Document 6528329

Transcription

Document 6528329

This document contains information relating to a transition to PARCC testing; however, at this time, there is
no plan to transition to PARCC at high school. Only End-of-Course (EOC) assessments will be administered for
high school courses. The multiple-choice and constructed-response items and annotations are still relevant.
Teachers are encouraged to use the sample items provided in this document as additional resources, but look
to the 2014-2015 Assessment Guidance Algebra I document for up-to-date EOC testing information.
SAMPLE TEST ITEMS
October 2013
Geometry
Louisiana State Board of Elementary and Secondary Education
Mr. Charles E. Roemer
Ms. Kira Orange Jones
Ms. Carolyn Hill
President
Second BESE District
Eighth BESE District
Ms. Lottie P. Beebe
Ms. Connie Bradford
Third BESE District
Member-at-Large
Mr. Walter Lee
Dr. Judy Miranti
Fourth BESE District
Member-at-Large
Secretary/Treasurer
Mr. Jay Guillot
Mr. Stephen Waguespack
Seventh BESE District
Fifth BESE District
Member-at-Large
Sixth BESE District
Mr. James D. Garvey, Jr.
Vice President
First BESE District
Ms. Holly Boffy
Ms. Heather Cope
Executive Director
The mission of the Louisiana Department of Education (LDOE) is to ensure equal access to education and to promote
equal excellence throughout the state. The LDOE is committed to providing Equal Employment Opportunities and is
committed to ensuring that all its programs and facilities are accessible to all members of the public. The LDOE does
not discriminate on the basis of age, color, disability, national origin, race, religion, sex, or genetic information. Inquiries
concerning the LDOE’s compliance with Title IX and other civil rights laws may be directed to the Attorney, LDOE, Office
of the General Counsel, P.O. Box 94064, Baton Rouge, LA 70804-9064; 877-453-2721 or customerservice@la.gov.
Information about the federal civil rights laws that apply to the LDOE and other educational institutions is available on
the website for the Office of Civil Rights, USDOE, at http://www.ed.gov/about/offices/list/ocr/.
This project is made possible through a grant awarded by the State Board of Elementary and Secondary Education
from the Louisiana Quality Education Support Fund—8(g).
This public document was published at a total cost of $6,000.00. This web-only document was published for
the Louisiana Department of Education, Office of Assessments, P.O. Box 94064, Baton Rouge, LA 70804-9064,
by Pacific Metrics Corporation, 1 Lower Ragsdale Drive, Building 1, Suite 150, Monterey, CA 93940. This material
was published in accordance with the standards for printing by state agencies established pursuant to R.S. 43:31
and in accordance with the provisions of Title 43 of the Louisiana Revised Statutes.
For further information, contact: Louisiana
Department of Education’s Help Desk
1-877-453-2721
Ask LDOE?
https://www.louisianabelieves.com/resources/ask-ldoe
© 2013 by Louisiana Department of Education
Table of Contents
Introduction .......................................................................................... 1
Purpose of This Document ....................................................................... 1
Geometry Administration ......................................................................... 1
EOC Achievement Levels ......................................................................... 2
EOC Achievement-Level Definitions ......................................................... 2
Geometry .............................................................................................. 3
Constructed-Response Rubric ................................................................. 3
Testing Materials and Online Tools .......................................................... 4
Multiple-Choice Items ........................................................................... 5
Constructed-Response Item ................................................................. 31
Constructed-Response Item ................................................................... 31
Scoring Information ............................................................................... 34
Sample Student Responses .................................................................... 35
Geometry Typing Help ........................................................................... 44
Geometry Reference Sheet ..................................................................... 46
i
Introduction
Louisiana Believes embraces the principle that all children can achieve at high
levels, as evidenced in Louisiana’s recent adoption of the Common Core State
Standards (CCSS). Louisiana Believes also promotes the idea that Louisiana’s
educators should be empowered to make decisions to support the success
of their students. In keeping with these values, the Department has created
documents with sample test items to help prepare teachers and students as
they transition to the CCSS. The documents reflect the State’s commitment to
consistent and rigorous assessments and provide educators and families with
clear information about expectations for student performance.
Purpose of This Document
Teachers are encouraged to use the sample items presented in this document
in a variety of ways to gauge student learning and to guide instruction and
development of classroom assessments and tasks. The document includes
multiple-choice and constructed-response items that exemplify how the
Common Core State Standards for Mathematics (CCSSM) will be assessed
on the End-of-Course (EOC) tests. A discussion of each item highlights the
knowledge and skills the item is intended to measure.
As Louisiana students and teachers transition to the CCSS and the Partnership
for Assessment of Readiness for College and Careers (PARCC) assessments, the
Geometry EOC assessment will include only items aligned to the CCSS.
As you review the items, it is important to remember that the sample items
included in this document represent only a portion of the body of knowledge
and skills measured by the EOC test.
Geometry Administration
The Geometry EOC test is administered to students who have completed one
of the following courses:
•
Geometry: course code 160323
•
Applied Geometry: course code 160332
Geometry EOC Sample Test Items—October 2013
1
EOC Achievement Levels
Student scores for the Geometry EOC test are reported at four achievement
levels: Excellent, Good, Fair, and Needs Improvement. General definitions of the
EOC achievement levels are shown below.
EOC Achievement-Level Definitions
Excellent: A student at this achievement level has demonstrated mastery
of course content beyond Good.
Good: A student at this achievement level has demonstrated mastery of
course content and is well prepared for the next level of coursework in the
subject area.
Fair: A student at this achievement level has demonstrated only the
fundamental knowledge and skills needed for the next level of coursework
in the subject area.
Needs Improvement: A student at this achievement level has not
demonstrated the fundamental knowledge and skills needed for the next level
of coursework in the subject area.
Because of the shift from grade-level expectations to the CCSS, this document
differs from the Released Test Item Documents. Many of the released items from
past test administrations may not be indicative of the types of items on the
upcoming December and May EOC transitional assessments. To better align
the transitional test to the content of the CCSS, new items were developed.
Therefore, this document includes sample items, rather than released
items. These sample items reflect the way the CCSS will be assessed and
represent the new items that students will encounter on the transitional EOC
assessments. Because these are not released items, item-specific information
about achievement levels is not included.
2
Geometry
The Geometry EOC test contains forty-six multiple-choice items and one
constructed-response item. In addition, some field test items are embedded.
Multiple-choice items assess knowledge, conceptual understanding, and
application of skills. They consist of an interrogatory stem followed by four
answer options and are scored as correct or incorrect.
Constructed-response items require students to compose an answer, and these
items generally require higher-order thinking. A typical constructed-response
item may require students to develop an idea, demonstrate a problem-solving
strategy, or justify an answer based on reasoning or evidence. The Geometry
constructed-response item is scored on a scale of 0 to 4 points. The general
constructed-response rubric, shown below, provides descriptors for each
score point.
Constructed-Response Rubric
Score
Description
•
4
The student’s response demonstrates in-depth understanding of
the relevant content and/or procedures.
• The student completes all important components of the task
accurately and communicates ideas effectively.
• Where appropriate, the student offers insightful interpretations
and/or extensions.
• Where appropriate, the student uses more sophisticated
reasoning and/or efficient procedures.
•
3
The student completes the most important aspects of the task
accurately and communicates clearly.
• The student’s response demonstrates an understanding of major
concepts and/or processes, although less important ideas or
details may be overlooked or misunderstood.
• The student’s logic and reasoning may contain minor flaws.
2
• The student completes some parts of the task successfully.
• The student’s response demonstrates gaps in conceptual
understanding.
1
0
•
The student completes only a small portion of the task and/or
shows minimal understanding of the concepts and/or processes.
•
The student’s response is incorrect, irrelevant, too brief to
evaluate, or blank.
3
Testing Materials and Online Tools
Students taking the Geometry EOC test have access to a number of resources.
Scratch paper, graph paper, and pencils are provided by test administrators and
can be used by students during all three sessions of the Geometry EOC test.
There are also buttons at the top of the screen that a student may click to open
the online tools. The list below identifies the online tools available for each
session. Please note that the rulers and/or protractor may not be available for
some items.
Session 1
•
Geometry Reference Sheet
•
protractor
•
inch ruler
•
centimeter ruler
Note: Students are NOT allowed to use calculators during session 1 unless
they have the approved accommodation Assistive Technology and are
allowed the use of a calculator.
Sessions 2 and 3
•
•
calculator
•
protractor
•
inch ruler
•
centimeter ruler
Also available in session 2, which contains the constructed-response item,
is the Geometry Typing Help (see pages 44 and 45). This online tool describes
how to enter special characters, symbols, and formatting into typed responses.
The graph paper, Geometry Reference Sheet, Geometry Typing Help, and
EOC Tests online calculator can be found on the EOC Tests homepage at
www.louisianaeoc.org under Test Coordinator Materials: Testing Materials.
4
Multiple-Choice Items
This section presents ten multiple-choice items selected to illustrate the type
of skills and knowledge students need in order to demonstrate understanding
of the CCSS in the Geometry course. These items also represent the skills
students need in order to meet performance expectations for Math Practices.
Information shown for each item includes
•
conceptual category,
•
domain,
•
cluster,
•
standard,
•
Math Practices,
•
calculator designation (calculator allowed or calculator not allowed),
•
tool restrictions,
•
correct answer,
•
commentary on the skills and knowledge associated with the standard
measured by the item,
•
commentary on the Math Practices linked with the item,
•
commentary on why the correct answer is correct including, in some cases,
how the answer is achieved, and
•
commentary on each answer choice, explaining why it is correct or
incorrect.
5
Conceptual Category: G—Geometry
Domain:
CO—Congruence
Cluster:
D—Make geometric constructions
Standard:
12—Make formal geometric constructions with a
variety of tools and methods (compass and
straightedge, string, reflective devices, paper
folding, dynamic geometric software, etc.).
Copying a segment; copying an angle; bisecting
a segment; bisecting an angle; constructing
perpendicular lines, including the perpendicular
bisector of a line segment; and constructing a
line parallel to a given line through a point not on
the line.
Math Practices:
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
Calculator:
Calculator allowed
Tool Restrictions:
None
6
Melanie wants to construct the perpendicular bisector of line segment AB
using a compass and straightedge.
B
A
Which diagram shows the first step(s) of the construction?
A.
B.
A
B
A
B
C.
D.
A
A
B
B
*correct answer
7
This item requires students to identify the initial steps of constructing a
perpendicular bisector. Constructions are typically difficult to assess in a
multiple-choice format. However, asking students to analyze which figure will
result in the given construction assesses reasoning about constructions without
the need for providing geometric tools during testing or having to manually score
each response.
This item is linked to three of the Math Practices.
•
Math Practice 5 (Use appropriate tools strategically.): Students must recognize
how to effectively use a compass to begin constructing a perpendicular
bisector of a line segment.
•
Math Practice 6 (Attend to precision.): Students must recognize that only
one method shown guarantees a precise construction.
• Math Practice 7 (Look for and make use of structure.): Students must use
the structure of the two congruent circles and recognize that the circles
intersect at points that are equidistant from the endpoints of the given
line segment. These points may be connected to construct a perpendicular
bisector.
Option D: This is the correct answer. The figure shows two circles that appear
to be congruent centered at the endpoints of the given line segment.
Connecting the intersection points of these two circles guarantees the bisecting
line created is perpendicular (at a 90 degree angle) to segment AB.
Option A: The student incorrectly draws two circles that appear to be congruent
and intersect at what appears to be the midpoint of segment AB; however, the
figure does not show how a perpendicular bisector could be constructed, as it
does not guarantee that a line drawn would intersect at a 90 degree angle.
Option B: The student incorrectly draws one circle that appears to be centered
at the midpoint of segment AB. The midpoint is where a bisector will intersect,
but this figure does not show how to ensure the bisector will be perpendicular.
Option C: The student incorrectly draws two circles with different radii centered
at the endpoints of the given line segment. Connecting the intersection points
of these two circles will result in a line perpendicular to segment AB that does
not bisect it.
8
Domain:
CO—Congruence
Cluster:
C—Prove geometric theorems
Standard:
11—Prove theorems about parallelograms.
Theorems include: opposite sides are congruent,
opposite angles are congruent, the diagonals of a
parallelogram bisect each other, and conversely,
rectangles are parallelograms with congruent
diagonals.
Math Practices:
1. Make sense of problems and persevere in
solving them.
3. Construct viable arguments and critique the
reasoning of others.
Calculator:
Calculator allowed
Tool Restrictions:
None
9
Use the figure to answer the question.
B
A
C
D
Missy is proving the theorem that states that opposite sides of a
parallelogram are congruent.
Given: Quadrilateral ABCD is a parallelogram.
Prove: AB ≅ CD and BC ≅ DA
Missy’s incomplete proof is shown.
Statement
Reason
1. Quadrilateral ABCD is a
parallelogram.
2. AB || CD; BC || DA
1. given
3.
?
3.
4.
AC ≅ AC
4. reflexive property
5.
∆ ABC ≅ ∆CDA
6.
AB ≅ CD and BC ≅ DA
5. angle-side-angle
congruence postulate
6. Corresponding parts of
congruent triangles are
congruent (CPCTC).
2. definition of parallelogram
?
Which statement and reason should Missy insert into the chart as step 3
to complete the proof?
A.
BD ≅ BD; reflexive property
B.
AB ≅ CD and BC ≅ DA ; reflexive property
C.
∠ABD ≅ ∠CDB and ∠ADB ≅ ∠CBD; When parallel lines are cut
by a transversal, alternate interior angles are congruent.
∠BAC ≅ ∠DCA and ∠BCA ≅ ∠DAC; When parallel lines are cut
by a transversal, alternate interior angles are congruent.
D.
*correct answer
10
This item requires students to provide the missing statement and reason
in a two-column proof of the theorem that says that the opposite sides of a
parallelogram are congruent. Assessing student’s abilities to write proofs in a
multiple-choice item is difficult. This item allows students to demonstrate their
reasoning skills by asking for one of the middle steps in the proof, including
both the statement and reason.
•
Math Practice 1 (Make sense of problems and persevere in solving them.):
Students must analyze the given information in the item and the constraints
of the details provided in the proof, and understand the goals of the problem.
•
Math Practice 3 (Construct viable arguments and critique the reasoning of
others.): Students must complete the logical progression presented in the
proof.
Option D: This is the correct answer. Students are given a set of corresponding
congruent sides in statement and reason 4. This leads into proving triangle
congruence using the angle-side-angle congruence postulate in statement
and reason 5. In order to reach statement and reason 5, students need to
provide two pairs of corresponding congruent angles, which include the given
corresponding congruent sides in statement and reason 4.
Option A: This answer is a true statement with a correct reason for the
statement based on the given information; however, it is not relevant to
completing the proof because it does not support statement and reason 5.
Option B: The student uses the final result of the proof as the statement and
provides an incorrect reason for the statement.
Option C: This response is also a true statement with a correct reason, but it
is not relevant to the given proof because it does not support statement and
reason 5.
11
Domain:
SRT—Similarity, Right Triangles, and Trigonometry
Cluster:
A—Understand similarity in terms of similarity
transformations
Standard:
1—Verify experimentally the properties of dilations
given by a center and a scale factor:
a. A dilation takes a line not passing through the
center of the dilation to a parallel line, and leaves
a line passing through the center unchanged.
Math Practices:
Calculator:
Calculator not allowed
Tool Restrictions:
None
12
Rosa graphs the line y = 3x + 5. Then she dilates the line by
a factor of 1 with (0, 7) as the center of dilation.
5
y
10
8
6
2
10
8
6
4
4
6
8
10
x
2
4
6
8
10
Which statement best describes the result of the dilation?
A.
1
The result is a different line the size of the
5
original line.
B.
The result is a different line with a slope of 3.
C.
1
The result is a different line with a slope of – .
3
D.
The result is the same line.
*correct answer
This item requires students to evaluate the effect of a dilation of a line not
passing through the center of the dilation. Specifically, students must recognize
that the dilated line will have the same slope as the original line.
This item is linked to one of the Math Practices.
•
Math Practice 7 (Look for and make use of structure.): Students must
recognize that the effect of the described dilation would not change the
slope of the line, and that it would not be the same line either.
13
Option B: This is the correct answer. The result of a dilation of a line not
passing through the center of the dilation is a line parallel to the original line.
Therefore, the slope will remain the same.
Option A: The student confuses the line with line segment. A dilation of a line
segment with a scale factor of 1/5 would result in a smaller line segment;
however, a dilation of a line results in a line.
Option C: The student incorrectly assumes that the result of the dilation is a
line that is perpendicular to the original line.
Option D: The student incorrectly assumes that the result of the dilation is the
original line; however, this is only true of lines that pass through the center of
the dilation.
14
Domain:
Cluster:
A—Understand similarity in terms of similarity
transformations
Standard:
2—Given two figures, use the definition of similarity
in terms of similarity transformations to decide
if they are similar; explain using similarity
transformations the meaning of similarity for
triangles as the equality of all corresponding
pairs of angles and the proportionality of all
corresponding pairs of sides.
Math Practices:
solving them.
Calculator:
Calculator allowed
Tool Restrictions:
None
15
Use the graph to answer the question.
y
10
P
B'
8
B
6
4
C'
A'
10
8
6
4
4
6
8
10
x
2
4
6
A
8
C
10
Kelly dilates triangle ABC using point P as the center of dilation and creates
triangle A'B'C'.
By comparing the slopes of AC and CB and A'C' and C'B', Kelly found that
∠ACB and ∠A'C'B' are right angles.
Which set of calculations could Kelly use to prove ∆ ABC is similar to ∆ A'B'C'?
A.
slope AB =
7 − (−7) 14
=
=2
2 − (−5)
7
slope Aꞌ Bꞌ =
B.
7−3
4
= =2
−3 − (−5) 2
AB2 = 72 + 142
AꞌBꞌ2 = 22 + 42
C.
tan ABC =
AC 7
=
BC 14
tan Aꞌ Bꞌ Cꞌ =
D.
Aꞌ C ꞌ 2
=
Bꞌ Cꞌ 4
∠ABC + ∠BCA + ∠CAB = 180
∠Aꞌ Bꞌ C ꞌ + ∠Bꞌ C ꞌAꞌ + ∠Cꞌ Aꞌ Bꞌ = 180
*correct answer
16
This item requires students to decide which evidence is sufficient for proving
that the given triangles are similar. Providing the calculations in the answer
choices eliminates the possibility of a student missing the item due to an
arithmetic error. This allows the item to focus on assessing the student’s
reasoning in determining the AA criterion for triangle similarity.
This item is linked to two of the Math Practices.
•
of the details provided, and understand the goals of the problem.
•
Math Practice 3 (Construct viable arguments and critique the reasoning
of others.): Students must determine which calculation is sufficient for
constructing a proof showing the given triangles are similar.
Option C: This is the correct answer. The tangents of the corresponding
angles ABC and A'B'C' are equal. This shows that the corresponding angles
are congruent and the corresponding sides have the same constant of
proportionality.
Option A: The student incorrectly assumes that congruent corresponding
slopes in the triangles imply that corresponding angles are congruent.
Using the slopes is a possible method, but additional steps are needed
to prove that the triangles are similar.
Option B: The student incorrectly selects a set of calculations that help
determine one pair of corresponding side lengths. This is not sufficient for
proving triangle similarity.
Option D: The student incorrectly assumes that because all triangles have
3 angles whose measures sum to 180 degrees, the triangles are similar.
17
Domain:
Cluster:
B—Prove theorems involving similarity
Standard:
5—Use congruence and similarity criteria for
triangles to solve problems and to prove
relationships in geometric figures.
Math Practices:
solving them.
Calculator:
Calculator allowed
Tool Restrictions:
None
Hector knows two angles in triangle A are congruent to two
angles in triangle B. What else does Hector need to know
to prove that triangles A and B are similar?
A.
Hector does not need to know anything else about
triangles A and B.
B.
Hector needs to know the length of any
corresponding side in both triangles.
C.
Hector needs to know all three angles in triangle
A are congruent to the corresponding angles in
triangle B.
D.
Hector needs to know the length of the side
between the corresponding angles on each triangle.
*correct answer
This item requires students to determine which information is needed to
complete an argument that two triangles are similar. Specifically, students
must recognize that two pairs of corresponding congruent angles are sufficient
for proving two triangles are similar.
•
18
•
others.): Students must identify which additional piece of information,
if any, is needed to construct an argument that establishes the similarity
of the given triangles.
Option A: This is the correct answer. Two pairs of corresponding congruent
angles are sufficient for proving two triangles are similar. Since the sum of the
interior angles of all triangles is 180°, the third corresponding pair of angles of
the two triangles must be the same measure.
Option B: The student incorrectly assumes that information about the
corresponding sides of two triangles is necessary for proving similarity.
Option C: The student incorrectly assumes that three pairs of corresponding
congruent angles are necessary to prove two triangles are similar, which is not
true because if two angles are known, then the third can be determined.
Option D: The student incorrectly uses the angle-side-angle (ASA) criterion for
congruence instead of AA criterion for similarity.
19
Domain:
C—Circles
Cluster:
A—Understand and apply theorems about circles
Standard:
2—Identify and describe relationships among
inscribed angles, radii, and chords. Include the
relationship between central, inscribed, and
circumscribed angles; inscribed angles on a
diameter are right angles; the radius of a circle
is perpendicular to the tangent where the radius
intersects the circle.
Math Practices:
solving them.
Calculator:
Calculator allowed
Tool Restrictions:
None
Use the figure to answer the question.
T
S
Q
R
Triangle STR is drawn such that segment ST is tangent to
circle Q at point T, and segment SR is tangent to circle Q at
point R. If given any triangle STR with these conditions, which
statement must be true?
A.
Side TR could pass through point Q.
B.
Angle S is always smaller than angles T and R.
C.
Triangle STR is always an isosceles triangle.
D.
Triangle STR can never be a right triangle.
*correct answer
20
This item requires students to identify properties of a triangle constructed using
a chord of a circle and two line segments tangent to the same circle. Specifically,
students must recognize that tangent segments to a circle from the same
external point are congruent.
•
•
others.): Students must determine which statement may be supported by
properties that exist based on the construction of the given triangle.
Option C: This is the correct answer. Tangent segments drawn from the same
fixed point outside of the circle to points of tangency on the circumference of
the circle are congruent. Therefore, the triangle is isosceles.
Option A: The student incorrectly assumes that a triangle could be created if
side TR passed through the center of the circle; however, if this were possible,
then the lines tangent to the circle at T and R would be perpendicular to
segment TR. Thus, a triangle could not be created.
Option B: The student incorrectly assumes that angle S must always be smaller
than angle T and angle R; however, as points T and R move closer to each other
on the circumference of the circle, point S moves closer to the circle. This
eventually allows angle S to grow larger than angle T and angle R.
Option D: The student incorrectly assumes that triangle STR cannot be a
right triangle. It is true that angles T and R can never be right angles, but it is
possible for angle S to be equal to 90°.
21
Domain:
GPE—Expressing Geometric Properties with
Equations
Cluster:
A—Translate between the geometric description and
the equation for a conic section
Standard:
1—Derive the equation of a circle of given center and
radius using the Pythagorean Theorem; complete
the square to find the center and radius of a
circle given by an equation.
Math Practices:
Calculator:
Calculator not allowed
Tool Restrictions:
None
A circle has this equation.
x 2 + y 2 + 4x – 10y = 7
What are the center and radius of the circle?
A.
center: (2, –5)
radius: 6
B.
center: ( –2, 5)
radius: 6
C.
center: (2, –5)
radius: 36
D.
center: ( –2, 5)
radius: 36
*correct answer
This item requires students to use the method of completing the square in
order to determine the center and radius of a circle given the equation.
•
Math Practice 7 (Look for and make use of structure.): Students must
complete the square for the equation to create an equivalent equation of the
form (x – h )2 + (y – k )2 = r 2. Then, they must use this structure to determine
that the center is (h, k ) and the radius is r.
22
Option B: This is the correct answer. Completing the square in the original
equation results in (x + 2)2 + (y – 5)2 = 62. Therefore, the center of the circle is
the point (–2, 5) and the radius is 6 units.
Given
Equation
x 2 + y 2 + 4x – 10y = 7
Step 1
x + 4x + y – 10y = 7
2
(x 2 + 4x +
2
Step 2
)= 7
+ 4
+4
+25 +25
2
2
(x + 4x + 4) + (y – 10y + 25) = 36
Step 3
(x + 2)2 + (y – 5)2 = 36
Step 4
Step 5
) + (y 2 – 10y +
(x – h )2 + (y – k )2 = r 2
(x – (–2))2 + (y – 5)2 = 62
h = –2
k =5 r= 6
36 = 6
Rearrange x-terms
together and y-terms
together.
Complete the square
for the expression
x 2 + 4x by adding 4 to
both sides because
4 2
= 4.
2
Complete the square
for the expression
y 2 –10y by adding 25
to both sides because
10 2
–
= 25.
2
Convert to factored
form.
Find center at (h, k ).
Find the radius.
Option A: This response results from a sign error when finding the center of
the circle in step 4.
Option C: This response is the result of a sign error when finding the center
of the circle in step 4, and not taking the square root of the constant term to
determine the radius in step 5.
Option D: This response results from not taking the square root of the constant
term to determine the radius in step 5.
23
Domain:
GMD—Geometric Measurement and Dimension
Cluster:
A—Explain volume formulas and use them to solve
problems
Standard:
3—Use volume formulas for cylinders, pyramids,
cones, and spheres to solve problems.
Math Practices:
4. Model with mathematics.
Calculator:
Calculator allowed
Tool Restrictions:
None
Use the pyramid to answer the question.
8 in.
6 in.
This right pentagonal pyramid has a height of 8 inches and a
base area of 61.94 square inches. To the nearest hundredth,
what is the volume of the pyramid?
A.
80.00 cubic inches
B.
165.17 cubic inches
C.
240.00 cubic inches
D.
495.52 cubic inches
*correct answer
24
This item requires students to calculate the volume of a right pentagonal
pyramid given the area of the base.
•
Math Practice 4 (Model with mathematics): Students must identify the
correct formula to calculate volume of a pentagonal pyramid. They must
apply the given information.
Option B: This is the correct answer. The formula for the volume of a
1
pyramid is V = Bh, where B is the area of the base and h is the height.
3
1
For this pyramid, V = (61.94)(8) = 165.17 cubic inches.
3
1
by the product of the
3
height, the given side length, and the number of sides of the base.
Option A: This response is the result of multiplying
Option C: This response results from computing the product of the height,
the given side length, and the number of sides of the base.
Option D: This answer results from failing to multiply
area of the base and the height.
1
by the product of the
3
25
Domain:
MG—Modeling with Geometry
Cluster:
A—Apply geometric concepts in modeling situations
Standard:
2—Apply concepts of density based on area and
volume in modeling situations (e.g., persons per
square mile, BTUs per cubic foot).
Math Practices:
solving them.
2. Reason abstractly and quantitatively.
Calculator:
Calculator allowed
Tool Restrictions:
None
Use the diagram to answer the question.
9 yards
12 yards
Aviary
5 yards
An aviary is an enclosure for keeping birds. There are 134 birds
in the aviary shown in the diagram.
What is the number of birds per cubic yard for this aviary?
Round your answer to the nearest hundredth.
A.
0.19 birds per cubic yard
B.
C.
D.
*correct answer
26
This item requires students to compute the volume of a rectangular prism.
Then, students must use this value to determine the density of birds in an
aviary.
•
Students must make sense of the given quantities and recognize that they
must first compute the volume to then find the quotient of the number of
birds and the total volume of the aviary.
•
Math Practice 2 (Reason abstractly and quantitatively.): Students must
decontextualize the given information and represent it in a logical
mathematical format. Then, they must manipulate this representation
and contextualize their results to answer the question.
•
Math Practice 4 (Model with mathematics.): Students must model the given
real-world situation using the equation for volume and the process to
determine density.
Option B: This is the correct answer. The volume of the aviary is given by the
product of 5 yards, 12 yards, and 9 yards. The density of birds per cubic yard
is given by the quotient of the number of birds and the volume of the aviary.
Option A: This response is the result of incorrectly using a sum of the given
dimensions instead of calculating the volume of the aviary.
Option C: This answer results from incorrectly computing the number of birds
per square yard of floor space.
Option D: This response is the result of calculating the number of cubic yards
per bird rather than the number of birds per cubic yard.
27
Domain:
MG—Modeling with Geometry (MG)
Cluster:
A—Apply geometric concepts in modeling situations
Standard:
3—Apply geometric methods to solve design
problems (e.g., designing an object or structure
to satisfy physical constraints or minimize cost;
working with typographic grid systems based on
ratios).
Math Practices:
solving them.
Calculator:
Calculator allowed
Tool Restrictions:
None
28
x
fenced-in
area
width
house
Note: not drawn to scale
Beth is going to enclose a rectangular area in back of her house.
The house wall will form one of the four sides of the fenced-in
area, so Beth will only need to construct three sides of fencing.
Beth has 48 feet of fencing. She wants to enclose the
maximum possible area.
What amount of fence should Beth use for the side labeled x ?
A.
12 feet
B.
16 feet
C.
24 feet
D.
32 feet
*correct answer
This item requires students to solve an optimization problem involving area and
perimeter. The most efficient pathway to solve the problem is to write equations
to represent the area and perimeter. Then, students may solve for one variable
in the linear perimeter formula and substitute this expression into the area
formula. This will lead to a quadratic equation representing the area in terms of
the length of one of the sides. Finding the maximum point of this parabola will
determine the side length that will maximize the area.
•
Students must make sense of the given quantities and recognize that they
must compute both area and perimeter.
29
•
decontextualize the given information and represent it in a logical
mathematical format. Then, they must manipulate this representation
and contextualize their results to answer the question.
•
Math Practice 4 (Model with mathematics.): Students must model the given
real-world situation with equations for the area and perimeter.
Option C: This is the correct answer.
x + 2w = 48, where w represents the width
A=x•w
48 – x
w=
2
Determine necessary
equations to use.
Step 3
48 – x
A=x•
2
Replace w in the Area
formula with equivalent
expression from step 2.
Step 4
A=–
Step 1
Step 2
Step 5
x=
1 2
x + 24x
2
−b
=
2a
−24
= 24
1
2 −
2
Solve perimeter formula
for w.
Simplify the expression.
Find the x-value of the
vertex of the parabola,
which represents where
the area is maximized.
Therefore, the area is maximized when x = 24 feet.
Option A: This answer is a result of finding the width, w, for the maximized area.
Step 1
Step 2
Step 3
x + 2w = 48, where w represents the width Determine necessary
A=x•w
equations to use.
Solve perimeter formula
x = 48 – 2w
for x.
Replace x with
A = (48 – 2w )w
equivalent expression
from Step 2.
Step 4
A = 48w – 2w 2
Step 5
b
−48
w =−
=
= 12
2a 2(−2)
Simplify expression.
Find the width.
(incorrect to stop at
this point)
Option B: The student incorrectly assumes that a square will maximize the
area of the fenced-in area without doing any calculations. This is true for
situations where all four sides of the enclosed area are being used.
Option D: The student incorrectly assumes that the largest given value for x
will result in the maximum area without doing any calculations.
30
Constructed-Response Item
This section presents a constructed-response item and samples of student
responses that received scores of 4, 3, 2, 1, 1 for minimal understanding, and 0.
This section also includes information used to score this constructed-response
item: an exemplary response, an explanation of how points are assigned, and a
specific scoring rubric. In addition to the online resources available for all test
questions, students have access to the Geometry Typing Help (pages 44 and 45),
which describes how to enter special characters, symbols, and formatting into
typed responses.
Constructed-Response Item
Domain:
Cluster:
C—Define trigonometric ratios and solve problems
involving right triangles
Standard:
8—Use trigonometric ratios and the Pythagorean
Theorem to solve right triangles in applied problems.
Math Practices:
solving them.
6. Attend to precision.
Calculator:
Calculator allowed
Tool Restrictions:
Protractor not allowed
Ruler not allowed
31
top of stairs
5 in.
12 in.
5 in.
12 in.
5 in.
base of bottom stair
Note: Not to scale
Leah needs to add a wheelchair ramp over her stairs. The ramp
will start at the top of the stairs. Each stair makes a right angle
with each riser.
Part A
1
. To the nearest
12
hundredth of a foot, what is the shortest length of ramp that Leah
can build and not exceed the maximum slope? Show your work
or explain your answer.
The ramp must have a maximum slope of
Length of ramp:
(student enters response in text box)
Λ
–
V
Work/Explanation:
Λ
–
V
Part B
Leah decides to build a ramp that starts at the top of the stairs
and ends 18 feet from the base of the bottom stair. To the nearest
hundredth of a foot, what is the length of the ramp?
Λ
–
V
32
Part C
To the nearest tenth of a degree, what is the measure of the angle
created by the ground and the ramp that Leah builds in part B?
Λ
–
V
This item requires students to reason about how slope, angle measures,
trigonometric ratios, and the Pythagorean Theorem relate mathematically.
Students must determine a horizontal measure for a leg of a right triangle
1
that results in a hypotenuse slope of no more than
. Students then use the
12
Pythagorean Theorem to find the measure of the hypotenuse in two different
right triangles. Finally, students must use a trigonometric ratio to determine
the measure of an angle in a right triangle.
This item is linked to five of the Math Practices.
•
Students must make sense of given information to determine how to
solve each part of the problem. The problem calls on multiple banks of
information and requires perseverance to solve each part.
•
decontextualize given measurements to manipulate the numbers using
equations. Then, they must contextualize to determine the reasonableness
of their answer within the given context.
•
others.): Students must construct an argument to justify their reasoning
behind how they calculated the minimum length of ramp that will satisfy
the given conditions in part A.
•
Math Practice 4 (Model with mathematics.): Students use the Pythagorean
Theorem to model the relationship between the legs and hypotenuse of
the right triangle used to calculate the length of the ramp. Students use
trigonometric ratios to model the relationship between side lengths and
angle measures of the right triangle used to calculate the measure of the
angle created by the ground and the ramp.
•
Math Practice 6 (Attend to precision.): Students must accurately convert
between measurement units, specify the correct units of measure in their
final responses, and use proper rounding techniques.
33
Scoring Information
Scoring Rubric
Score
Description
4
The student’s response earns 4 points.
3
2
1
1 point or minimal understanding of using Pythagorean
Theorem and trigonometric ratios.
0
The student’s response is incorrect, irrelevant, too brief to
evaluate, or blank.
Exemplary Response
Part A
Length of ramp: 15.05 feet
Work/Explanation: First find length of base of triangle (x)
1.25ft/x = 1/12
1.25*12 = 1*x
x = 15
Then find hypotenuse/ramp (c)
1.25^2 + 15^2 = c^2
226.5625 = c^2
c =* 15.05
Part B
20.04 feet
Part C
3.6 degrees
Points Assigned
Part A: 1 point for correct length
Part A: 1 point for complete and accurate work or explanation
Part B: 1 point for correct length
Part C: 1 point for correct degree measure
Note: The student may receive credit for part C for correctly finding the angle
measure based on an incorrect value from part B.
34
Sample Student Responses
Score Point 4
The following authentic student responses show the work of two students who
each earned a score of 4. A score of 4 is received when a student completes all
required components of the task and communicates his or her ideas effectively.
The response should demonstrate in-depth understanding of the content
objectives, and all required components of the task should be complete.
Score Point 4, Student 1
Part A
15.05
To find the shortest length of the ramp, I first set the stairs up as a right trianlge.
The height of the triangle(base to top of stairs) was 15 inches so I used the maximum
slope to set it up a proportion to find the length of the bottom of the triangle. I found
the length would be 180 inches. I then used the pythagorean theorem to find the
hypotenuse(length of the ramp) of the triangle. This answer was about 180.6324
inches. However, the length of the ramp length needed to be in feet so I converted
it from inches to feet and got the shortest ramp possible would be 15.05 feet.
Part B
20.04
Part C
3.6 degrees
This student response is correct and well-reasoned. The student calculates
the correct length and provides a correct and complete explanation of his or her
work for part A. The student also provides the correct length of the ramp in
part B and the correct angle measure in part C.
35
Part A
15.05 feet
1/12 = 1.25/15
(1.25 x 1.25) + (15 x 15) = 226.56
sqrt(226.56 = 15.052 feet
Part B
20.04 feet
Part C
3.6
This response receives full credit. The student calculates the correct length
and provides a correct and complete explanation of his or her work for part A.
The student correctly calculates the length of the ramp in part B and the
correct angle measure in part C.
36
Score Point 3
each earned a score of 3 for their responses. A score of 3 is received when a
student completes three of the four required components of the task. There may
be simple errors in calculations or some confusion with communicating his or
her ideas effectively.
Part A
15.05 feet
The shortest length of ramp that Leah can build and not exceed the maximum slope
is 15.05 feet. In order to get the length of the ramp, I first used the slope (1/12) as
a ratio to find the length of the stairs. By multiplying the total heights of the risers,
15 inches, by 12, I found that the total length of the stairs would be 180 inches.
By using the Pythagorean Theorum to find the length of the hypotenuse, I found that
15^2 + 180^2 =* 180.62^2. I then divided the hypotenuse by 12 to get the length of
the ramp in feet. 180.62 / 12 = 15.05 feet.
Part B
19.04 feet
Part C
3.8 degrees
This student calculates the correct length and provides a correct and
complete explanation of his or her work for part A. The student does not
provide the correct length of the ramp in part B; however, the student’s
response in part C is correct based on this incorrect value in part B.
37
Part A
12.04 feet
I added the rise of each step to find the length of the short leg. For each inch of the
short leg I need 12 inches on the long leg. Then I used the pythagorean theorum to
find the hypotenuse to find the length of the ramp. Last, I converted it to feet.
Part B
20.04 feet
Part C
3.6
In this response, the student provides an incorrect length of the ramp in
part A. However, the student received partial credit of 1 point for providing a
correct and complete explanation of how to determine the length of the ramp.
The student correctly calculates the length of the ramp in part B and the
correct angle measure in part C.
38
Score Point 2
each earned a score of 2 for their responses. A score of 2 is received when a
student completes two of the four required components of the task. There may
be simple errors in calculations, one or two missing responses, or unclear or
incorrect communications of his or her ideas.
Part A
15.05
I used the pythagorem thereom to get find that , with a slope of 1/12, each in down
would take 12.04 inches. With this I could find the length of the ramp by multiplying
it by the height,15 inches. Which came out to be 180.6 inches. Next, I needed to
convert the answer to feet, which gave me 15.05 feet.
Part B
18.04
Part C
6.90%
This student provides the correct length of the ramp in part A and correctly
reasons that each vertical rise of 1 inch will result in a hypotenuse length of
12.04 inches and multiplies that length by the height. The student does not
provide the correct length of the ramp in part B or the correct angle measure
in part C.
Part A
15.05 feet
I used a ratio and Pythagorean Theorem. Then I found how many feet.
Part B
20.04 feet
Part C
30 degrees
The student provides the correct length for the ramp; however, the explanation
is incomplete. There is not enough detail to receive full credit for their reasoning.
The student provides the correct length of the ramp in part B, but the angle
measure provided in part C is incorrect.
39
Score Point 1
each earned a score of 1 for their responses, and two students who each earned
a score of 1 for demonstrating minimal understanding. A score of 1 is received
when a student correctly addresses one of the four required components, or
demonstrates at least minimal understanding of the key concepts.
Part A
180.62
First I started out with what I know is going to be the height of the triangle we are
going to make with the ramp and the floor. Since slope = rise/run I then figured out
that the base of our tiangle would have to be 180in for the ratio to match 1/12. After
that I simply used the pathagoream therom to figue out the hypotenuse or in this
case, our ramp.
Part B
216.52
Part C
4
This student provides the correct length of the ramp in part A, but expresses
the measurement in the wrong unit. The item specifically asks for the length
of the ramp in feet; however, the reasoning provided is correct and complete,
so the student earned 1 point. The student does not provide the correct length
of the ramp in part B or the correct angle measure in part C.
40
Part A
15 feet
To the top of the stairs is 15 inches. So every inch you go up you have to go over one
foot. Which would make the ramp 15 feet long.
Part B
20 feet
Part C
3.6 degrees
The student provides an incorrect ramp length. While the student correctly
explains how to determine the measure of the long leg of the right triangle,
the response is incomplete because the student does not provide a method
for calculating the length of the hypotenuse. The student does not provide the
correct length of the ramp in part B. However, the student’s response in part C
is correct based on this incorrect value in part B.
Score Point 1 (minimal understanding), Student 3
Part A
180.62
Because the ramp cannot exeed a slope of (1/12), the angle of the ramp cannot
exeed 4.76 degrees. Using trigonometry, you can find the length of the ramp.
Part B
44.6
Part C
19.65
This student provides the correct length of the ramp in part A in inches,
but does not provide a unit of measure. The explanation is also incomplete
as it does not provide a method for computing the hypotenuse of the right
triangle. The student does not provide the correct ramp length in part B or
the correct angle measure in part C. This response does not earn any points;
however, one point is awarded for minimal understanding of the Pythagorean
Theorem for providing the correct length of the ramp in part A using the wrong
measurement unit.
41
Score Point 1 (minimal understanding), Student 4
Part A
39 in.
A squared plus B squared equals C squared. 5 is A. 12 is B. 5 squared equals 25.
12 squared equals 144. 25 plus 144 equals 169. The square root of 169 is 13.
There are three 5 inch drops. Multiply 13 in by 3 and that will equal 39 in.
Part B
240.46 in.
Part C
3.5
This student does not provide the correct ramp length in part A. The explanation
does not take into account the slope of the ramp. While the Pythagorean Theorem
is used correctly, it is applied using the wrong numbers. The student provides
the correct ramp length in part B, but uses incorrect units of measure.
The student provides an angle measure in part C that is a result of truncating
the correct response instead of rounding to the nearest tenth. This response
does not earn any points; however, one point is awarded for minimal
understanding of using trigonometric ratios to solve problems for part C.
One point for minimal understanding could also be awarded for part B;
however, only one minimal understanding point may be awarded per item.
42
Score Point 0
The following samples show the work of two students who each earned a score
of 0. A score of 0 is received when a student response is incorrect, irrelevant,
too brief to evaluate, or blank.
Part A
12/13
12^2 + 5^2= 169 sqrt(169)= 13 rise/run 24/26 = 12/13
Part B
51 feet
Part C
90 degrees
This student does not provide the correct length of the ramp in part A.
The reasoning provided uses the Pythagorean Theorem, but it is applied using
the wrong numbers. There is no mention of the slope of the ramp. The student
does not provide the correct ramp length in part B or the correct angle measure
in part C.
Part A
0.0ft
If you divide 12 into 1 it comes out to be 0.083. round that to the nearest hundredth of
a foot it comes out to be 0.1ft., anything less then 0.ft is 0.0ft.
Part B
1.5ft
Part C
2ft
This student does not provide the correct length of the ramp in part A.
The reasoning provides no useful information for calculating the length of
the ramp. The student does not provide the correct ramp length in part B.
The answer provided for the missing angle in part C is incorrect in value and
is expressed in the wrong unit of measure.
43
Geometry Typing Help
As of July 2014, the Geometry Typing Help has been updated to include typing
complex roots and the trigonometric functions. A current version can be found
at https://www.louisianaeoc.org/Documents/GeometryTypingHelp.pdf
Keystrokes for Special Symbols
1. If the Response
Includes:
×
multiplication symbol
2. Type This Instead:
x letter
x OR
*
asterisk (SHIFT + 8)
÷
division symbol
mixed number
32
exponent
≤
“less than or equal to”
square root
about equal to
45º
degree symbol
3 * 4 = 12
12 / 3 = 4
/
forward slash
(12 – 7)/(3 – 1)
Note: Parentheses
are required.
space between whole
number and fraction;
forward slash to separate
numerator and
denominator of fraction
2 3/4
^
“caret” (SHIFT + 6)
3^2 = 9
(pi)
Area = 9(pi)
square inches
pi symbol
≥
“greater than or equal to”
3 x 4 = 12
/
forward slash
fraction or ratio
3
2
4
3. Example:
>=
greater than sign, followed
by equals sign
<=
less than sign, followed by
equals sign
sqrt()
the letters sqrt, with the
radicand in parentheses
=*
equals sign, followed by
asterisk (SHIFT + 8)
degrees
(spell out the word)
y >= 13
y <= 13
sqrt(4) = 2
(pi) =* 3.14
The angle measures 45
degrees.
44
Keystrokes for Special Symbols (continued)
angle
Angle PQR is a right angle.
triangle
Triangle ABC is a right
triangle.
m q
perpendicular symbol
perpendicular
Line m is perpendicular to
line q.
m || q
parallel symbol
||
two “pipes” (SHIFT +
backslash)
OR parallel
m || q
OR
Line m is parallel to line q.
∆ ABC
∆ RST
congruence symbol
congruent
Triangle ABC is congruent
to triangle RST.
∆STU ˷ ∆VWX
similarity symbol
similar
Triangle STU is similar to
triangle VWX.
line segment
line segment
(spell out the words)
Line segment AB bisects
line segment CD.
line
line
Line AB is parallel to
line CD.
ray
ray
Ray AB goes through
point P.
angle symbol
∆ ABC
triangle symbol
45
Use the information below to answer
questions on the Geometry test.
Pythagorean Theorem
a2 + b2 = c2
c
a
b
Sphere
π ≈ 3.14
r
Surface Area = 4π r
4
Volume = π r 3
3
2
Cone
Surface Area
h s
πrs
πr 2
1
Volume = π r 2 h
3
r
Cylinder
r
Surface Area = 2π r 2 + 2π rh
Volume = π r 2 h
h
Regular Pyramid
B = area of base
L = area of lateral surfaces
s
Surface Area = B + L
h
b
Volume = 1 Bh
3
Rectangular Solid
h
l
w
Surface Area = 2wl + 2lh + 2wh
Volume = lwh
46
Circle
Area = π r 2
Circumference = 2π r
r
Rectangle
w
l
Area = lw
Perimeter = 2l
2w
Trapezoid
b1
h
Area =
1
h ( b1+ b 2 )
2
Area =
1
bh
2
b2
Triangle
h
b
Parallelogram
Area = bh
h
b
Cartesian Distance
Formula
Point 1: ( x1 , y1 )
Point 2 : ( x 2, y2 )
d=
( x2 – x1 ) 2 + ( y2 – y1 ) 2
47
Louisiana Department of Education
Office of Assessments

Document 6528329

Transcription

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