the theorem is true for all triangles. Even
though we work with a specific triangle, the argument we develop must apply to any
triangle. So it’s not enough just to verify for one specific triangle that the angles add
to 180◦ .
2. (a) The turner faces all directions once.
(b) The turner made a full rotation of 360◦ .
3., 4., & 6. See the related practice exercise.
Class Activity 10I: Angle Problems
I like to tell students that these angle problems are their “Sudoku fix for the day”!
1. x = a + b. If the other interior angle of the triangle is called c, then we know
a + b + c = 180◦. But c + x = 180◦ , so x must be equal to a + b. We can
summarize this result as saying that the sum of two angles in a triangle is equal
to the exterior angle at the other vertex.
2. There are many ways to solve this problem. One is to draw additional lines
that are parallel to the two parallel lines through the vertices at angles b and
c. Then by applying the parallel postulate several times, the sum a + b + c + d
consists of three pairs of angles that add to 180◦ , so the sum is 3 · 180◦ = 540◦ .
Another method is to extend the line segment between angles b and c so as to
make two triangles with the two parallel lines and add all the angles. On the
one hand, this full sum is a + b + c + d + e plus the sum of the angles in two
triangles, and on the other hand, this full sum is made from 4 straight angles
and two angles that add to 180◦ because of the parallel lines.
Two more methods involve connecting the two parallel lines with a perpendicular line and then considering the pentagon or hexagon created that way, or to
connect the vertices at angles a and d and consider the quadrilateral created
that way.
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3. One method is to divide the pentagon into 5 triangles that have a common vertex
in the middle of the pentagon. Adding all the angles, one gets 5 × 180◦ = 900◦ ,
but from that, we must take away 360◦ in the middle, leaving 540◦ . Another
option is to subdivide the pentagon into 3 triangles.
4. Extending the reasoning of the previous part, the sum of the angles in an N-gon
is (N − 2) · 180. (Technically, for the arguments above to extend, one should
really assume that the N-gon is convex, i.e., that the line segment connecting
two vertices lies entirely within the N-gon.)
Class Activity 10J: Students’ Ideas and Questions About Angles
1. Harry’s idea applies to area, not to angles. The area of the circle is less than
the area of the triangle.
2. Sam is right that you could put in more and more smaller and smaller wedges,
but the angles in these wedges would become less than 1 degree. Kaia is wrong
because you can have angles of 12 of a degree, for example.
10.3
Angles and Phenomena in the World
Angles of Sun Rays
How the Tilt of the Earth Causes Seasons,
The Special Shape of Satellite Dishes.
You can safely skip this section if you are low on time or prefer to focus on other
material.
Class Activity 10K: Eratosthenes’s Method for Determining
the Circumference of the Earth
This activity is designed to fit with the From the Field children’s literature recommendation, The Librarian Who Measured the Earth, by Kathryn Lasky.
Because the sun rays are (essentially) parallel, the Parallel Postulate together
with the theorem about vertically opposite angles formed by two lines crossing tells
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us that the angle at C, is also 7.2◦ . To form the full circumference of the earth we
could repeat the 7.2◦ sector to form a full circle. Because 360 ÷ 7.2 = 50, the sector
should be repeated 50 times to make the full circle. This means that the 500 mile
distance is repeated 50 times to make the full circumference of the earth. Based on
these measurements, the circumference of the earth is 50 · 500 = 25, 000 miles, which
is pretty close to the actual distance!
Class Activity 10L: How Big Is the Reflection of Your Face in
a Mirror?
This is a fun activity and quite surprising.
In Section 9.3 there is a follow up problem (problem 2 on page 491) in which
students will use angle-side-angle congruence to explain the phenomena observed.
1. The measured size of the reflected face in the mirror, using a ruler held on the
mirror, does not change, even though the reflected face does appear smaller
when the mirror is held farther away.
The length of the reflected face in the mirror is always
the face.
1
2
of the actual length of
2. At point A in the mirror, the person will see something that is back behind
them (perhaps something on a wall). At point B, the person will see her bangs.
At point C, the person will see her chest.
3. In the mirror, the place where the person sees the top of her head is opposite
the location that is half way between the person’s (actual) eyes and the top
of her (actual) head. Similarly, the place where the person sees the bottom
of her chin in the mirror is opposite the location that is half way between the
person’s (actual) eyes and the bottom of her (actual) chin. Thus the length of
the reflected face in the mirror is half the length of the actual face.
Class Activity 10M: Why Do Spoons Reflect Upside Down?
Remind the students that the lines drawn are normal lines to the spoon, not lines of
sight. (At point B the normal line is also the line of sight.)
At point A in the spoon, the person sees her chin. At point B she sees her eye.
At point C she sees her hair.
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10.4
Circles and Spheres
More on Circle Designs.
Class Activity 10N: Points That Are a Fixed Distance From
a Given Point
It is important for teachers to understand the mathematical definition of circle and
sphere and not just see these shapes as “nice and round”.
1. Students should gradually see a circle being formed.
2. The points form a sphere.
Class Activity 10O: Using Circles
1. A compass draws a circle because the pencil point stays a fixed distance away
from the point of the compass. The radius of the circle is the distance between
the point of the compass and the pencil point.
2. Tie the string to the pencil. Tie the other end of the string to a tack, or hold it
on a piece of paper. Keeping the string taut and keeping the non-pencil-end of
the string at a fixed point on a piece of paper, swing the pencil around so that
it draws a circle on the paper.
3. In 1 hour and 45 minutes the prisoner can go at most 7 miles. So the prisoner
will be inside a circle of radius 7 miles on the map. Thus the prisoner could be
in the woods or in the swamp.
4. Draw a circle centered at X of radius 30 feet on the map and draw another
circle centered at O of radius 50 feet on the map. The treasure could be buried
at one of two locations: one in the cactus patch and the other in the snake pit.
As an extension, and as a preview to some of the practice exercises and problems,
ask students where the treasure could be if it was less than 30 feet from X and
less than 50 feet from O.
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Class Activity 10P: The Global Positioning System (GPS)
5. The locations form a circle which is the intersection of two spheres.
7. If three spheres intersect, it will generally be at 2 points.
Class Activity 10Q: Circle Designs
Part 1 implicitly contains the important construction for equilateral triangles as well
as the construction for regular hexagons. These constructions will come up later, but
you can preview the ideas here. Some students might be interested in looking for
other designs based on circle designs.
1. The compass should stay fixed at the same opening throughout. Once you draw
one circle, you pick a point on it, put the compass point there, and draw another
circle of the same radius as the first. Then at the two points where that circle
meets the first circle, put the compass point in there and draw two more circles.
Keep putting the compass point at a place where two (or more) circles meet
and keep drawing circles of the same radius.
10.5
Triangles, Quadrilaterals, and Other Polygons
Relating the Kinds of Quadrilaterals,
Constructing Quadrilaterals with Geometer’s Sketchpad.
Class Activity 10R: What Shape Is It?
The examples show that it’s not reliable to consider only prototypical shapes oriented
in the usual way when considering what type a shape is. To identify shapes reliably
we must attend to attributes.
Class Activity 10S: What Properties Do These Shapes Have?
1. In each case: there are 4 sides and all 4 sides have the same length. In each
case there are 4 angles and opposite angles are the same size. Adjacent angles
add to 180◦ . (Another property is that the diagonals are perpendicular.)
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2. Orientation, length or width, and color are not relevant for deciding what type
a shape is.
Class Activity 10T: Classify Shapes into Categories Based on
Their Properties [core]
1. Category 1: C, E.
Category 2: C, D, E, M.
Category 3: same as 2.
Category 4: same as 2 and 3.
Category 5: C, E, G, I, J
Category 7: same and 5 and 6.
Category 8: C, D, E, G, H, I, J, M.
Category 10: B, C, D, E, G, H, I, J, K, M.
Category 11: B, K.
Category 12: all but N.
2. Category 1 consists of the squares.
Categories 2, 3, 4 have the same shapes. They are rectangles.
Categories 5 and 6 have the same shapes. They are rhombuses.
Categories 8 and 9 have the same shapes. They are parallelograms.
Category 10 consists of trapezoids according to one definition, whereas category
11 consists of trapezoids according to a different definition.
Category 12 is quadrilaterals.
3. Category 1 is a subcategory of Category 2 (and 3 and 4). So squares are a
subcategory of rectangles.
Category 2 is a subcategory of Category 8 (and 9). So rectangles are a subcategory of parallelograms.
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Category 5 (and 6 and 7) is a subcategory of Category 8 (and 9). So rhombuses
are a subcategory of parallelograms.
Category 8 (and 9) is a subcategory of Category 10. So parallelograms form a
subcategory of trapezoids (according to one definition of trapezoid).
Category 10 is a subcategory of Category 12.
Class Activity 10U: Classify Triangles Based on Their Properties
Category 1: A, E, J
Category 3: A, B, E, G, H, I, J, L, M.
Category 5: A, B, E, J, L, N.
Category 6: C, G, H, K.
Category 7: D, F, I, M.
Category 1 (and 2) is a subcategory of Category 3 (and 4).
Every triangle falls into one and only one category among Categories 5, 6, and 7.
Class Activity 10V: Venn Diagrams Relating Quadrilaterals
1. Show the set of squares entirely within the set of rectangles.
2. See the answer to the related practice exercise.
3. Show the set of parallelograms entirely within the set of trapezoids. See the
related practice exercise.
4. In this case, the set of parallelograms and the set of trapezoids would not meet
at all. See the related practice exercise.
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Class Activity 10W: Using a Compass to Construct Triangles
and Quadrilaterals
Note that the last part of this activity relates to the triangle inequality, which is
discussed briefly in Chapter 14.
1. Draw a point and use a compass to draw a circle with center at this point.
Choose any two distinct points on this cirle. These two pints and the center are
the vertices of an isosceles triangle.
2. It’s a good idea to let students try this without using a compass first because
many will approach it this way anyway and will soon discover how awkward it
is.
3. See the appropriate practice problem.
4. Draw a circle with center C that passes through A and another circle with center
B that passes through A. Let D be the point other than A where the two circles
meet. Since AB and AC have the same length, the circles have the same radius.
Therefore CD and BD have the same length as AC and AB and so the resulting
quadrilateral is a rhombus.
5. Draw two circles, one centered at A, one centered at B, one of radius 3 inches
and one of radius 2 inches. Then proceed as in part 4 of the previous activity.
6. Draw a line segment of length 6 inches. Draw circles centered at the endpoints
of the line segment. One circle should be of radius 5 inches, the other of radius
3 inches. Then proceed as in part 1.
7. No it is not possible because if you draw a line segment of length 6 inches and
you make circles centered at the two endpoints, one of radius 3 inches and one
of radius 2 inches, the circles will not intersect.
Ask students if there is a general relationship among the lengths of the sides
that must hold for any triangle. There is, namely the triangle inequality, which
states that in a triangle, the length of any side is less than the sum of the lengths
of the other two sides.
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Class Activity 10X: Making Shapes by Folding Paper
You might like to use this activity to discuss that shapes have properties and characteristics beyond the defining properties and characteristics of the shape.
1. When you cut and then unfold you create a triangle with two sides corresponding
to the cut line and hence equal.
The “base angles” of the triangle are congruent. The triangle consists of two
copies of a right triangle.
2. When you cut, all 4 edges are created to be the same length. Therefore the
resulting shape is a quadrilateral with 4 sides of the same length, and so is a
rhombus.
The diagonals of the rhombus are perpendicular, bisect each other, and bisect
the angles of the rhombus.
3. The angles on either side of a fold are equal and add to 180◦ . Therefore each
such angle is 90◦ . Since the resulting quadrilateral has 4 right angles, it is a
rectangle, by definition.
4. Opposite sides are created to be parallel by the converse of the Parallel Postulate
because the angles are kept the same during sliding. Therefore the resulting
shape is a quadrilateral with opposite sides parallel, and so is a parallelogram.
Opposite sides have the same length.
Class Activity 10Y: Making Shapes by Walking and Turning
Along Routes
1. See the related Practice Exercise. In each case the total turning is 360◦. The
robot will turn the exterior angles of the shape, which we can find by dividing
360◦ by the number of turns (which is the same as the number of sides). Walking
around the triangle, each turn is 120◦ ; around the pentagon each turn is 72◦ ;
around the hexagon each turn is 60◦ .
2. See the similar Practice Exercise.
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Measurement
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CHAPTER 11
The Class Activities in this chapter are generally quicker and less involved than
activities in some of the other chapters. If you are used to allowing a lot of time
for discussion, take care to keep things moving. But even though this material is
less involved, it’s important for teachers to be well aware of the fundamental ideas of
measurement and know about some of the misconceptions their students may have.
11.1
Fundamentals of Measurement
Concerning the material on Systems of Measurement on pages 483 – 487, I generally
only spend very little class time on this material, expecting students to learn this basic
factual information mainly through their reading. I do like to discuss briefly how the
metric system uses prefixes and I also like to bring up the relationship between the
units of length, capacity, and mass described in Table 11.4.
Class Activity 11A: The Biggest Tree in the World
In order to say which tree is biggest, you have to specify a measurable attribute that
you are using first.
The next section will go into the distinction between one-dimensional, two-dimensional,
and three-dimensional attributes.
Class Activity 11B: What Does “6 Square Inches” Mean?
Use this activity to emphasize that a measurement that is described as “A units”
means an amount that is made from A of those units.
If you have cubic inch blocks you could also ask students what 6 cubic inches
means and have them show you with the blocks. They should show a shape made
from 6 blocks.
1. 6 square inches means the area formed by 6 non-overlapping 1-inch-by-1-inch
squares (which can be cut apart and rearranged).
A 6-inch-by-6-inch square has area 36 square inches.
2. (a), (c), and (e) describe the same area.
(b) and (d) describe the area, namely 16 square inches.
(f) is different from all the rest.
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3. Not every shape is a rectangle. For shapes that aren’t rectangles, the area is
not just “length times width”. If something has an area of A square inches,
that means that it can be thought of as made from A squares, each 1 inch by 1
inch.
Class Activity 11C: Using a Ruler
1. When we read the number on a tick mark of an inches ruler we are reading the
number of inch long strips (or objects) that could be placed end-to-end along
the ruler from its begining (0 tick mark) to the tick mark we are reading.
2. The child should focus on the 1 cm intervals, not just the numbers at the end
of each interval or the number of tick marks that the object touches. We can
measure by starting at a tick mark other than 0, but then we have to subtract
to get the correct length.
3. Some children just look at the endpoint, to get 8 cm. Others count the tick
marks, yielding 7 cm (this is essentially the same problem as in part 2).
4. Each of the small intervals are
11.2
1
8
inch long, not
1
10
inch long.
Length, Area, Volume, and Dimension
Class Activity 11D: Dimension and Size
1. One-dimensional: distance across, distance around, depth. Two-dimensional:
surface area. Three-dimensional: volume of water in the lake.
2. One-dimensional: length of the front, distance to the curb, height, length of
gutters. Two-dimensional: floor area, surface area of siding. Three-dimensional:
volume of air in the house.
11.3
Error and Precision in Measurements
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Class Activity 11E: Reporting and Interpreting Measurements
1. (a) Assuming that the food label is following the usual conventions, it means
that the amount of trans fat in a serving is 0 grams when rounded to the
nearest whole number. This means that there could be up to 0.5 grams of
trans fat in the food.
(b) Even in this case there could still be some trans fat in the food. The
amount of trans fat, when rounded to the nearest tenth of a gram, is 0.0.
This means there could be up to 0.05 grams of trans fat in the food.
2. Since 92,960,000 miles rounds to 93,000,000 miles, both can be considered correct.
11.4
Converting from One Unit of Measurement
to Another
This section includes material on area and volume conversions. If you feel that your
students may need more time to work with area and volume first, you might choose
to delay that part of the section.
Class Activity 11F: Conversions: When Do We Multiply?
When Do We Divide?
1. (a) As Julie could see from a yardstick each yard is a group of 3 feet. Julie
should then realize that if she has 6 yards she has 6 groups of 3 feet. The
meaning of multiplication then tells her that all together she has 6×3 = 18
feet.
(b) The bigger the unit, the fewer that are needed to measure the object.
The smaller the unit, the more are needed to measure the object. A
measurement reported in feet will be a larger number than a measurement
reported in yards, so she should multiply by 3 to convert yards to feet.
(c) You could ask Julie to determine her height in feet and yards and to then
compare the two measurements.
2. You might show Nate a meter stick with centimeter markings. Nate should then
see that each meter is a group of 100 centimeters.
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Class Activity 11G: Conversion Problems
1. Every group of 12 inches is 1 foot, so we have to see how many groups of 12 are
in 57 in order to see how many feet are in 57 inches. This means that we must
calculate 57 ÷ 12. Since 57 ÷ 12 = 4, remainder 9, we can say that Shaquila is
4 feet, 9 inches tall. We can also say that 57 ÷ 12 = 4 34 or 4.75, which means
that Shaquila is 4 34 or 4.75 feet tall. The first answer is the most common way
we describe people’s heights. Since 9 inches is 43 of a foot or .75 of a foot, these
answers all mean the same thing.
2. One way would be to ask how many groups of 5 paperclips (2 rods) are in a
group of 35 paperclips. The answer is 35 ÷ 5 = 7. The since each group of 5
paperclips exactly corresponds to a group of 2 rods, there are 7 × 2 = 14 rods
in a group of 35 paperclips.
3. One method: 25 people will drink 25 × 8 = 200 fluid ounces. Since a gallon is
16 cups, a half-gallon is 8 cups. Since each cup is 8 fluid ounces, a half-gallon is
8 × 8 = 64 fluid ounces. To see how many half-gallon containers we must buy,
we must see how many groups of 64 are in 200, so we must divide 200 ÷ 64.
Since 200 ÷ 64 = 3, remainder 8, this tells us that 3 half-gallon containers would
leave us 8 fluid ounces short, so we should get 4 half-gallon containers to be
sure to have enough.
Another method: Instead of calculating in fluid ounces, work in cups, pints,
gallons, or half-gallons instead.
Class Activity 11H: Using Dimensional Analysis to Convert
Measurements
1. (a) The two methods are basically just different ways of writing the same thing.
Method B fits better with standard mathematical notation, but Method
A seems to be in common use and well known to many students.
(b) The fractions are chosen so that the units will cancel and you’ll be left
with the units you want. Each fraction is equal to 1.
(c) Each fraction is equal to 1, so when you multiply all the fractions, you
have not changed the amount, you’ve just changed how it is expressed.
Therefore the beginning and end amount are the same, just expressed in
terms of different units.
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CHAPTER 11
(d) It’s the exact same calculations.
2.
1 mile ×
5280 ft 12 in 2.54 cm
1m
1 km
×
×
×
×
= 1.6 . . . km
1 mile
1ft
1 in
100 cm 1000 m
Class Activity 11I: Area and Volume Conversions
1. No, 1 square yard can be broken in to 3 rows with 3 square feet in each row.
So 1 square yard is 3 × 3 = 9 square feet.
2. The rug is 15 feet long and 12 feet wide, so its area is 15 × 12 = 180 square feet.
Another way: the rug is 5 × 4 = 20 square yards in area. Each square yard is 9
square feet, so the area is 20 × 9 = 180 square feet.
3. 35 × 9 = 315 square feet.
4. No the volume of the compost pile is 8 cubic yards.
5. The compost pile is 6 feet by 6 feet by 6 feet and so has volume 6 × 6 × 6 = 216
cubic feet. Another way: each cubic yard is 27 cubic feet. Since there are 8
cubic yards, the compost pile is 8 × 27 = 216 cubic feet.
Class Activity 11J: Area and Volume Conversions: Which Are
Correct and Which Are Not?
1. (a) Incorrect, because the calculations are done with linear units, not square
units.
(b) Correct.
(c) The first part, converting 25 meters to feet is correct. But the next part is
not correct because it would have to be 252 square meters, not 25 square
meters.
(d) Correct
2. One way: convert 1 foot to meters, use that to convert 1 cubic foot to cubic
meters and then multiply the result by 27. Another way: convert 1 foot to
meters, multiply by 3 to find 3 feet in meters. Cube that result to find 27 cubic
feet in cubic meters.
201
Class Activity 11K: Problem Solving with Conversions
1. If each person takes a 10 minute shower, flushes the toilet several times, and
uses some more water for hand washing, cooking and drinking, that might add
up to 150 liters or so. So 100,000 people would use 15 million liters. One cubic
meter is 100·100·100 = 1 million cubic centimeters. Since each cubic centimeter
is 1 milliliter, 1 million cubic centimeters is 1000 liters. So 1 cubic meter is 1000
liters. So 15 million liters is 15,000 cubic meters, which is a volume of a “box”
that is 100 meters by 100 meters by 1.5 meters. Such a “box” would cover a
football field and be about 5 feet tall.
2. In a 1 mile by 1 mile plot planted with these trees, you’d have about 528×528 =
278, 784 trees. So for 1,000,000 trees you’d need about 4 square miles of land,
which might be the size of a large park. For 1 billion trees, you’d need about
3,600 square miles of land, which is the amount of land in a 60 mile by 60 mile
square plot, and is larger than Rhode Island and Delaware. For 1 trillion trees,
you’d need about 3.6 million square miles of land, which is about the same as
the total land area of the U.S.
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Area of Shapes
203
204
12.1
CHAPTER 12
What Area Is and Areas of Rectangles Revisited
Class Activity 12A: Units of Length and Area in the Area
Formula for Rectangles
1. The shape can be covered with 15 square centimeter squares, allowing that some
squares may need to be cut apart and rearranged.
2. Cover the rectangle with square centimeter tiles and count that it took 15 of
them.
3. The squares covering the rectangle can be viewed as arranged in 3 groups, each
group containing 5 squares. Therefore there are 3 × 5 squares total, each of area
1 square centimeter, so the total area is 3 × 5 square centimeters.
4. Each of the 3 centimeter units of length along a 3 cm side of the rectangle can
be viewed as giving rise to a group of squares. But the 3 cm length itself is
one-dimensional and should not be described as made from the 3 squares along
the edge. Similarly, a 5 cm side of the rectangle gives rise to a group of 5
squares, but the 5 cm length itself is a one-dimensional attribute and should
not be described as made from the 5 squares along an edge.
5. Note that the side lengths are 7 and 9 tenths of a unit whereas the area of each
small square is 1 hundredth of a square unit. So here we really see why we
need to distinguish area and length measures—the squares really don’t stand
for lengths. In part (b), students should see that there are 63 squares shaded,
each of area 1 hundredth of a square unit.
12.2
The Moving and Additivity Principles About
Area
More on Using the Moving and Additivity Principles.
205
Class Activity 12B: Different Shapes with the Same Area
You might like to prepare the small rectangular pieces of paper for your students.
Make sure they are rectangles that aren’t squares. An ordinary piece of paper can be
cut into eight 4 14 in by 2 34 in pieces (cut the paper in half lengthwise first).
2. Three sides of the triangles can each be joined in two different ways (both right
side up, or with one triangle flipped over). When the hypotenuses are joined,
you create the original rectangle or a kite. When one of the other two sides are
joined, you create a triangle one way, and a parallelogram the other way.
3. In each case, the area of the shape is the same as that of the original rectangle.
This is because of the moving and additivity principles. If we call the area of
the triangles T1 and T2 , then in each case, the area of the shape is T1 + T2
because the areas are not changed when the pieces are moved around. (In fact,
T1 = T2 because the triangles are identical.)
4. The area of the new shape is the same as that of the original rectangle for the
same reason as before.
Have students marvel at the fact that they can determine the area of such a
complicated shape!
5. The areas are all the same but the perimeters, widths, and lengths can all be
different.
Class Activity 12C: Using the Moving and Additivity Principles
You may wish to give students paper cutouts of enlarged versions of the shapes in
parts 1 and 2 of this activity. See the blackline masters on pages 284 and 285.
1. (a) Dividing the L-shape into two rectangles, the area is 6 × 5 + 4 × 5 = 50
square units or 2 × 5 + 4 × 10 = 50 square units. (b) View the shape as a
rectangle with a smaller rectangle removed from the top right. The area is
6 × 10 − 2 × 5 = 50 square units. (c) Cut the top half of the left portion of
the L that sticks up off and reattach it to the right to make a 5 by 10 rectangle
whose area is 5 × 10 = 50 square units. (d) Put two copies of the shape together
to make a 10 by 10 square. This square’s area is twice the area of the L-shape
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because it’s made of two copies of the L, so the L-shape has area 10×10÷2 = 50
square units.
Students should be aware that even in the case where they are subtracting areas,
they are still using the additivity principle because of the link between addition
and subtraction. In other words, the additivity principle can be used in the
case of an unknown addend, i.e., a situation involving an equation of the form
B+? = A, where ? is found by subtracting B from A.
2. One way: The shaded square can be subdivided into 4 right triangles, 2 of which
can be moved and reattached to make a 1 unit by 2 unit rectangle. Therefore
the shaded square has area 2 square units. Another way: By cutting off the
white triangles surrounding the shaded square and joining those white triangles
together, you can see that the large square is made of two copies of the shaded
square, so the shaded square has half the area of the 2 unit by 2 unit square.
12.3
Areas of Triangles
Some curricula develop the parallelogram area formula before the triangle area formula. One reason to discuss parallelograms first is because they are generalizations
of rectangles, so their area formula can be seen as a generalization of the rectangle
area formula. On the other hand, the area formula for triangles is probably more important and provides a contrasting case to rectangles. As long as the presentation is
logical and coherent, I think it is fine to discuss areas of triangles and parallelograms
in either order.
Class Activity 12D: Determining Areas of Triangles in Progressively Sophisticated Ways
You may wish to copy pages 287 and 288 onto transparencies so students can share
their methods with the class. You may also wish to make copies of pages 286 for
students to work on. You might also like to bring scissors to class if you’d like to let
students cut the triangles apart.
1. Emphasize that teachers must know how to use the principles at various levels
of sophistication to determine area.
(a) The least sophisticated way is to subdivide into very small pieces and recombine those pieces to make many individual squares, and count up that there
207
are 12 squares. Students should understand and appreciate this simple method
because it relies very directly on the meaning of area. However, this method is
limited to cases where there is a grid and a small enough number of squares to
be practical to count.
(b) More sophisticated is to move bigger chunks. A tip of the triangle on the
left can be cut off and moved to form either a 2 by 6 rectangle or a 4 by 3
rectangle. For the triangle on the right, the tip on the right can be cut off, then
divided into two pieces, one which can be rotated to the top, the other which
can be rotated to the bottom to form a 4 by 3 rectangle.
(c) Probably most sophisticated is to see the triangles as enclosed in a 4 by
6 rectangle. In each case, the rectangle can be thought of as made from two
copies of the triangle. This is pretty clear for the triangle on the left. For the
triangle on the right, you could take another copy of it, and cut it into 2 pieces
in such a way that the two pieces are the two extra triangles inside the 4 by 6
rectangle. Since the rectangle enclosing the triangle is made from two copies of
the triangle, its area is twice the area of the triangle, so the area of the triangle
is (4 × 6) ÷ 2.
2. For “moving chunks,” solving problems using the associative properties of addition and multiplication involves similar reasoning. For example, consider the
make-a-ten method in addition. For “putting inside a rectangle and taking
half,” consider multiplication by 5, which can be viewed as multiplication by
10, but then you take half. For “putting inside a rectangle and taking away,”
consider a multiplication problem such as 38 × 7, which can be solved by calculating 40 × 7 and then subtracting 2 × 7.
Class Activity 12E: Choosing the Base and Height of Triangles
See Figure 12.1.
Class Activity 12F: Explaining Why the Area Formula for
Triangles Is Valid
If there are scissors available for students to use, you may wish to give each student
4 small rectangular pieces of paper to cut apart and experiment with: two for the
right triangle in part 1, and two for the triangle in part 2. Alternatively, you could
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base
height
base
height
height
base
Figure 12.1: Three Ways to Choose the Base and Height
209
make copies of pages 289 and 290 and give each student three or four copies of each
triangle to cut apart. I like to use colored paper so that when students put two copies
of triangles together, the two copies are nicely distinguished.
You might ask students to find the approximate area of the triangle in square
inches or square centimeters, but emphasize that the point is to understand and
explain the general formula.
1. You can put two copies of the triangle together to make a b by h rectangle. So
according to the moving and additivity principles, the area of the triangle is
1
(b × h). Or you can cut off part of the triangle and rearrange it to make a
2
1
b by h rectangle of area ( 12 b) × h. Or you can cut off part of the triangle and
2
rearrange it to make a b by 12 h rectangle of area b × ( 21 h).
According to the associative and commutative properties of multiplication, all
these formulas are equivalent.
2. You can cut apart and rearrange two copies of the triangle to form a b by h
rectangle, so that according to the moving and additivity principles, the area
of the triangle is
1
(b × h)
2
Or you can cut off the top half of the triangle, half way up the height, cut that
top apart, and arrange the two pieces on either side of the remaining portion of
the triangle to form a b by 12 h rectangle, which has area
1
b × ( h)
2
By the moving and additivity principles, this rectangle has the same area as
the triangle.
3. There are several incorrect statements in the given reasoning. The area of the
indicated rectangle is not b × h square units because the base is longer than
b. The triangle ABC takes up less than half of the area of the rectangle. The
triangle AEC has half the area of the rectangle.
4. Use area of triangle ACB equals area of triangle ACE minus area of triangle
BCE.
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Class Activity 12G: Area Problem Solving
I like to joke that when we do these problems we have to think outside the box to
think inside the box.
1. The triangle has area 10 square units. The triangle is surrounded by a 3 by
8 rectangle, and there are three unshaded right triangles inside the rectangle,
in addition to the shaded triangle. By the additivity principle, the area of the
rectangle is the sum of the areas of all four of those triangles. The combined
area of the three right triangles, in square units, is
1
1
1
(2 · 4) + (4 · 3) + (1 · 8) = 4 + 6 + 4 = 14
2
2
2
Subtracting this from the 24 square units of area of the rectangle leaves 10
square units for the area of the shaded triangle.
2. Area of shaded triangle equals area of rectangle minus area of upper triangle
minus area of lower triangle = 160 − 12 × 16 × 7 − 12 × 16 × 10 = 24.
Alternatively we can use the
1
2
× b × h formula with b = 10 − 7 = 3 and h = 16.
3. Area of shaded triangle equals area of combined triangle minus area of unshaded
triangle = 12 × (7 + 5) × 9 − 21 × 7 × 9 = 22 21 .
Alternatively we can use the
1
2
× b × h formula with b = 5 and h = 9.
4. The combined areas of the left and right lower unshaded triangles is 21 ×14×8 =
56 square units. The area of the upper unshaded triangle is 21 × 14 × 3 = 21
square units. So the unshaded area is 77 square units and therefore the shaded
triangle has area 14 · 8 − 77 = 35 square units.
12.4
Areas of Parallelograms and Other Polygons
Class Activity 12H: Do Side Lengths Determine the Area of
a Parallelogram?
Another way to do this activity is as follows: make a rectangle by threading 4 pieces
of straws onto string and tying it tight (as in the Is This Shearing? activity in the
next section. Demonstrate to the students how you can collapse the sides down to a
line segment. Ask them what remains the same and what changes. They should see
211
that the lengths of the sides remains the same, but the area changes. Therefore there
cannot be a formula for the area that is only in terms of the side lengths. I like to
hold the straw parallelogram behind my back and ask them if they can tell me the
area without seeing it.
1. The first parallelogram (which is a rectangle) has area 21 cm2 .
The second parallelogram has area 14 cm2 .
The third parallelogram has area 7 cm2 .
Since all three parallelograms have sides of the same lengths but have different
areas, there can be no formula for areas of parallelograms only in terms of their
side lengths.
2. The area of a parallelogram is its base times its height.
Class Activity 12I: Explaining Why the Area Formula for Parallelograms Is Valid
Here is a really cool demonstration you can do for part 2: Take a toilet paper roll
or paper towel roll and cut it along its seams. When you then unroll it, you will
have a parallelogram that is “very oblique” like the one in part 2. Take another
toilet paper roll or paper towel roll and cut it lengthwise, so that you can unroll it
to make a rectangle (or just make a cylinder the same size out of paper and have
another rectangular piece of paper that would make a cylinder of the same size).
Show the students how to go from the parallelogram to a cylinder and then to a
rectangular paper. Have them appreciate that each has the same area of paper and
that the rectangular piece of paper has base b and height h, just like the original
parallelogram.
You may want to use the blackline master on page 291 in the appendix so you
can give students several parallelograms to cut apart.
1. Cut along the height line and move the right triangle left so that its hypotenuse
coincides with the left edge of the parallelogram. You now have a rectangle
with area = b × h.
2. See the text for method (a). To use this method, students will need to name
the additional length along the base of the rectangle. If they are stuck, suggest
that they consider using a letter, such as a, for an unknown length. To use
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method (b), draw a line segment from the right end of the base on the bottom
to the left end of the base on the top, thus creating two triangles, each with
base b and height h.
Class Activity 12J: Finding and Explaining a Trapezoid Area
Formula
You may want to use the blackline master on page 292 in the appendix so you can
give students several trapezoids to cut apart.
Two copies of the trapezoid fit together to make a parallelogram of base a + b and
height h. The area of the trapezoid is therefore half the area of that parallelogram,
and so is 21 (a + b)h square units. See the figure in the problems in the text that shows
how to divide the trapezoid into two triangles, one of base a and one of base b. The
trapezoid therefore has area 12 a · h + 21 b · h which is equal to the other formula by the
distributive property.
You could show students how to relate this formula to the triangle area formula
by letting a go to zero. You can also relate the formula to the parallelogram area
formula by letting a go to b (and changing the trapezoid into a parallelogram of base
b. The trapezoid area formula fits with both of those as “limiting cases.”
12.5
Shearing: Changing Shapes Without Changing Area
You may wish to mention the link to calculus if some of your students have had the
course.
If you are able to show Geometer’s Sketchpad sketches in class, you might like to
create triangles and parallelograms that you can shear in class. Have the area visible
in the sketch so students see that it doesn’t change.
Area and perimeter will be contrasted in Section 12.8, but note that shearing
provides examples of shapes that have the same area but different perimeter.
Class Activity 12K: Is This Shearing?
During shearing, the area does not change, the height of the stack of toothpicks does
not change, the length of the top and bottom sides do not change, but the lengths of
the sides on the left and right do change.
213
During “squashing” (which is not a standard mathematical term—I made this up
to describe the situation), the lengths of the sides do not change but the area of the
area and the vertical height of the straw figure do change.
Therefore shearing and “squashing” are not the same.
Class Activity 12L: Shearing Parallelograms and Triangles
To extend this activity, give students graph paper and ask them to shear some triangles whose bases are not horizontal or vertical. They will need to think about what
characterizes a line that is parallel to a given line.
1. CD will move along the horizontal line through C and D which is parallel to A
and B.
2. On the left: You could shear the parallelogram so that the side parallel to the
base is directly above the base. The height above the base should remain the
same.
On the right: You could shear the parallelogram so that the side parallel to the
base is directly to its right, 5 units away.
3. The point C can move to any location on the line through C parallel to AB.
4. During shearing, the vertex that is not on the base must remain the same
distance from the base, so must move parallel to the base.
12.6
Areas of Circles and the Number Pi
Class Activity 12M: How Big Is the Number π?
The circumference of the circle is visibly greater than the perimeter of the hexagon
formed by the 6 triangles, which is 3 units (because each side of the triangle has
length 21 unit), and less than the perimeter of the large square surrounding the circle,
which is 4 units. So we can say that π is between 3 and 4.
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CHAPTER 12
Class Activity 12N: A Hands-On Method to Approximate the
Value of π
You may wish to copy page 293 for your students instead of having them draw their
own circles. The short and long strip should be taped together and students should
verify that the (combined) strip is the length of the circumference. The markings
along the diameter help students see that pi really is about 3.14.
3. Students should find that “3 and a little more” diameters fit into the circumference strip. Therefore the circumference is “3 and a little more” times the
diameter of the circle. Since π is the value of the circumference divided by the
diameter, this means that π should be approximately “3 and a little more.”
Class Activity 12O: Why the Area Formula for Circles Makes
Sense
Emphasize that the moving and additivity principles are again used to explain an
area formula.
You may wish to copy page 294 for your students so that they can decompose the
circle more easily.
Another really fun hands-on way to see why the area formula for circles makes
sense uses the kind of gum that consists of a very long tape rolled up into a circle.
Cut the roll of gum from the center out to the circumference. You will then be able
to open up the gum roll to make a triangle — each “wind” in the tape roll unrolls
to become a line segment, so the triangle consists of a bunch of parallel segments of
tape. The segments vary in length from tiny to the circumference of the gum roll.
The base of the triangle is the circumference of the gum roll and the height of the
triangle is the radius of the gum roll. Applying the formula for areas of triangles you
get the formula for the area of the circle! I would like to thank my student, Michael
Ziegler, for showing me this. Instead of a roll of gum, a long play-doh “snake” coiled
up also works nicely. Have students make the snake as long and narrow as possible.
See practice exercise 2.
Class Activity 12P: Area Problems
1. The area of the pool is
(
π
− 1) · 302
2
215
square feet, which is about 514 square feet. There are several ways students can
reason to solve this problem. One way is to draw a diagonal through the middle
of the pool from one corner of the square that the pool touches to the other
corner of the square that the pool touches. Now find the area of the triangle
on one side of this diagonal. If you subtract this area from a quarter-circle of
radius 30 feet, then you will have found half the area of the pool.
Another way to solve the problem is to think of the pool as the location where
two pieces of tissue paper in the shape of quarter-circles of radius 30 feet overlap.
If you add the areas of these two quarter-circles and subtract the area of the
square, it’s like subtracting one layer of the tissue paper, leaving a single layer
over the pool.
2. If Melchior’s chain is attached at a corner, then he can roam over 34 of a circular
region of radius 15 meters plus 41 of a circular region of radius 5 meters, so the
area of grass he can eat is
3
1
π152 + π52
4
4
square meters. If his chain is attached to the middle of the long side, then
Melchior can roam over half of a circular region of radius 15 plus two quarters
of a circular region of radius 7 meters. If Melchior’s chain is attached to the
middle of the shorter side, then he can roam over half of a circular region of
radius 15 meters plus two quarters of a circular region of radius 10 meters.
Attaching the chain to the corner of the shed provides Melchior with the largest
area of grass to graze on.
3. The same kind of reasoning as in finding the area of a circle can be used.
If the octagon is subdivided into 16 triangles, they can be placed alternately
so as to make 8 rectangles, each of base P/16 and height r, so the area is
8 × (P/16) × r = 21 P r. Because a circle’s area can be described as 12 Cr, where
C is the circumference, the formula for the area of the octagon is similar to the
formula for the area of a circle.
12.7
Approximating Areas of Irregular Shapes
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Class Activity 12Q: Determining the Area of an Irregular
Shape
Instead of having the students work with the lake shown in the manual, you might
prefer to give them copies of page 295 (in the appendix), which has graph paper on
it (the heavy grid lines are 1 inch apart).
1. You can trace the lake onto graph paper, cut the tracing up into squares and
partial squares, and combine partial squares to make full squares (more or less),
to determine how many square inches the map covers.
You can cover the lake with squares—more or less—compensating for squares
that “stick out” by leaving some parts of the lake uncovered. The number of
squares is the approximate area of the lake in square miles.
You can trace the lake onto paper, cut out this paper lake and draw it onto
heavy card stock. Weight a full sheet of card stock, weigh the card stock lake,
and the fraction of the card stock that is the lake is the fraction of the area of
the card stock that is the paper lake’s area.
You can spread modeling dough over the lake and then reform this modeling
dough as a rectangle on graph paper—taking care to use the same thickness.
The number of square inches the modeling dough covers is the area of the lake
in square miles.
You can’t just take a length of string that goes around the lake and reform that
string as a rectangle on graph paper, finding the area of the rectangle. This issue
will be explored in Section 12.8. When this comes up in class I ask students
to “hold that thought, we’ll be getting back to it soon.” Some may have even
done some initial explorations that will fit lead into the ideas of that section.
(It turns out that the perimeter of the lake is about 16 inches. If students form
that into a 4 inch by 4 inch square, it has area 16 square inches, but the actual
area is smaller, more like 7 square inches.)
I like to have students discuss these strategies generally, rather than have students carry them out in detail. Then we spend time on parts 2 – 4 where
students can get into the details of carrying out the methods. And finally we
go directly to the next section to confront the issue of perimeter versus area.
2. The area of the map of the state is
108 ·
1 1
3
· =6
4 4
4
217
square inches. Since 1 inch represents 100 miles, 1 square inch represents 100 ·
100 = 10, 000 square miles. So the area of the state is about 67,500 square
miles. See the next part for other methods.
3. (a) This calculation finds the area in square miles that each graph paper square
represents. The number 625 can then be multiplied by the number of graph
paper squares to find the total area of the state.
(b) This calculation might be based on a misconception because it might be
confusing length units and area units. However, the student could also be
thinking of putting the 108 graph paper squares in a row. This will form a
rectangle that is 14 inch by 27 inches. In real life, the 27 inch side becomes
2700 miles long. The 41 inch side becomes 25 miles long. The area can
then be found by multiplying 25 · 2700.
(c) This calculation shows that each 1 inch by 1 inch square consists of 16
graph paper squares. Therefore the 108 graph paper squares have an area
of 6.75 square inches. Each square inch represents 100 · 100 square miles.
So the state has area 6.75 · 100 · 100.
1
square
(d) This calculation shows that each graph paper square has area 16
1
inches. Therefore the 108 graph paper squares have area 108 · 16
square
inches, which can then be multiplied by 100 · 100 to find the area of the
state in square miles.
Numerical expression for the area of the state (based on the last part):
108 ·
1 1
· · 100 · 100
4 4
We can get the various calculations described above by evaluating this
expression in different ways.
4. The area of the county on the map is
3
1 3
2 · 3 = 9 9.375
2 4
8
square inches. Since 1 inch represents 15 miles, 1 square inch represents 15·15 =
225 square miles, so the area of the county is
9.375 · 225 = 2109.375
square miles or about 2100 square miles.
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6
of the area of the paper. Since the area
5. The area of the map of the state is 10
of the paper is
1
1
8 · 11 = 93
2
2
square inches, the area of the map of the state is
0.6 · 93.5 = 56.1
square inches. Since 1 inch represents 50 miles, 1 square inch represents 50·50 =
2500 square miles, so the area of the state is
56.1 · 2500 = 140, 250
square miles, or about 140,000 square miles.
12.8
Contrasting and Relating the Perimeter and
Area of Shapes
Calculating Perimeters.
If you did the section on shearing, then you could refer back to those examples to
see how shapes can have the same area but different perimeters.
Class Activity 12R: Perimeter Misconceptions
1. No, the squares at outer corners have two edges that should be counted, but
only get counted once. Squares at inner corners don’t contribute anything to
the perimeter and shouldn’t be counted.
2. (a) The diagonal line segment is
theorem), or about 1.4 cm.
√
2 cm long (according to the Pythagorean
(b) A rectangle that is 4 cm long and 3 12 cm wide has perimeter 15 cm.
219
Class Activity 12S: How Are Perimeter and Area Related for
Rectangles?
You may wish to hand out several pieces of the large graph paper on page 296 in
the appendix for students to make large versions of their rectangles in part 2. These
rectangles could be taped onto the board, ordered by their area.
1. Different expressions: A + B + A + B, 2A + 2B, 2(A + B). Of course all are
equivalent.
2. Students could use a strategy they develop from thinking about part 1. A
useful strategy is to find numbers for the length and width that add to half the
perimeter.
3. You could have students draw various rectangles of perimeter 20 units on another piece of graph paper (such as the graph paper on page 296), label the
rectangles with their area, and cut them out. Then students can put their
rectangles in order from larger to smaller area to see how shape and area are
related.
Otherwise, I like to write students’ results up on the board for all to see and
compare. First, ask what the largest area they got was and write the dimensions
on the board on the far left. Then ask for something smaller. Get a few students
to tell you their answers and put some up on the board, squeezing “in between”
answers in between ones you already have. Draw rough sketches of the rectangles
so students can see how shape and area are related.
More square-like rectangles have larger area, more elongated rectangles have
smaller area.
4. The largest possible area occurs when the rectangle is a square, which will have
sides of length 5 cm, and area 25 cm2 . Every positive number less than 25 is the
area of some rectangle of perimeter 20 cm. There is no smallest area rectangle of
perimeter 20 cm because you can keep making narrower and narrower rectangles
that will have smaller and smaller areas.
Class Activity 12T: How Are Perimeter and Area Related for
All Shapes?
1. No, this is not a valid method as we have seen from the previous activity.
Given a fixed perimeter, many different areas are possible, so perimeter does
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not determine area.
2. The largest area that can occur is that of a circle of circumference 200 miles,
whose area is
200 2 10, 000
) =
π·(
2π
π
square miles, which is about 3183 square miles. Every positive number less than
or equal to this is the area of some shape of perimeter 200 miles. So, although
perimeter does not determine area, it does constrain which areas can occur.
12.9
Using the Moving and Additivity Principles
to Prove the Pythagorean Theorem
Using the Pythagorean Theorem,
Can We Prove the Pythagorean Theorem by Checking Examples?.
Class Activity 12U: Side Lengths of Squares Inside Squares
Internet video opportunity. You might like to have students watch some of the clips
at http://www.mmmproject.org/ls/mainframeS.htm
from a Grade 8 Connected Math lesson, “Looking for Squares.” These video clips
can also be found by going to http://www.mmmproject.org/ls/mainframeS.htm
and to http://www.mmmproject.org/video_matrixS.htm.
1. Square A: The four outer unshaded triangles combine to make two 3 by 4
rectangles, so have a total area of 24 square units. Because the big square has
area 72 = 49 square units, the tilted square has area 25 square units. Therefore
its sides have length 5 units.
Square B: Similar reasoning shows
√ that the tilted square has area 5 square units.
Therefore each side has length 5 units.
2. Students could draw a square by connecting the midpoints
√ of each side. That
square has area 8 square units, so each side has length 8 units. They could
also draw a square that connects points that are 14 of the way across each side.
√
This square has area 10 square units, so each side length is 10 units long.
221
Class Activity 12V: A Proof of the Pythagorean Theorem
If you do not want to ask students to download the page for the activity, then you
might like to copy page 297 for your students. It has two sets of the shapes needed
for the activity, but does not include the squares that have sides of length a + b.
A different cool proof of the Pythagorean Theorem that uses the moving and additivity principles and is shown in an animated fashion can be found online at:
http://themetapicture.com/media/why-couldnt-i-have-been-shown-this-in-maths-class.gif
1. & 2. See Practice Exercise 6 and its answer. Encourage your students to find several
different proofs. In addition to the one given in the solution to the practice
exercise, students could just work with the one way of subdividing the a + b by
a + b square that involves the c by c square and describe the full area of the
square in two ways: as (a + b)2 and as c2 + 2ab. After some algebra, they should
deduce the theorem. The explanation given in the practice exercise can also be
carried out in a more algebraic way.
3. See Practice Exercise 7. At the places where two copies of the triangle and a
square meet at a point, the three angles are: a right angle (from the square)
and the two non-right angles from the triangle. Since the sum of the angles in
a triangle is 180◦ , the two non-right angles in a right triangle add to 90◦ . So
all three angles add to 180◦ and there really is no bend at the point, in other
words, the edge that appears to be a straight line really is a straight line.
4. See Practice Exercise 8.
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Solid Shapes and their Volume and
Surface Area
223
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CHAPTER 13
13.1
Polyhedra and Other Solid Shapes
Why Are There No Other Platonic Solids?,
Relating the Numbers of Faces, Edges, and Vertices of Polyhedra.
Class Activity 13A: Making Prisms and Pyramids
This is a lot of fun and something that teachers often do in elementary school.
Class Activity 13B: Analyzing Prisms and Pyramids
1. A pentagonal prism has 7 faces total, 5 are triangles and 2 (the bases) are
pentagons. A pentagonal prism has 15 edges: 5 around one base, 5 around the
other base, and 5 connecting vertices on the bases. A pentagonal prism has 10
vertices, 5 around each base.
2. A pyramid with a hexagon base has 7 faces total, 6 are triangles and 1 (the
base) is a hexagon. A pyramid with a hexagon base has 12 edges: 6 from the
lateral faces and 6 along the base. It has 7 vertices: 6 around the base and 1
at the top of the pyramid.
3. A rectangle is a flat, two-dimensional shape. A rectangular prism is a solid,
three-dimensional shape which has rectangles as faces.
You might also bring up the case of cubes versus squares, which are commonly
confused.
Class Activity 13C: What’s Inside the Magic 8 Ball?
You may wish to copy the triangles, squares, and pentagons on pages 298, 299, and
300, for students to cut out and make shapes with.
You can do part of this activity even if you don’t have a Magic 8 Ball. Most
students will be familiar with this toy. So you can start with part 2.
I like to use this activity to motivate the Platonic solids: the shape inside the
Magic 8 Ball should “be the same all the way around,” and “perfectly symmetrical,”
and this essentially describes what the Platonic solids are.
One option is to split this activity across 2 days. Show students a Magic 8 Ball
in class and have them think about part 1 of the activity. Then for homework, have
225
students do the Magic 8 Ball problem in the text before the next class. You could
modify the homework problem and ask students to make two shapes — one that they
think could possibly be in the Magic 8 Ball and one that they think couldn’t be, and
they should say why or why not. Discuss their answers in class and use the discussion
as a springboard to the Platonic solids.
2. The shapes with the bulleted properties in part 2 are the Platonic solids.
3. There is a single solid shape inside the Magic 8 Ball. If you see what answers it
provides, you will discover that it appears to have at least 20 faces. All the faces
are triangles and the shape should be “the same all the way around” otherwise
some answers would be more likely to come up than others. The only such
shape is an icosahedron. In fact, one of my students once broke one open and
found that the shape really is an icosahedron.
Class Activity 13D: Making Platonic Solids
As students are making these shapes, or immediately afterwards, you might also
discuss the activity in the online bank, Why Are There No Other Platonic Solids?, if
not in whole, then perhaps just informally and in part.
The tetrahedron has 4 faces, 6 edges, and 4 vertices.
The cube has 6 faces, 12 edges, and 8 vertices.
The octahedron has 8 faces, 12 edges, and 6 vertices.
The dodecahedron has 12 faces, 30 edges, and 20 vertices.
The icosahedron has 20 faces, 30 edges, and 12 vertices.
Notice that the numbers of faces, edges, and vertices, are reversed for the cube
and the octahedron, and they are reversed for the dodecahedron and the icosahedron.
This is because those pairs of shapes are “dual” to each other. That is, if we take
one of the shapes, put a vertex in the center of each face, and connect vertices that
lie on adjacent faces, then we will get the other shape in the pair.
There are relationships between the number of faces and the number of edges,
depending on the types of faces. For the shapes made of triangles, 3 times the
number of faces is equal to 2 times the number of edges. This is because if you think
of putting dots on each face right near each edge, then you will put 3 dots on each
face, and each edge will have two dots next to it. Similarly, and for the same kind
of reason, for the shape made of squares (the cube), 4 times the number of faces
is equal to 2 times the number of edges, and for the shape made of pentagons (the
dodecahedron), 5 times the number of faces is equal to 2 times the number of edges.
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Also, there are relationships between the number of faces and the number of
vertices, depending on the number of faces that come together at each vertex and the
number of vertices on each face. If you think of putting dots on each face right near
each vertex, then if N is the number of faces coming together at each vertex and if M
is the number of vertices on each face, then (number of vertices) · (number of faces
combing together at each vertex) = (number of faces) · (number of vertices on each
face). For example, for the dodecahedron, this equation becomes 20 · 3 = 12 · 5.
Students might not discover these relationships on their own, but you could have
them put dots on models as described above and then ask them if they can find
relationships.
13.2
Patterns and Surface Area
More on What Shapes Do These Patterns Make?,
More on Patterns and Surface Area for Cylinders,
Shadows of Solid Shapes.
Class Activity 13E: What Shapes Do These Patterns Make?
You might follow this activity by giving students models of shapes and having them
draw patterns for the shapes by putting the model face down on blank paper, tracing
around the outside of the face that is on the paper, and then turning the shape to
adjacent faces/surfaces so as to make adjacent pieces of the pattern.
Pattern 1 makes a triangular prism. Pattern 2 makes a triangular (triangle base)
pyramid. Pattern 3 makes a cylinder. Pattern 4 makes a cone.
Note the similarity between Patterns 1 (prism) and 3 (cylinder) and between
Patterns 2 (pyramid) and 4 (cone). These correspond to the similarities between
prisms and cylinders and between pyramids and cones. Note also that both patterns
1 and 2 will have triangular bases, although bases are not included in the patterns.
Patterns 5 – 8 included bases. Pattern 5 makes a triangular prism and Pattern 7
makes a prism with an octagon base. Pattern 6 makes a pyramid with a square base
and Pattern 8 makes a pyramid with a hexagon base.
Ask students how patterns 3 and 4 would need to be modified to include their
bases. Their bases will be circles, which are attached to the main portion of the
pattern at only a single point (instead of along a whole edge).
227
Class Activity 13F: Patterns and Surface Area for Prisms and
Pyramids
You might warm up to this activity by giving students models of shapes and having
them draw patterns for the shapes by putting the model face down on blank paper,
tracing around the outside of the face that is on the paper, and then turning the
shape to adjacent faces/surfaces so as to make adjacent pieces of the pattern.
1. The formula for the surface area, S, of the shape (in square centimeters) is
S = 2W L + 2W H + 2LH because there are two W by L faces, two W by H
faces, and two L by H faces.
2. This problem is trickier that it may seem at first because you need the Pythagorean
Theorem to find the size of the triangular faces of the pyramid. For the pyramid
to be 4 cm tall over the base, the triangular faces should be isosceles triangles
that have a base of 6 cm and height of 5 cm (not 4 cm!). To see why, think
about a vertical cross-section of the pyramid through the apex and straight
down, cutting through the middle of two sides of the square. This cross-section
is an isosceles triangle that has base 6 cm and height 4 cm. By the Pythagorean
theorem, the two other sides of the isosceles triangle are 5 cm.
The easiest way to make the pattern will be to make it look like the pattern 2
in Practice Exercise 1.
The surface area of the pyramid is therefore
1
(6 · 6) + 4 · ( · 6 · 5)
2
square centimeters, (because it is the area of the base plus the area of the four
triangular faces), which is 96 square centimeters.
Class Activity 13G: Patterns and Surface Area for Cylinders
1. The area of the lateral side of the cylinder (not including the bases) is the same
as the area of the rectangular piece of paper (by the moving principle), which
is 93.5 square inches.
If the cylinder is created as shown in the picture, then each base is a circle of
2
= 4.25
, which has area 4.25
. So the total surface area of the cylinder
radius 8.5
2π
π
π
is
4.252
93.5 + 2 ·
π
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CHAPTER 13
square inches which is about 105 square inches.
2. The label should be a 3π = 9.4 inch by 4 inch rectangle.
Class Activity 13H: Patterns and Surface Area for Cones
I recommend demonstrating for students how a sector of a circle becomes a cone
(by attaching radii) and highlighting how the part of the circumference of the circle
sector wraps all the way around the circumference of the base. (Note that here I am
using “circumference” not just to denote a length but also to denote the actual circle
perimeter or portion of it.)
1. Students should recognize that patterns for cones are made by taking a sector
of a circle (a “pie piece” or “pac-man”) and joining the cut radii.
2. The radius of the circular base will be 2 21 inches. This is because the halfperimeter of the circle of radius 5 inches is 21 · 2π · 5 = 5π. This half-perimeter
wraps around the base of the cone. So the circumference of the base of the cone
is therefore 5π. The radius of this base therefore satisfies the equation
2πr = 5π
so r = 2.5.
3. The circumference part of the half-circle (the half-perimeter of the circle) must
wrap completely around the circumference of the base. Therefore the circumference part of the half-circle must be 4π inches and so the circumference of
the whole circle that the half-circle is taken from is 8π inches. Therefore the
this circle, from which the half-circle is taken, has radius 8π ÷ 2π = 4. The
total surface area of is the sum of the area of the half-circle making the slanted
portion of the cone and the area of the base. Therefore the surface area of the
cone is
1
π · 42 + π · 22 = 12π
2
square inches.
4. The circumference part of the circle portion making the lateral (non-base) part
of the cone must wrap completely around the circumference of the base. Since
229
the circumference of the base is 4π and the circumference of a circle of radius 6
inches is 12π, and
1
4π
=
12π
3
you should use a third of a the 6-inch circle to make the lateral portion of the
cone. You can do this by measuring a 120◦ central angle in the 6-inch circle.
The total surface area of the cone is
1
π · 62 + π · 22 = 16π
3
square inches.
Class Activity 13I: Cross-Sections of a Pyramid
I think it’s best to run this activity as an exploration and not to get too concerned
about detailing exactly which shapes can and cannot occur.
Horizontal planes will create squares. Vertical planes can create triangles (by
slicing through the top of the pyramid or by slicing off the “bottom corners”), a
trapezoid, or another quadrilateral. Other planes can create triangles, quadrilaterals,
and even pentagons. When I do this activity I don’t insist that everyone find all these
shapes.
You cannot slice a Pryamid and get a plane shape that has a curved boundary
such as a circle. You also can’t get a polygon with more than 5 sides since the pyramid
only has 5 faces.
Class Activity 13J: Cross-Sections of a Long Rectangular Prism
You may need to remind students that the instructions say the cuts are not to go
through the ends of the board.
As with the previous activity, I think it’s best to run this activity as an exploration
and not to get too concerned about detailing exactly which shapes can and cannot
occur.
1. Yes, using a suitably angled plane, it is possible to get a square.
2. Yes, using a suitably positioned plane it is possible to get a variety of parallelograms.
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4. As with a pyramid, you cannot get a plane shape that has a curved boundary
such as a circle. The only cross-sections you can get from slices that don’t
go through the ends are quadrilaterals, because the slice will only meet 4 sides.
Furthermore, the quadrilaterals arising this way must be parallelograms because
if not, then a pair of opposite sides would meet somewhere, and then the planes
that those sides came from would also meet, meaning those planes would not
be parallel.
13.3
Volumes of Solid Shapes
Filling Boxes and Jars,
The Volume of a Rhombic Dodecahedron.
Class Activity 13K: Why the Volume Formula for Prisms and
Cylinders Makes Sense
As a warmup, it would be good to ask students what it means to say that a solid
shape has a volume of 12 cubic inches? It means that the shape can be made from
12 1-inch-by-1-inch-by-1-inch cubes, allowing for cubes to be cut apart and pieces to
be moved if necessary (or think of the cubes as made of clay, for example).
You can demonstrate shearing in part 3 with a stack of index cards.
1. The volume is the total number of 1-inch cubes it takes to make the shape. By
dividing the shape into layers, we obtain the formula.
2. The volume of each such layer of cubes is (area of the base) cubic units, because
the number of cubes needed to cover the base is given by the area of the base.
(Each portion of the base must be covered by a face of a cube, or a portion of a
face of a cube.) There are (height) layers needed to make the prism or cylinder.
Therefore the total volume of the prism or cylinder is (height) × (area of the
base) cubic units.
Each unit of length in the height of the prism gives rise to a layer. The area
of the base tells you how many cubes are in each layer even though area itself
is a two-dimensional attribute and the number of cubes in the layer gives the
volume of the layer. Each square unit of area of the base gives rise to a cubic
unit of volume in a layer.
231
3. When you shear the shape, the base, height, and volume do not change, so the
volume of the oblique prism or cylinder is still (height) × (area of base) cubic
units. The height is measured perpendicular to the base, not on the slant,
because this is the height of the prism or cylinder when it is sheared into a right
prism or cylinder.
Class Activity 13L: Comparing the Volume of a Pyramid With
the Volume of a Rectangular Prism
You may wish to make a class set of pyramids and prisms from the patterns on
pages 301 and 302. If so, use as heavy a card stock as you can. Beware that if the
shapes are made from paper or card stock, they bow out a bit when you fill them
with beans and they hold a little more than they actually should.
3.
volume of prism = 3 · volume of pyramid
volume of pyramid =
Class Activity 13M: The
mids and Cones
1
3
1
· volume of prism
3
in the Volume Formula for Pyra-
You might like to make some sturdy sets of oblique pyramids out of card stock to
re-use every time you teach this material. If so, Figure A.26 on page 303 (of this
manual) makes 3 oblique pyramids.
2. Three oblique pyramids fit together “nose to nose to nose” to make a cube.
3. The oblique pyramid can be sheared to form a right pyramid. According to
Cavalieri’s Principle, the right pyramid has the same volume as the oblique
one.
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CHAPTER 13
Class Activity 13N: Volume Problem Solving
1. Since the circumference of the base of the cone is 10π, the radius of the base
= 5 cm. Since the slant height of the cone equals 10 cm, by the
equals 10π
2π
Pythagorian Theorem √
we have that h2 + 52 = 102 which implies that the height
h of the cone
follows that the volume of the cone equals
√ equals 3 · 5. It
1
2
3
·
π
·
5
·
3
·
5
=
226.72
.
.
.
cm
.
3
You could make the cone and fill it with dry rice or beans which you can then
pour into a measuring cup to determine the volume (only very approximately).
2. The volume formula for a prism can be applied, viewing the block as a prism
lying on its side: the prism has a height of 5 inches and a triangular base of
area 12 · 2 · 2 = 2 square inches, so the volume is 5 × 2 = 10 cubic inches. Some
students may erroneously think that the pyramid formula can be applied, using
a 2.8 by 5 inch rectangular base. Point out that pyramids have a single point
at their apex, but this shape has a whole edge along the top.
Another way to determine the volume is to put two such blocks together to
make a 2 inch by 2 inch by 5 inch rectangular prism of volume 20 cubic inches.
A single block therefore has volume 10 cubic inches.
3. One way to determine the volume is to see the staircase as a prism (turned on
its side) of height 1 meter = 100 cm, and a stairstep base that turns out to
have area 2400 cm2 (which you can see by viewing it as made of 6 squares each
20 cm by 20 cm, for example). The volume is therefore 240,000 cm3 , which is
about 14 of a cubic meter.
Another way to determine the volume is to put two staircases together to make
a 60 cm by 80 cm by 100 cm rectangular prism, which has volume 480,000 cm3 .
The staircase has half the volume, 240,000 cm3 .
Class Activity 13O: Deriving the Volume of a Sphere
1. Because of the Pythagorean Theorem, h2 + s2 = r 2 , so s2 = r 2 − h2 . The crosssection of the half-sphere at height h is a circular region that has area πr 2 .
The cross-section of the cylinder minus the cone at height h has area πr 2 − πh2
because it’s a circular region of radius r “minus” a circular region of radius h.
Therefore the cross-sections have the same area.
233
2. By Cavalieri’s Principle, the half-sphere and the part of the cylinder that is
outside of the cone have the same volume. The cylinder has volume πr 2 ·r = πr 3 .
The cone has volume 31 · πr 2 · r = 13 πr 3 . Therefore the cylinder “minus” the cone
has volume πr 3 − 13 πr 3 = 32 πr 3 . A whole sphere therefore has twice as much
volume, namely 43 πr 3 .
Class Activity 13P: Volume Versus Surface Area and Height
You might want to bring cubic inch or cubic centimeter blocks for students to use in
part 2 when they discuss the distinction between surface area and volume.
1. There are many possible such boxes. For example, a 5 inch tall box with a 2
inch by 2 inch base has a smaller volume than a 2 inch tall box with a 3 inch
by 4 inch base.
2. If students made boxes in part 1, you could encourage them to use these props
as well as blocks to discuss the distinction between surface area (the total area
of the outer surface of the shape) and volume (how many cubic unit blocks the
shape could be made of).
3. There actually is no cylinder of largest volume that can be made from 1 piece
of paper. By cutting narrower and narrower strips and joining them together
to make a very long strip, the volumes can be made larger and larger. The
winning cylinder will have the smallest height, but all cylinders make from a
full sheet of paper will have the same lateral surface area (not including the
bases), namely the area of the sheet of paper.
13.4
Volume of Submersed Objects Versus Weight
of Floating Objects
Determining Volumes by Submersing in Water.
Class Activity 13Q: Underwater Volume Problems
1.
1
3
of a liter or 333 13 cubic centimeters.
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2. The volume of the tank = 50 × 80 × 30 = 120000 cubic centimeters or 120 liters.
Since the volume of the water is 80 liters and the volume of 34 of the tank is 90
1
liters, the volume of the rocks is 10 liters or 100
of a cubic meter.
3. Since the volume of 23 − 12 = 61 of the tank is 30 liters, the volume of the whole
tank is 6 × 30 = 180 liters, or 180000 cm3 . Next, using the height times area of
base formula we see that the height equals 180000 ÷ 3000 = 60 cm.
Class Activity 13R: Floating Versus Sinking: Archimedes’s
Principle
1. No, it turns out that the increase in water level doesn’t tell you the volume of
the clay, although you should leave this as an open question for now. When
students submerse the clay in part 2, that will tell them the volume of the clay.
2. The water level should go down when you sink the boat (unless you have a
non-standard kind of clay). See part 3.
4. A quarter weighs about 5 12 grams. It’s a good idea to float a number of quarters
in a cup in order to displace enough water to be able to measure. 10 quarters
would be a good number.
As an extension, ask students why people can float in salt water, such as in the
Great Salt Lake in Utah.
Salt water is more dense than ordinary water. That is, given a fixed volume of
ordinary water and the same volume of salt water, the salt water weighs more.
So you have to displace a smaller volume of salt water in order to displace your
weight in salt water than in ordinary water. The weight of your body’s volume
in salt water is more than your weight, therefore you float in salt water.
Geometry of Motion and Change
235
236
14.1
CHAPTER 14
Reflections, Translations, and Rotations
Exploring Reflections with Geometer’s Sketchpad,
Exploring Translations with Geometer’s Sketchpad,
Exploring Rotations with Geometer’s Sketchpad.
To extend this section, you could discuss that reflections, translations, and rotations provide many examples of non-commutativity. For example, the net effect of
rotating by 90 degrees counterclockwise about the origin and then reflecting across
the x-axis is not the same as the net effect of first reflecting across the x-axis and
then rotating 90 degrees counterclockwise. Students will see this example in the last
problem of this section.
Class Activity 14A: Exploring Reflections, Rotations, and Translations with Transparencies
When discussing how the location of a shape changes after a rotation or a reflection,
you might ask students to draw three line segments on their coordinate grids as well
as on their transparencies: one from the origin to a point on the shape, a second one
perpendicular to the x-axis from the x-axis to a point on the shape, and a third one
perpendicular to the y-axis from the y-axis to a point on the shape.
1. The points A and A′ are the same distance from the x-axis, but on opposite
sides. A line segment between A and A′ is perpendicular to the x-axis and is
divided in half by the x-axis. (Same for the other points.)
2. The points A and A′ are the same distance from the y-axis, but on opposite
sides. A line segment between A and A′ is perpendicular to the y-axis and is
divided in half by the y-axis. (Same for the other points.)
3. The points A and A′ and the origin lie on a straight line, with the origin exactly
half way between A and A′ . (Sam for the other points.)
4. The line segment between A and the origin is perpendicular to the line segment
between A′ and the origin. The distance from A to the x-axis is the same as
the distance from A′ to the y-axis. The distance from A to the y-axis is the
same as the distance from A′ to the x-axis. These last two statements make
sense because a 90◦ rotation takes the x-axis to the y-axis and the y-axis to the
x-axis.
237
5. Same as previous part.
6. The coordinates of A′ are obtained from the coordinates of A by adding the
coordinates of the chosen arrow. An arrow drawn from A to A′ looks just like
the chosen arrow, except that its tail is at A instead of at the origin.
Class Activity 14B: Reflections, Rotations, and Translations
in a Coordinate Plane
You could extend this activity to discuss why it is that the product of the slopes of
perpendicular lines is −1. To do so, consider a line segment in the plane that has
“run” b and “rise” a (i.e., to get from one endpoint to the other you must go b units
to the right and a units up). When the line segment is rotated by 90 degrees, the
new line segment created that way will be perpendicular to the original one and its
run and rise become −b and a respectively. See also the problems in this section in
which students are asked to describe the location of a point after rotating it by 90
degrees about the origin.
1. Translate each vertex of the shape 4 grid lines to the right and 2 grid lines up.
Then you can connect these translated points to make the translated shape.
2. Determine where the vertices of each shape should go when they are reflected
by considering their location relative to the line of reflection. For example, if
a point is 4 grid lines to the left of the line of reflection, then after reflection,
that point will be 4 grid lines to the right of the line of reflection. After finding
the locations of the reflected corner points, they can be connected to form the
reflected shape.
3. Determine where the corner points of each shape should go when they are
rotated by considering their locations relative to axes and by considering where
various points on the axes go when they are rotated. For example, the point on
the y-axis that is 4 units up from the origin will rotate to the point on the same
line that is 4 units below the origin. The vertex that is 4 units up from the
origin and 1 unit to the left of the y-axis will rotate to the point that is 4 units
below the origin and 1 unit to the right of the y-axis. After finding the locations
of the rotated vertices, you can connect them to form the rotated shape.
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4. Determine where the corner points of each shape should go when they are
rotated by considering their locations relative to the two axes and by considering
where various points on the axes go when they are rotated. For example, the
point on the y-axis that is 3 units up from the origin will rotate to the point on
the x-axis that is 3 units to the left of the origin. So the point that is 3 units
up and 2 units to the left of the origin rotates to the point that is 3 units to the
left and 2 units down from the origin. After finding the locations of the rotated
vertices, you can connect them to form the rotated shape.
Class Activity 14C: Which Transformation Is It?
1. Left: reflection; Right: 180◦ rotation. We can tell because on the left, the line
segments connecting A and A′ ; B and B ′ , C and C ′ , and D and D ′ are all
perpendicular to and cut in half by a single line (the line of reflection), whereas
on the right, those line segments all meet in one point (the point of rotation).
2. Left: 90◦ rotation; Right: reflection. We can tell because on the right, the line
segments connecting A and A′ , B and B ′ , and C and C ′ are all perpendicular
to and cut in half by a single line (the line of reflection), whereas on the left,
the line connecting A and C is perpendicular to the line connecting A′ and C ′ .
3. No, it doesn’t because the arrow does not connect corresponding parts of the
shape. It does show a translation by an arrow that connects a point to a
corresponding point.
14.2
Symmetry
Frieze Patterns,
Creating Symmetrical Designs with Geometer’s Sketchpad,
Creating Symmetrical Designs (alternate).
I like to introduce this section by showing students a bunch of pictures by the
artist M. C. Escher, including some pictures that have only rotation or translation
symmetry and asking students if the pictures are symmetrical. Students usually do
feel that the pictures are symmetrical in nature, but they have been used to thinking
of symmetry only as reflection symmetry.
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Class Activity 14D: Checking for Symmetry
When I do this activity I also like to bring in other designs I have that have symmetry
such as tablecloths and prints by Escher. Students are used to thinking about symmetry only as reflection symmetry. If you show students other symmetrical designs
and then ask them to notice the regularity in these designs, they will be receptive to
thinking about symmetry beyond just reflection symmetry.
The first design on the left has reflection symmetry. Proceeding to the right,
the next design has 4-fold rotation symmetry. The next has both 3-fold rotation
symmetry and reflection symmetry. The 4th design, on the far right, has 2-fold
rotation symmetry and reflection symmetry. The design on the bottom, if we imagine
that it continues forever to the left and right, has translation symmetry.
Make sure that students understand that only infinitely long designs can possibly
have true translation symmetry.
After students have a chance to explore the symmetries ask them about the distinction between a type of transformation such as rotation and the corresponding
type of symmetry, namely rotation symmetry. Students should understand that rotation symmetry is a property that some shapes or designs have, whereas a rotation
transforms a plane by moving shapes that are in the plane.
Class Activity 14E: Traditional Quilt Designs
1. First reflect each of the 4 patch designs across the middle vertical grid line. Next
reflect across the middle horizontal grid line. The resulting design is symmetric
across the middle horizontal and vertical grid lines.
2. Perform 3 90◦ rotations on each of the 4 patch designs about the center grid
point. The resulting design has 4-fold rotation symmetry about the center point
Class Activity 14F: Creating Interlocking Symmetrical Designs
14.3
Congruence
Triangles with an Angle, a Side, and an Angle Specified,
Investigating Diagonals of Quadrilaterals with Geometer’s Sketchpad,
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Investigating Diagonals of Quadrilaterals (alternate).
At some point during this section (and/or the next), you might want to give
students guidance on the logical “flow” of congruence problems.
1. Find sides/angles of the same size from the situation.
2. Explain that a congruence criterion (or in the next section, a similarity criterion)
can be applied because of those sides or angles that are the same size.
3. Now use the fact that the triangles are congruent to deduce that some other
sides or angles are the same size. (Or in the next section use the fact that the
triangles are similar to deduce that sides are related by a common scale factor
or internal factor.)
Class Activity 14G: Motivating a Definition of Congruence
Use this activity to motivate the definition of congruence. We want congruence to
mean “identical copy of.” When we check to see if two shapes are congruent (or
not), we move them around and try to match them up. The moving process uses the
transformations discussed in Section 14.1.
The two triangles are congruent. One needs to use a reflection (which is an upside
down flip with the paper) to see that they match up identically.
Class Activity 14H: Triangles and Quadrilaterals of Specified
Side Lengths
Some students do not have the intuition in advance that the triangle will be rigid but
the quadrilateral will be floppy. It is therefore important for students to experience
this distinction first hand.
Use this activity as a springboard to discussing side-side-side congruence.
Class Activity 14I: What Information Specifies a Triangle?
Use this activity to continue to the discussion of side-side-side congruence and to
introduce the other congruence criteria.
There is only one triangle with the description of Triangle 1. Students may not
recognize right away that the side-side-side congruence criterion is relevant for Triangle 1.
241
There is no triangle with the description of Triangle 2. Students can see this This
leads to the Triangle Inequality, which states that the length of one side of a triangle
is less than the sum of the lengths of the other two sides.
There is only one triangle with the description of Triangle 3. This illustrates the
side-angle-side congruence criterion.
There is only one triangle with the description of Triangle 4. This illustrates the
angle-side-angle congruence criterion.
There are two different triangles with the description of Triangle 5 because the
circled centered at B with radius 4 cm intersects the ray that makes a 20◦ angle with
the line segment AB at 2 distinct points. Hence there are 2 possible triangles. Thus
there is no angle-side-side congruence criterion in general. Note, however, that the
hypotenuse-leg theorem does provide such a criterion in the special case where the
specified angle is 90 degrees. This theorem is not covered in the text.
There are many triangles with the description of Triangle 6. It turns out that all
of them are similar to each other. There is no angle-angle-angle congruence criterion.
Class Activity 14J: Explaining Why Isosceles Triangles and
Rhombuses Decompose into Congruent Right Triangles
1. If you add the line segment from A to the midpoint M of BC, then you can
apply SSS congruence to triangles AMC and AMB to conclude that they are
congruent. The angles CMA and BMA must therefore be the same size. Since
they form a straight angle, they must each be a right angle. So the isosceles
triangle has been decomposed into two congruent right triangles.
If you add the line segment that divides the angle at A in half, then you can
apply SAS congruence to conclude the those two smaller triangles are congruent
and continue as in the previous argument. Or you can argue that this line
segment is a line of symmetry.
2. Here’s the problem: how do you know that this line segment satisfies all those
conditions simultaneously?
3. Draw the diagonal BD and apply SSS congruence to the triangles DAB and
DCB, which are isosceles triangles. Now apply the result of part 1 to those
isosceles triangles to decompose them each into two congruent right triangles.
We can see that all four are congruent by applying SSS to the isosceles triangles
ADC and ABC and decomposing them into two congruent right triangles each,
which includes one triangle from each of the other two pairs of isosceles triangles.
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Class Activity 14K: Sewing Boxes and Congruence
Draw a diagonal from A to C and apply SSS congruence to triangles ADC and CBA.
Now view that diagonal as a transversal to AB and CD. Because ADC and CBA
are congruent, angles BAC and DCA are the same size. Therefore by the converse
of the parallel postulate, AB and CD are parallel.
14.4
Constructions With Straightedge and Compass
Class Activity 14L: Relating the Constructions to Properties
of Rhombuses
1. & 2. See practice exercise 3.
3. Let S be the point where the two circles meet (other than P). Then PRSQ is
a rhombus for the following reason. PQ and QS are both a radius of one of
the circles drawn, PR and RS are both a radius of another circle drawn, and
PR and PQ are a radius of the first circle drawn in the construction. Therefore
PR, RS, SQ, and QP all have the same length and so PRSQ is a rhombus, by
definition.
4. Rhombuses have the special property that a diagonal divides the angle at either
endpoint in half. The diagonal of the rhombus in part 3 is PS, so this property
of rhombuses guarantees that the line segment PS will divide the angle at P,
namely QPR, in half.
Class Activity 14M: Constructing a Square and an Octagon
with Straightedge and Compass
1. Students may be stuck and first and may only be able to think about bisecting
the line segment AB. Ask them where they want a perpendicular line to be. If
they want the perpendicular line to be at A, then they will need a line segment
such that A is the midpoint of that line segment. Can they use a circle to make
such a line segment?
Here’s how to make such a line segment: extend the line segment through A.
Use a compass to make a circle centered at A of radius AB. Let C be the other
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point where the circle intersects the line through AB. Then A is the midpoint of
CB because CA and AB are radii of the circle. Now construct the perpendicular
bisector to CB. This line will pass through A and will be perpendicular to AB.
To construct a square: construct a perpendicular line through B. Then construct
circles centered at A and B of radius AB. Construct the points E, F where these
circles meet the perpendiculars to AB. Then ABFE is a square.
2. If students are stuck, ask them what they can say about the line segments
from the center of the circle to the vertices of the octagon. Students should
recognize that these line segments will make 45◦ angles. So the octagon can
be constructed by making 45◦ angles, which can be created by first making 90◦
angles and then bisecting those angles.
To create these angles: draw a line through the center of the circle. Construct
the perpendicular to this line through the center of the circle. Then either bisect
the 90◦ angles formed at the center of the circle, or construct the perpendicular bisectors to the line segments formed by connecting the points where the
constructed lines meet the circle. The points on the circle that result can be
connected to form an octagon.
14.5
Similarity
Using Scaling to Understand Astronomical Distances,
More Scaling Problems.
If you have an overhead projector in the room, you could illustrate similarity by
projecting a shape and moving the projector closer and farther from the screen to show
similar shapes. Or if you are able to project from a computer that has Geometer’s
Sketchpad on it, you could show similar shapes using that software. You can even
show similar shapes by shining a flashlight at the wall, placing your hand between
the flashlight and the wall, and moving your hand closer and farther from the wall,
keeping it parallel to the wall.
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Class Activity 14N: Mathematical Similarity Versus Similarity
in Everyday Language
In the top left shape that is similar to the original, the width and height and all
the lengths within the shape are twice as long as in the original. In the top right
shape, the width and height and all lengths within the shape are half as long as in
the original. In the other shapes there is no such consistency in the way the lengths
of the shape compare to the lengths in the original. In the bottom left shape the
width is the same as in the original, but the height is half that of the original. In the
bottom right shape, it’s the other way around: the height is the same but the width
is half that of the original.
Class Activity 14O: A First Look at Solving Scaling Problems
You can use this activity to get students thinking about scaling before you discuss it
in greater detail.
You might also ask students to think about incorrect solution methods that students might come up with. If so, students may come up with reasoning like that in
Class Activity 14Q.
The larger poster is to be 3 times as wide as the original, so it should also be 3
times as long and therefore 3 × 4 = 12 feet long.
The original poster is twice as long as it is wide, so the large poster should also
be twice as long as it is wide and should therefore be 2 × 6 = 12 feet long.
You can set up and solve a proportion such as
2
6
=
4
x
You can draw a picture showing copies of the small poster inside the larger one.
Class Activity 14P: Using the Scale Factor and Internal Factor
Methods
Be sure to connect the reasoning used here to the ways of reasoning about proportions
discussed in Chapter 7.
= 5 times as wide as the postcard, so it should
1. Scale factor: The paper is 20
4
also be 5 times as long. Therefore you should cut the paper
5 × 6 = 30
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inches long.
Internal factor: The postcard is 64 = 32 times as long as it is wide, so the paper
should be cut to be 23 times as long as it is wide also. Therefore the paper
should be cut
3
× 20 = 30
2
inches long.
Set up a proportion: Let x be the length to cut the paper. Then
x
6
=
20
4
or
or
or
x
20
=
6
4
4
20
=
x
6
4
6
=
x
20
Note that even when we set up one of these proportions, we can solve the proportion without cross-multiplying as illustrated in the text (and as was discussed
in Chapter 7). In this way, you can connect the proportion to the scale factor
and internal factor methods.
2. Since the enlarged bunny is 3 times as tall as the original, it will be easy to
use the scale factor method to determine that the enlarged bunny is 3 · 6 = 18
inches wide. If we used the relative sizes method we would have to divide 33
by 1 65 which is harder than multiplying 3 · 6.
3. Since the actual car is 2 times as long as it is wide, the scale model will be also,
and so the scale model will be 2 · 2 12 = 5 inches long. To use the internal factor
method we would have to divide 2 21 by 6 and then multiply that number by 12
(and this is also going between feet and inches – if we wanted to work with like
units there would be even more calculations involved).
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Class Activity 14Q: Addressing a Common Misconception About
Scaling
1. No, the reasoning is not valid. If you draw a scale picture of both posters you will
see that this can’t be correct. Johnny is reasoning additively but multiplicative
reasoning is needed in this case.
Compare to Class Activity 7A, in which students might also mistakenly believe
the drink mixtures taste the same by using this additive reasoning.
2. The points of posters that are similar to the 4-foot-wide, 6-foot-long one all lie
on a line that goes to the origin. The point corresponding to Johnny’s proposed
poster does not lie on that line.
You could also ask the students to show the posters on the graph in such a way
that the top right corner of the poster is the point that is plotted. They will
see how the similar posters line up.
Class Activity 14R: Applying a Dilation
The dilated figure will be twice as wide and twice as tall as the original and will be
above and to the right of the original. The dilation doubles all distances, including
ones within the figure. It also preserves the sizes of angles. Discuss also that the
dilation doubles the coordinates of a point.
Class Activity 14S: Measuring Distances by “Sighting”
See the text discussion (especially “thumb sighting”) and also the practice exercises
for this approach.
Class Activity 14T: Using a Shadow or a Mirror to Determine
the Height of a Tree
Students will need to know the laws of reflection in order to use the method of part
2 (see section 10.3).
First method:
3. The triangles in Figure 14.1 are similar because the angle that the Sun’s rays
make with horizontal ground is the same for both triangles and both triangles
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have a right angle. Therefore all three angles of the triangles are the same, and
so the triangles are similar.
tree
person
tree's shadow
person's shadow
Figure 14.1: Similar Triangles Created by the Sun’s Rays
Second method:
2. The triangle formed by the person’s eye, the point in the middle of the mirror
where the tree is seen, and the person’s feet is similar to the triangle formed by
the tip of the tree, the middle of the mirror, and the base of the tree. This is
because the two triangles have the same angles because of the laws of reflection
and because the tree and the person both make a 90 degree angle with the
ground.
14.6
Areas, Volumes, and Scaling
Surface Areas and Volumes of Similar Cylinders.
Class Activity 14U: Surface Areas and Volumes of Similar
Boxes
In this activity, have students see the size relationships by comparing the physical
boxes and also by examining the formulas for surface area and volume.
2. The surface area of the big box is 4 times as large as the surface area of the
small box because each rectangular portion making the surface of the larger
box is 4 times as large as the corresponding rectangular portion of the smaller
box. A note of caution: ask students to use correct wording. The phrase “4
times larger than” really means “5 times as large as”, just as “100% more than”
means “200% of”.
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3. The volume of the larger box is 8 times the volume of the smaller box. You
might want to have students explain this with the models: that you could fit
2 layers in the larger box, each layer consisting of 4 of the smaller box. Ask
students to explain the “8 times as large as” using the formulas as well.
4. The surface area of the larger box would be 3 × 3 = 9 times as large as the
surface area of the smaller box, as you can see from drawing patterns and
comparing corresponding rectangular portions. The volume of the larger box
would be 3 × 3 × 3 = 27 times as large as the smaller box, as you can see from
the formula for the volumes, or from imagining filling the larger box with copies
of the smaller box: there will be 3 layers, with 3 × 3 small boxes in each layer.
5.
Size of big box compared to small box
length, width, depth 2 times 3 times 5 times
2.7 times
k times
surface area
22 = 4 32 = 9 52 = 25
2.72 = 7.29
k2
times
times
times
times
times
3
3
3
3
volume
2 = 8 3 = 27 5 = 125 2.7 = 19.683
k3
times
times
times
times
times
Class Activity 14V: Determining Surface Areas and Volumes
of Similar Objects
1. The surface area of the larger blimp would be 1.52 = 2.25 times the surface area
of the small blimp. The volume of the large blimp would be 1.52 = 3.375 times
the volume of the small blimp.
2. Since weight is generally proportional to volume, we should see how the volume
of the dinosaur compares to the volume of the alligator. The dinosaur was twice
as long, deep, and wide as the alligator, therefore its volume was 8 times the
volume of the alligator. The dinosaur should therefore have weighed about 8
times as much as the alligator, namely 8 × 475 = 3800 pounds.
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Class Activity 14W: A Scaling Proof of the Pythagorean Theorem
1. The two smaller right triangles each have a right angle and share a common
angle with the larger right triangle. Hence all three triangles have the same
angles and are therefore similar.
2. The area of the original triangle is some percentage, say P % of the area of
the square on its hypotenuse. When the original triangle is scaled to make the
smaller, similar triangles, the area of the triangle and the area of the square on
the hypotenuse both scale by the square of the scale factor, so the percentage of
the area of the square that is taken up by the triangle remains the same, P %.
Therefore the areas of the three triangles are P % · c2 , P % · a2 , and P % · b2 .
3. The equation P %a2 + P %b2 = P %c2 follows from the previous part and the
additivity principle, which tells us that the area of the triangle with hypotenuse
a plus the area of the triangle with hypotenuse b equals the area of the triangle
with hypotenuse c.
4. Factor out and then cancel P % on both sides of the equation P %a2 + P %b2 =
P %c2 .
Class Activity 14X: Area and Volume Problem Solving
1. The picture is a square that is twice as wide and twice as high as the unshaded
square inside it. The full outer square therefore has 4 times the area of the inner
unshaded square. The area of the shaded region is the area of the outer square
minus the area of the inner unshaded square. Therefore the shaded region has
area 3 times the area of the inner unshaded square. Students can also reach
this conclusion by subdividing the shaded region into pieces and recombining
the pieces to make 3 copies of the inner unshaded square. Or they could work
with measurements of the squares and calculate that way.
2. Assuming that the shaded region was created by taking a shape and removing
a scaled down copy of the shape, scaled down by a scale factor of 12 , the same
reasoning described for the previous part shows that the shaded region has area
3 times the area of the inner unshaded shape.
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3. The volume of the cup equals the volume of the cone with height h = 20 in and
radius r = 4 in minus the volume of the cone with height h = 10 in and radius
r = 2 in. Hence the volume of the cup equals π3 · 42 · 20 − π3 · 22 · 10 = 293.22
cubic inches. You could also view the cup as having 87 of the volume of a cone
of radius 4 in and height 20 in. The fraction 78 comes about because the cup
is a cone with a cone of half the height and half the radius inside it removed.
The cone that is removed has 81 of the volume of the full cone, so the remaining
portion has volume 1 − 81 = 78 of the full cone.
It might be interesting to work with measurements from an actual paper cup
and then check the volume of the cup (only very approximately) by filling the
cup with dry rice or beans and then pouring these into a measuring cup. If so,
it would be better to use metric measurements.
Statistics
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15.1
Formulating Statistical Questions, Gathering
Data, and Using Samples
Challenges in Formulating Survey Questions.
Class Activity 15A: Statistical Questions Versus Other Questions
1. (a), (c), and (f) are not statistical questions, but (b), (d), (e), and (g) are
statistical questions because they anticipate variability and addressing them
could involve collecting data.
For (g), the circle can cover different numbers of dots depending on where it is
placed (however it turns out that the average number approximates the area of
the circle).
Class Activity 15B: Choosing a Sample
This activity provides an opportunity for students to think about some of the issues
involved in designing a plan to collect data.
Class Activity 15C: Using Random Samples
This Class Activity provides the opportunity to reason about proportional relationships beyond simply setting up and solving proportions.
Problem 2 gives “germs” of different ways to reason. I think it’s good practice for
prospective teachers to elaborate on someone else’s initial idea. Some of these ways
may be similar to ideas students come up with in Problem 1. I put the “germs” of
solution methods second, not first, so that students would have a chance to come up
with some of their own ideas first.
1. 3% of the chips tested were defective. Since this was a random sample, it is
likely to be representative, and so about 3% of 5000 or
3% · 5000 = 150
chips in the whole batch are likely to be defective.
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Instead of a percentage calculation, we could solve the problem with a proportion by solving
x
3
=
100
5000
for x, where x stands for the approximate number of defective computer chips
in the whole batch.
Instead of working explicitly with a percentage or a proportion, students could
reason that about 3 out of every 100 chips are likely to be defective. There are
5000 ÷ 100 = 50 groups of 100 in 5000, so there should be about 50 · 3 = 150
defective chips in the whole batch.
See Problem 2 for more ideas.
2. (a) Out of every group of 20 plastic squares, about 4 should be yellow. If we
keep making groups of 20 until all 160 squares are used, we will make 8
groups of 20. So there should be a total of about 8 ×4 = 32 yellow squares.
(b) Just like part (a) except now counting by 20s to 160 is shown. This is
essentially a “ratio table” (see Section 7.6).
(c) In the sample of 20 squares that were picked, there were 15 as many yellow
squares as squares. Therefore in the whole bag of squares there should be
about 15 as many yellow squares as squares. Thus there should be about
1
× 160 = 32 yellow squares total.
5
(d) Since Taryn picked 18 of the squares in the bag she must have picked 18
of the yellow squares in the bag. Thus there should be about 8 × 4 = 32
yellow squares total.
(e) This is an equivalent fraction formulation of (a) and (b).
(f) Each column of 5 small boxes holds 20 plastic squares with the unshaded
box at the bottom holding 4 yellow squares. Since there are a total of 160
squares, there must be 160 ÷ 20 = 8 columns. Thus there are 8 lightly
shaded boxes for a total of 8 × 4 = 32 yellow squares.
4
x
(g & h) Two different ways of solving the proportion 20
= 160
are illustrated. (g)
is closely related to (d) and (h) is closely related to (c). See also Sections
7.6, Ratio and Proportion, and Section 9.4, Similarity, for these ways of
reasoning.
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Class Activity 15D: Using Random Samples to Estimate Population Size by Marking (Capture-Recapture)
This Class Activity provides the opportunity to reason about proportional relationships beyond simply setting up and solving proportions.
I like to run this activity as a whole class demonstration. I use a large bag of dry
beans and fish out 75 – 100 beans, which I distribute to a bunch of students to mark
and throw back. Then I fish out 75 or so, distribute to various students and have
them count up how many are marked. If time, it’s good to repeat the “recapture”
part. It also works well for students to work in groups with smaller baggies filled with
beans or cut-up index cards.
See the comments on the previous Class Activity.
3. See the next part for some methods students could use.
4. (a) Out of every group of 40 fish, about 5 should be marked. If we keep making
groups of 40 until we get to 30 marked fish, counting 5 marked fish at a
time, we see that we need 6 groups. So there should be about 6 × 40 = 240
fish total.
(b) Just like part (a) except now counting by 5s to 30 is shown.
(c) In the sample 40 fish that were caught, there were 8 times as many fish
as marked fish. Therefore in the whole population of fish there should be
about 8 times as many fish total as marked fish. Thus there should be
about 8 × 30 = 240 fish total.
(d) Since 61 of the marked fish were choosen about 16 of all the fish were choosen.
Thus there should be about 6 × 40 = 240 fish total.
(e) This is an equivalent fraction formulation of (a) and (b).
(f) Each row of small rectangles represents a group of 40 fish divided evenly
into 8 groups of 5 with the group of 5 marked fish represented by the 1
small shaded rectangle. Since there are a total of 30 marked fish, there
must be 30 ÷ 5 = 6 small shaded rectangles and hence 6 rows of small
rectangles and thus a total of 6 × 40 = 240 fish.
5
(g & h) Two different ways of solving the proportion 40
= 30
are illustrated. (g)
x
(or 7) is closely related to (d) and (h) (or 8) is closely related to (c).
15.2
255
Displaying Data and Interpreting Data Displays
More on the Length of a Pendulum and the Time It Takes to Swing.
Class Activity 15E: What Is Wrong With These Displays or
Their Interpretation?
You might like to use this activity in conjunction with introducing the different kinds
of graphs. If so, start by introducing a real graph by using the idea of a scoop of
plastic animals, as described in part 1. Discuss pictographs in conjunction with part
1. Then ask how the graph in part 1 could be modified to display the same data,
which will lead to bar graphs. Then examine parts 2 and 3 to reinforce bar graphs
and their proper use.
1. This pictograph does not facilitate accurate comparisons because, for example,
even though there are more crocodiles than giraffes, the stack of giraffes is taller
than the stack of crocodiles.
Some teachers provide students with pre-drawn bars that have been separated
into squares. Within each bar, students can fill one square for each item they
have for that category, either with a small picture, or by coloring the square.
2. A line graph should not be used to display these data since there is no logical
reason to connect grains to vegetables, vegetables to fruits, etc. Instead, a bar
graph should be used.
3. Because the pictures represent 3-dimensional objects (smokestacks) they exaggerate the distinctions between the carbon dioxide emissions of the countries.
For example, the US emits about 2 and a quarter times as much carbon dioxide
per person as does the EU, but the volume of the US smokestack and smoke
plume as depicted is more than 8 times the volume of the EU smokestack and
plume. A plain bar graph would provide a more accurate representation.
4. Even though the percentages add to 100%, the data should not be displayed in a
single pie graph because the different categories do not represent distinct, nonoverlapping groups of children. For example, among the 16% of children aged
4–6 who meet the dietary recommendations for vegetables, there are probably
many who also meet the dietary recommendations for fruits.
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5. The student has listed the frequencies with which the various data occur, not
the data themselves. The list of data is
27, 29, 29, 29, 30, 30, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 31, 32, 32, 32, 32, 32, 33
6. No, because the vertical axis begins at 6, not at 0. The graph is not incorrect,
but caution should be used in interpreting line graphs. In particular, the way
the axes are labeled should be checked carefully!
Class Activity 15F: Three Levels of Questions about Graphs
It’s best to view the three levels as approximate. Some questions may legitimately fit
at two different levels. So I think it’s best not to worry about identifying the exact
level of a question. Rather, encourage students to think about a variety of different
kinds of questions might be interesting to pose. A range of questions will help them
determine if their students are understanding the information that is in the graph.
1. (a) How many times were 6 yellow tiles picked? Answer: 9.
What number of yellow tiles got picked 5 times? Answer: 7.
(b) What were the most and least number of yellow tiles that people got?
Answer: 10 and 2.
How many times were 10 tiles picked? Answer: 30 (the total number of
dots).
(c) How many yellow tiles do you think are in the box? Answer: since most
of the dots are on or near 6, it should be that somewhere around 60% of
the tiles are yellow. So that would be about 60 tiles.
If we did this experiment again, would the dot plot come out exactly the
same? Answer: probably not.
2. A few examples of questions:
What percent of the population was male and age 20 – 29? (Read the data)
Answer: 7%.
What percent of the population was under age 20? (Read between the data)
Answer: 7% + 8% + 7% + 7% = 29%.
Why is the percentage of males and females not the same for each age group?
(Read beyond the data)
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Class Activity 15G: Display These Data about Pets
2. (a) The axis should be labeled 0 through 12. Three dots are over 0 because 3
students have no pets, 5 dots are over 1, 3 over 2, 2 over 3, 1 over 6, and
1 dot is over 12.
This display will be useful for answering questions 2 and 3.
(b) & (c) The categories could be each of the different kinds of animals, or some of
the animals could be put together in a category. For example, “very small
mammals” or “reptiles, amphibians, and fish.” Note that some students
will be plotted in two or more categories if they have more than one pet.
These displays will be useful for answering question 1.
(d) In this case the 12 fish will each count as an entry in the appropriate
category (“fish,” or “reptiles, amphibians, and fish”).
These displays will be useful for answering question 4 as well as a question
about what the most common animal is as a pet.
Class Activity 15H: Investigating Small Bags of Candies
You may wish to have sticky notes and small paper plates on hand for this activity
(see below).
If it does not arise, encourage a question about how many candies are in each
bag. A dot plot is good for answering this question. You can have students write the
number of candies in each bag on a sticky note and then create a class dot plot on
the chalkboard.
You can have students make pie graphs showing the color break-down of their
candies by organizing the candies by color and then placing them around the rim of
the paper plate (see text for an example).
Class Activity 15I: The Length of a Pendulum and the Time
It Takes to Swing
Sample “Read The Data” Question: If the length of the string is 5 inches how long
does 10 swings take? Answer: 3.96 seconds.
Sample “Read Between The Data” Question: How much longer does it take the
pendulum to swing 10 times if the length of the string is 10 inches instead of 5?
Answer: 5.13 − 3.96 = 1.19 seconds.
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Sample “Read Beyond The Data” Question: The data points almost form a
straight line. Would they exactly lie on a straight line if the time measurements
were exact?
Sample “Read Beyond the Data” Question: How long a length of string do you
predict you would need so that one swing would take 1 second? Answer: If the string
is about 30.5 inches long, then 10 swings should take 10 seconds and 1 swing should
take 1 second. (See the activity in the online activity bank for more.)
Class Activity 15J: Balancing a Mobile
This activity gives students the opportunity to conduct an experiment, collect data,
formulate questions, and attempt answers.
My data: straw length 19.5 cm.
1 clip: string is 9.7 cm from each side.
2 clips: string is 6.7 cm from 2-clip side (12.8 from other side).
6 clips: string is 2.5 cm from 6 clip side (16.9 from other side)
In general, the product of the weight on one side with the length is equal to the
product of the weight on the other side with its length.
15.3
The Center of Data: Mean, Median, and Mode
Class Activity 15K: The Mean as “Making Even” or “Leveling
Out”
2. List 1: mean 2.
List 2: mean 4.
List 3: mean 3 15 .
List 4: mean 2 34 .
3. See text.
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Class Activity 15L: Solving Problems about the Mean
1. To have an average of 4 blocks in all 5 towers you will need a total of 5 × 4 =
20 blocks. Since you have already used 11, you need 9 more, which can be
distributed any way you like among the 5 towers. Another way to solve the
problem is to think about leveling out the towers so that there will be 4 blocks
in each. After distributing 2 blocks from the 6 block tower to make the 3 block
towers into 4 block towers you will still need 2 more blocks for the 3rd tower,
and 4 blocks for each of the 4th and 5th towers.
2. If you think about solving the problem in terms of “leveling out”, then you will
need 1 more mile for each of the first 5 days and you need 4 miles for the 6th
day, which is a total of 5 + 4 = 9 miles. You can also solve this by figuring that
you will need to run a total of 6 × 4 = 24 miles over the 6 days but you have
already run 15 miles, so you need to run 9 more.
3. If you think in terms of leveling out, you can take 3 from the 41 and give 1 to
each of the three 37s to make them each 38. The fourth number is now also 38
because 3 was taken from 41. So the average is 38.
4. The numbers can be leveled out to become 80 each.
5. You can calculate the average by calculating
3+4+4
2
=3
3
3
or calculating
2
7·3+7·4+7·4
=3
7·3
3
Class Activity 15M: The Average as “Balance Point”
1. For each number in the list, students should plot one dot above that number
on the axis provided for the dot plot.
2. Be on the lookout for a common error in calculating the mean that is discussed
in the upcoming Class Activity on Errors with the Median and the Mean.
For the first dot plot, the mean is 4.
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For the second dot plot, the mean is 2.8. You might want to begin discuss that
without the 10, the mean would be 2, but that data that are far from the mean
exert a significant pull on the mean. In contrast, the median will not exert such
a strong pull.
For the third dot plot the mean is 7.2.
Class Activity 15N: Same Median, Different Average
1. To keep the median 6, half the pennies must be at or above 6 and half must be
at or below 6. To make the average less than 6, move pennies down, keeping
the constraint just mentioned. For example, one of the pennies at 6 could be
moved to 3 and the pennies at 4 and 5 could also be moved to 3. The pennies
at 7 and 8 could be moved to 6.
2. For example, one of the pennies at 6 could be moved to 9 and the pennies at 7
and 8 could be moved to 9. The pennies at 3 and 4 could be moved to 6.
Class Activity 15O: Can More than Half Be Above Average?
1. For dot plot 1, the median is 8 (there are 25 dots and the 13th from the right or
the left lies over 8). But the mean must be less than 8 because without the data
at 1, 4, and 5, the dot plot would balance at 8, but those additional pieces of
data will pull the balance point to the left. So the mean is less than the median
in this case.
For dot plot 2, the median is 3 (once again, there are 25 dots and the 13th from
the left or right lies over 3). If the dots over 7, 9, and 10 were not there, and if
the dot at 6 was moved to 5, then the mean would be at 3 because the graph
would be perfectly symmetrical around 3. Including the data at 7, 9, and 10
and moving the dot back from 5 to 6 all make the balance point greater than
3. Therefore the mean is greater than the median in this case.
2. Yes, it is possible that 90% of the class scores above average. For example, if
the test is out of 10 points and 18 children score 10 but 2 children score 9, then
the average is less than 10 and 90% of the class scores above average. Reverse
the situation to create a situation where 90% of the class scores below average.
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3. If we only consider the average scores of all the children in the town on some
common test they took, then it’s not possible for all the children to be above
average. But if the children are compared to children nationwide, then it’s
possible that all the children are above average.
Class Activity 15P: Errors with the Mean and the Median
1. Eddie’s method is a common error. He viewed each stack of dots as a piece
of data (like the towers in the activity on “leveling out” to find the mean).
However, each dot represents a piece of data that consists of the number the
dot lies over. So the 4 dots over 2 do not represent the datum 4 but rather
4 pieces of data, each of which is 2. Note that Eddie does come out with the
correct answer, but not for the right reason! You could use this phenomenon
to make the point that teachers need to look for reasoning, not just correct
answers to problems.
2. In the case of error 1, the student forgot to put the data in order before determining the median.
Error 2 occurred because the student applied the median to categorical data,
which doesn’t make sense because the categories could be put in a different
order, resulting in a different “median.”
In the 3rd case, the student found the median of the frequency counts. But
the median of these frequency counts is not appropriately applied to deduce a
median for the categorical data on votes for types of trees.
15.4
Summarizing, Describing, and Comparing Data
Distributions
Class Activity 15Q: What Does the Shape of a Data Distribution Tell Us About the Data?
1. One example of a “read the data” question is: what percent of the households
have an income of between $40,000 and $49,999? This income level is represented in the 5th bar from the left and is 9% of the population in the U.S. 5%
in hypothetical country A, and 3% in hypothetical country B.
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An example of a “read between the data” question is: what percent of the
households had incomes under $50,000? In the U.S. it is 49%, in hypothetical
country A it is 53%, and in hypothetical country B it is 7%.
2. You can tell how the data are distributed across the different incomes, which
you would not be able to tell from the means or medians alone.
3. The U.S. could claim that it has fewer poor people than country A and more
very wealthy people than country B. Country A could claim that it has more
wealthy people than the U.S. and country B. Country C could claim that it has
fewer poor people than the other two countries and that it has a higher mean
income than the other two countries.
Class Activity 15R: Distributions of Random Samples
1, 2, &3 As the number of samples increases one should expect the dot plot to approach
a normal bell-shaped curve centered over the 40% mark.
4 Both bar graphs are centered about 40%, but the bar graph for the smaller
sample size is much more spread out and much less resembles a bell-shaped
curve than the bar graph for the larger sample size.
5 One should expect the bar graph for the sample size of 200 to be less spread
out than the one for 100, but more spread out than the one for the sample size
of 1000. Also, the bar graph for the sample size of 200 should more closely
resemble a bell-shaped curve than the one for 100 but not as much as the one
for 1000.
Class Activity 15S: Comparing Distributions: Mercury in Fish
You can use this activity to motivate the need for ways of summarizing data that
include more than just the mean or median.
1. The mean of the samples for hypothetical fish A is around 0.88, whereas the
mean for fish B is around 0.84. But although the samples for fish A have a
higher mean, none go above the hazardous level of 1.00. On the other hand,
some of the samples for fish B do go above the hazardous level. The amount of
mercury found in fish B varies much more than it does for fish A.
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2. If we were only given the means, we wouldn’t know about the variability and
whether some of the samples contained hazardous amounts of mercury.
Class Activity 15T: Using Medians and Interquartile Ranges
to Compare Data
Notes: Here we focus on determining the value of a given percentile but students may
wonder how a percentile is determined for a given test score (e.g., when they receive
their score on a test as a percentile). After all, if you examine the second data set
in this activity, the 75th percentile is 9, but so is the 80th percentile as well as the
70th percentile. So which percentile is a student who scored a 9 at? A typical way to
determine this percentile is to take the percentile in the middle of all the scores that
are the same. Note also that there are different definitions of percentile in common
use. Also, in order to understand the concept of percentile, we work with small data
sets here, but in general, percentiles are most useful for large data sets.
1. First dot plot: 25th percentile: 4.5;
50th percentile: 6;
75th percentile: 7.5.
Second dot plot: 25th percentile: 5.5;
50th percentile: 7;
75th percentile: 9;
Third dot plot: 25th percentile: 2;
50th percentile: 6.5;
75th percentile: 9.
2. Knowing the percentiles for the first dot plot we could say that
(a) 25% of the data values are ≤ 4.5 and 75% are ≥ 4.5.
(b) 50% of the data values are ≤ 6 and 75% are ≥ 6.
(c) 75% of the data values are ≤ 7.5 and 75% are ≥ 7.5.
Similiar statements can be made for the other dot plots.
In a more impressionistic vein, we can tell from the percentile information that
most of the data in the second dot plot are higher than in the other two data
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sets, that the third data set is more widely dispersed than the other two. We
can’t tell the exact shape of the data sets from the percentile information alone,
but the percentile information does indicate that the second data set could be
skewed and that the third data set might be bimodal.
Class Activity 15U: Using Box Plots to Compare Data
If you wish to discuss the interquartile range, you might like to do so at this point.
The interquartile range is the difference between the 75th and 25th percentiles and
gives us a sense of how widely dispersed the data are. In a box plot, the interquartile
range is the distance between the right and left endpoints of the box. The greater
the interquartile range, the more widely dispersed the middle 50% of the data are.
1. See Figure 15.1.
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
Figure 15.1: Box Plots
2. The first data set is more concentrated around its median than the the second
or third, and the second is more concentrated around its median than the third.
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Class Activity 15V: Percentiles Versus Percent Correct
1. If a student scores at the 90th percentile on a test then that student has scored
better than or the same as 90% of other students taking the test. This does
not tell us how many questions the student answered correctly on the test, only
how the student did in comparison to other students. If the student gets 90%
correct on a test then that does not tell us anything about how the student did
in comparison to other students taking the test.
Class Activity 15W: Comparing Paper Airplanes
The data in the dot plots are from actual paper airplanes. The first paper airplane
was an ordinary one made just out of paper. The second paper airplane was the same
but with a paperclip attached to its nose. The third paper airplane was the same as
the first two but with a paperclip attached to both the nose and the tail.
If you have time, you might get students to bring in their own paper airplane
models and try them out.
1. For airplane 1 the MAD is 2. For airplane 2 the MAD is 1.45. For airplane 3
the MAD is 1.
2. Although airplane 1 flew the farthest in a single trial, most people would probably conclude that airplane 3 is the best flyer overall because it has the highest
mean distance and is most consistent, as shown by its MAD being the smaller
than the others.
Notice also that the difference in the means (about 1 or 2 feet) is roughly the
same size as the MADs, so this indicates that there is a real difference in the
flying capacity of the paper airplanes. Had the means differed by much less
than the MADs, we would be less confident in saying that the plane with the
higher mean is a better flyer.
3. Based on the means, Airplane 4 flies farther than Airplane 2 but not as far
as Airplane 3. We can be pretty confident that Airplane 4 flies farther than
Airplane 2 because the difference in the means are roughly around the same
size as the average variability as expressed by the MAD. But when it comes
to comparing Airplane 4 with Airplane 3 we should be less confident in saying
that Airplane 3 is not as good a flyer because the difference in means is small
compared to the variability (as described by the MADs).
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Class Activity 15X: Using Dot Paper to Estimate Area
Answers will vary for part 1. In parts 2 and 3, students should see that they may not
be able to determine differences in the area based on their data. If two data sets have
different means, but if the difference in the means is small compared to the amount of
variability, then one can’t say confidently that the data indicate an actual difference.
On the other hand, if two data sets have different means and the difference in the
means is more than the MAD (or so), then one has some confidence that the data
indicate an actual difference. See CCSS Standard 7.SP.3. In part 4, students should
be able to use their data to determine that the heart-shaped leaf does have a smaller
area than the circle or oval.
Probability
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16.1
Basic Principles of Probability
Class Activity 16A: Probabilities with Spinners
1. The paperclips are equally likely to land in the shaded area of each spinner
because the shaded regions occupy the same area of each circle. (Actually, it’s
not so much area that’s relevant but rather the angles.)
2. The paperclips are equally likely to land in the shaded area of each spinner
because the central angle covered by the shaded portions are the same in each
case.
3. The red portion could be made as 1 piece. Then the green portion will consist
of 2 pieces of that size, and the yellow portion will also consist of 2 pieces of that
size. The blue will need to consist of more than 2 pieces of that size. If the blue
is made of 3 pieces, then there will be 8 pieces altogether and the probabilities
will be 81 , 28 = 14 , 41 , and 38 . But the blue could be made of 4 pieces, for example,
and then each piece would be one ninth.
Class Activity 16B: Some Probability Misconceptions
You can use part 3 to begin discussing empirical (experimental) probability and to
lead into the next activities.
1. No, the problem is that the two outcomes are not equally likely.
4
2. The probability of picking a red block is 26 = 13 from bag 1 but only 20
= 51 from
bag 2, so Eva is more likely to win if she picks from bag 1 even though bag 2
has more red blocks.
3. No, but you will probably win close to 3 times out of 1000 tries. The probability
3
stands for the portion of times you should win the game “in the ideal” as
1000
opposed to in actuality. Moreover you can expect that as the number of tries
gets larger and larger the ratio of wins to tries should get closer and closer to
3
.
1000
If students still think that you will win 3 times, ask them what the probability
of flipping heads on a coin is. It is 12 . Does this mean that if you flip a coin twice
you will get 1 head? In this case, they know they may or may not. Point out
3
that the case where the probability is 1000
and there are 1000 tries is similar.
269
Class Activity 16C: Empirical Versus Theoretical Probability:
Picking Cubes From a Bag
You can use this activity to reinforce the fact that an empirical probability is often
not exactly equal to the theoretical probability although it’s usually close.
3. Put the number of possible reds (or blues) picked from 0 to 10 along the axis
of the dot plot. The data should cluster around 3 reds (or 7 blues).
Class Activity 16D: Using Empirical Probability to Make Predictions
If you have a bag and bears available you might try the activity with the class.
The percentage of yellow bears picked is the empirical probability of picking a
yellow bear. This empirical probability is our best estimate for the theoretical probability. The theoretical probability of picking a yellow bear is the fraction of yellow
4
, or 40%. So students can use the empirical probability,
bears in the box, which is 10
rounded to the nearest ten percent to make a good guess for the number of bears. For
example, if the empirical probability turns out to be 34%, students would figure that
3 out of the 10 bears in the bag are yellow. We expect the empirical probabilities to
get closer and closer to 40% as the night goes on.
One modification might be to keep a running list of the fraction of yellow bears
picked so far and to write these fractions as percentages. Another modification would
be to put a different number of bears in the bag, such as 100.
Class Activity 16E: If You Flip 10 Pennies, Should Half Come
Up Heads?
1. Many students will probably guess that the probability of getting 5 heads is 21 .
2. A dot plot, with the number of possible heads from 0 to 10 listed along the axis,
is a good way to display the data.
3. The theoretical probability of getting exactly 5 heads from 10 coins can be
calculated using a binomial coefficient. (There are 10 choose 5 ways to get
exactly 5 heads out of 210 possibilities.) The probability works out to be about
24.6%.
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One way to interpret this result is that if we flip a coin 10 times, then we are
actually unlikely to get exactly 5 heads (even though we are likely to get close
to 5 heads). Once again, this indicates that the experimental probability is
usually not exactly equal to the theoretical probability, although it is usually
close.
16.2
Counting the Number of Outcomes
This section briefly reviews ordered pair problems from Section 4.1, but it’s best if
students have previously studied that section.
Class Activity 16F: How Many Keys Are There?
1. For each of the 3 possible depths of the first notch there are 3 possible depths for
the second notch. This makes 3 × 3 = 9 possible outcomes. Some students may
be tempted to say the number of keys is 2 × 3. They should make an organized
list to see that it really is 9, not 6. Note that if a student counts the keys as
“notch 1 is D M or S, notch 2 is D M or S” this doesn’t actually describe any
keys. Each key that is made has to have a description for how both the first
and second notch are notched.
3. There are 3 × 3 × 3 × 3 = 81 possible keys that can be made.
4. There are 510 = 9, 765, 625 keys that can be made.
Class Activity 16G: Counting Outcomes: Independent Versus
Dependent
1. There are 4 × 4 × 4 = 64 codes that can be made.
2. The first letter can be any of the 4 letters. Once the first letter is chosen, only 3
letters remain for the second slot. Then once the first two slots are filled, only
2 letters remain for the third slot. So there are 4 × 3 × 2 = 24 codes.
16.3
Calculating Probabilities in Multi-Stage Experiments
271
Class Activity 16H: Number Cube Rolling Game
1. The numbers around 7 will probably come up more often than other numbers.
2. See practice problem 2 and its answer.
There are 6 outcomes for which the total number of dots is 7. There are 3
outcomes for which the total number of dots is 10. There is 1 way 12 can occur.
There are 3 ways the total of 4 can occur.
6
= 61 .
36
3
1
10 is 36
= 12
.
1
12 is 36
.
1
3
= 12
.
4 is 36
The probability of rolling a total of 7 is
The probability of rolling a total of
Class Activity 16I: Picking Two Marbles From a Bag of 1
Black and 3 Red Marbles
1. Many students guess that the probability is
black.
1
4
because 1 out of 4 marbles is
Another incorrect guess is 31 . Here is the reasoning a student once proposed:
When you pulled out 2 marbles, one has to be red. There are 3 other marbles
to choose from and 1 of them is black, so the probability of getting the black
marble is 13 .
I recommend not discussing which guesses are correct and which aren’t at this
point, but proceeding with the activity and then returning to their guesses at
the end of the activity.
3. Use the tree diagram in Figure 16.1.
(a) 12.
(b) 6.
(c) The probability is
6
12
= 12 .
4. By making all outcomes equally likely we can calculate the probability just by
seeing in how many out of the total possible outcomes a black marble is picked.
5. (a) There are
4·3
2
= 6 ways of picking a pair of marbles.
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first
pick
second
pick
R2
R1
R3
B
R1
R2
R3
B
R1
R3
R2
B
R1
B
R2
R3 1000 1pt 1pt
Figure 16.1: A Tree Diagram
273
(b) 3 pairs contain the black marble.
(c) Since all pairs of marbles are equally likely to be picked, the probability of
picking a pair that includes the black marble is 36 = 12 .
6. The difference between part 3 and part 5 is ordered pairs (part 3) versus unordered pairs (part 5). There are twice as many ordered pairs as unordered
pairs.
If time, go back to students’ guesses in Problem 1, discuss reasoning they may
have had for their guesses, and discuss why the incorrect ways of reasoning are
not correct.
Class Activity 16J: More Probability Misconceptions
1. The probability of flipping a head is always 21 no matter what has been flipped
before. The previous flips can’t influence the next flip in any way. So we say
that the next flip is independent of the previous flips.
2. No. The outcomes are not all equally likely. If one coin is a penny and the
other is a nickel, then you can get 2 heads, 2 tails, a head on the penny and
a tail on the nickel, or a tail on the penny and a head on the nickel. That’s 4
different equally likely outcomes, 2 of which have one head and one tail. So the
probability of one head and one tail is 24 = 12 .
Class Activity 16K: Expected Earnings from the Fall Festival
Technically, expected value is the average amount expected (in the ideal over the
long term). This activity, and the practice exercises and problems don’t work with
the average, but instead work with an expected total amount. I think the expected
total amount is easier to understand and work with and is a natural precursor to the
standard concept of expected value.
1. There are 10 green bear outcomes on the first draw and given one of these
outcomes there are 2 red bear outcomes on the 2nd draw for a total of 20 greenred outcomes on the 1st two draws. Overall there are 15 × 14 = 210 possible
outcomes for the 1st two draws. Thus the probability of a green-red outcome
20
on the 1st two draws = 210
= 9.52%. It follows that Ms. Wilkins should expect
to give out approximately 9.52% × 300 ≈ 29 prizes.
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2. The School should should take in 300×$.50 = $150 and should pay out 29×$2 =
$58. Thus the school should make a profit of $150 − $58 = $92.
16.4
Using Fraction Arithmetic to Calculate Probabilities
This section assumes that students have studied Section 5.1 on fraction multiplication
and Section 3.4 on adding and subtracting fractions.
Class Activity 16L: Using the Meaning of Fraction Multiplication to Calculate a Probability
If you do Problem 2, you could collect the whole class’s data to determine the experimental probability.
3. In the ideal, in half of the times Jill runs her experiment of spinning twice the
first spin will be blue.
In the ideal, in 14 of those times when the first spin is blue, the second spin will
be red. Therefore, in the ideal, in 41 of 12 of the times when Jill spins the spinner
twice, the outcome will be blue followed by red. Since 14 of 21 is 14 · 21 = 18 , the
probability of spinning blue followed by red in 2 spins is 18 . We can also see this
probability visually: the rectangle represents all the double spins (in the ideal).
First shade 12 of the rectangle. Then further shade 41 of 12 of the rectangle, which
is 18 of the rectangle.
Class Activity 16M: Using Fraction Multiplication and Addition to Calculate a Probability
If you do Problem 2, you could collect the whole class’s data to determine the experimental probability.
3. (a) In the ideal, the spin should be blue 21 of the time.
In the ideal, when the spin is blue the tile that is choosen should be blue 51
of the time. Therfore, in the ideal, in 15 of 12 of the time both the spinner
1
and the tile will be blue. Since 51 of 12 is 51 · 21 = 10
, the probability that
1
both the spinner and the tile are blue is 10 .
275
On the rectangle, first shade 21 of the rectangle. Then further shade
that 12 of the rectangle. See Figure 16.2.
1
5
of
(b) In the ideal, the spin should be red 14 of the time.
In the ideal, when the spin is red the tile that is choosen should be red 35
of the time. Therfore, in the ideal, in 35 of 14 of the time both the spinner
3
and the tile will be red. Since 53 of 14 is 35 · 14 = 20
, the probability that
3
both the spinner and the tile are red is 20 .
On the same rectangle as before, shade another (different than before) 14
of it. Then further shade 35 of this 41 .
(c) Since you win the game with blue-blue or red-red outcomes, and since
these outcomes cannot ocurr simultaneously, in the ideal, these outcomes
3
1
+ 20
= 41 of the time. Hence the probability of winning = 25%.
will ocurr 10
For a shaded rectangle that corresponds to these calculations see Figure 16.2.
Red
Blue
Figure 16.2: Probability as fraction of a rectangle
Copyright 276
CHAPTER
Copyright Appendix A
Blackline Masters
277
278
CHAPTER A
0
1
2
3
4
5
6
7
Long ticks:
whole numbers
Long ticks:
tenths
Long ticks:
hundredths
Long ticks:
thousandths
Figure A.1: Zoom In For Class Activity 1F “Zooming In on Number Lines”
Copyright 8
SECTION A.0
279
square 1
square 1
square 2
square 2
square 3
square 4
square 3
square 4
Figure A.2: Squares For Class Activity 3S “What Fraction is Shaded?”
280
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40
CHAPTER A
1
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40
Figure A.3: Strips for introducing the LCM in Section 8.2
Copyright SECTION A.0
281
Figure A.4: Spirograph designs for Class Activity 8N on flower designs
Copyright 282
CHAPTER A
Expression 1:
Expression 2:
_____________________________
_____________________________
Your expression:
Explain this expression:
_____________________
10 • 8 - 2 • 3²
Explain this expression:
2 • (7 • 5) - 4 • 2
Figure A.5: Dot and Star Designs for Class Activity 9A on Expressions
283
Expression 1:
Expression 2:
_____________________
_____________________
Figure A.6: Dot and Stick Designs for Class Activity 9A on Expressions
Copyright 284
CHAPTER A
5
5
2
5
6
2
5
6
4
4
10 units
5
10 units
5
2
5
6
2
5
6
4
4
10 units
5
10 units
5
2
5
6
2
5
6
4
4
10 units
5
10 units
5
2
5
6
2
5
6
4
10 units
4
10 units
Figure A.7: L-shapes for Class Activity 12C on the moving and additivity principles
285
Figure A.8: A square in a square for Class Activity 12C on the moving and additivity
principles
Copyright 286
CHAPTER A
Figure A.9: To copy for Class Activity 12D on triangle areas
287
Figure A.10: To copy onto transparencies for Class Activity 12D on triangle areas
Copyright 288
CHAPTER A
Figure A.11: To copy onto transparencies for Class Activity 12D on triangle areas
289
b
h
h
b
b
h
h
b
b
h
h
b
b
h
h
b
Figure A.12: To copy for Class Activity 12F on the triangle area formula
Copyright CHAPTER A
b
b
h
h
290
b
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h
b
b
h
h
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b
b
h
h
Figure A.13: To copy for Class Activity 12F on the triangle area formula
291
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a
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h h
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h
a
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h h
c
b
a
Figure A.14: To copy for Class Activity 12I on the parallelogram area formula
h
292
CHAPTER A
a
a
h
h
h
h
b
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a
a
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a
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a
a
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h
h
b
Figure A.15: To copy for Class Activity 12J on the trapezoid area formula
SECTION A.0
293
294
CHAPTER A
Figure A.17: To copy for Class Activity 12O on the circle area formula
295
1 inch on map
1 mile
heavy grid lines are 1 inch apart
Figure A.18: To copy for Class Activity 12Q on the area of irregular shapes
Copyright 296
CHAPTER A
Figure A.19: To copy for Class Activity 12S on the perimeter versus area for rectangles
297
b
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c
a
b
c
Figure A.20: To copy for Class Activity 12V on the Pythagorean Theorem
Copyright 298
CHAPTER A
Figure A.21: To copy for making Platonic solids, such as for Class Activity 13C on
the Magic 8 Ball
299
the Magic 8 Ball
Copyright 300
CHAPTER A
the Magic 8 Ball
301
Figure A.24: Prism Patterns for Class Activity 13L on Comparing the Volumes of
Pyramids and Prisms
Copyright 302
CHAPTER A
Figure A.25: Prism Pattern for Class Activity 13L on Comparing the Volumes of
Pyramids and Prisms
303
Figure A.26: Pieces for 3 Oblique Pyramids For Class Activity 13M on the Volume
Formula for Pyramids
Copyright

I ’ R M

Transcription

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