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 [email protected] 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. Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 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 • Geometry Reference Sheet • 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. Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 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 Geometry EOC Sample Test Items—October 2013 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 Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 8 Conceptual Category: G—Geometry 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 Geometry EOC Sample Test Items—October 2013 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 Geometry EOC Sample Test Items—October 2013 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. This item is linked to three of the Math Practices. • 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. Geometry EOC Sample Test Items—October 2013 11 Conceptual Category: G—Geometry 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: 7. Look for and make use of structure. Calculator: Calculator not allowed Tool Restrictions: None Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 14 Conceptual Category: G—Geometry Domain: SRT—Similarity, Right Triangles, and Trigonometry 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: 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 Geometry EOC Sample Test Items—October 2013 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 Geometry EOC Sample Test Items—October 2013 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. • 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, 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. Geometry EOC Sample Test Items—October 2013 17 Conceptual Category: G—Geometry Domain: SRT—Similarity, Right Triangles, and Trigonometry 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: 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 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. This item is linked to two of the Math Practices. • 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, and understand the goals of the problem. Geometry EOC Sample Test Items—October 2013 18 • Math Practice 3 (Construct viable arguments and critique the reasoning of 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. Geometry EOC Sample Test Items—October 2013 19 Conceptual Category: G—Geometry 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: 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 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 Geometry EOC Sample Test Items—October 2013 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. This item is linked to two of the Math Practices. • 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, and understand the goals of the problem. • Math Practice 3 (Construct viable arguments and critique the reasoning of 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°. Geometry EOC Sample Test Items—October 2013 21 Conceptual Category: G—Geometry 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: 7. Look for and make use of structure. 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. This item is linked to one of the Math Practices. • 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. Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 23 Conceptual Category: G—Geometry 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 Geometry EOC Sample Test Items—October 2013 24 This item requires students to calculate the volume of a right pentagonal pyramid given the area of the base. This item is linked to one of the Math Practices. • 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. Geometry EOC Sample Test Items—October 2013 1 by the product of the 3 25 Conceptual Category: G—Geometry 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: 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 4. Model with mathematics. 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. 0.25 birds per cubic yard C. 1.24 birds per cubic yard D. 4.03 birds per cubic yard *correct answer Geometry EOC Sample Test Items—October 2013 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. This item is linked to three of the Math Practices. • Math Practice 1 (Make sense of problems and persevere in solving them.): 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. Geometry EOC Sample Test Items—October 2013 27 Conceptual Category: G—Geometry 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: 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 4. Model with mathematics. Calculator: Calculator allowed Tool Restrictions: None Geometry EOC Sample Test Items—October 2013 28 Use the diagram to answer the question. 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. This item is linked to three of the Math Practices. • Math Practice 1 (Make sense of problems and persevere in solving them.): Students must make sense of the given quantities and recognize that they must compute both area and perimeter. Geometry EOC Sample Test Items—October 2013 29 • 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 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. Geometry EOC Sample Test Items—October 2013 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 Conceptual Category: G—Geometry Domain: SRT—Similarity, Right Triangles, and Trigonometry 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: 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 6. Attend to precision. Calculator: Calculator allowed Tool Restrictions: Protractor not allowed Ruler not allowed Geometry EOC Sample Test Items—October 2013 31 Use the diagram to answer the question. 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 (student enters response in text box) – 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? (student enters response in text box) Geometry EOC Sample Test Items—October 2013 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? (student enters response in text box) 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. • Math Practice 1 (Make sense of problems and persevere in solving them.): 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. • Math Practice 2 (Reason abstractly and quantitatively.): Students must decontextualize given measurements to manipulate the numbers using equations. Then, they must contextualize to determine the reasonableness of their answer within the given context. • Math Practice 3 (Construct viable arguments and critique the reasoning of 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. Geometry EOC Sample Test Items—October 2013 33 Scoring Information Scoring Rubric Score Description 4 The student’s response earns 4 points. 3 The student’s response earns 3 points. 2 The student’s response earns 2 points. 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. Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 35 Score Point 4, Student 2 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. Geometry EOC Sample Test Items—October 2013 36 Score Point 3 The following authentic student responses show the work of two students who 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. Score Point 3, Student 1 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. Geometry EOC Sample Test Items—October 2013 37 Score Point 3, Student 2 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. Geometry EOC Sample Test Items—October 2013 38 Score Point 2 The following authentic student responses show the work of two students who 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. Score Point 2, Student 1 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. Score Point 2, Student 2 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. Geometry EOC Sample Test Items—October 2013 39 Score Point 1 The following authentic student responses show the work of two students who 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. Score Point 1, Student 1 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. Geometry EOC Sample Test Items—October 2013 40 Score Point 1, Student 2 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. Geometry EOC Sample Test Items—October 2013 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. Geometry EOC Sample Test Items—October 2013 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. Score Point 0, Student 1 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. Score Point 0, Student 2 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. Geometry EOC Sample Test Items—October 2013 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) Geometry EOC Sample Test Items—October 2013 y >= 13 y <= 13 sqrt(4) = 2 (pi) =* 3.14 The angle measures 45 degrees. 44 Keystrokes for Special Symbols (continued) angle (spell out the word) Angle PQR is a right angle. triangle (spell out the word) Triangle ABC is a right triangle. m q perpendicular symbol perpendicular (spell out the word) Line m is perpendicular to line q. m || q parallel symbol || two “pipes” (SHIFT + backslash) OR parallel (spell out the word) m || q OR Line m is parallel to line q. ∆ ABC ∆ RST congruence symbol congruent (spell out the word) Triangle ABC is congruent to triangle RST. ∆STU ˷ ∆VWX similarity symbol similar (spell out the word) Triangle STU is similar to triangle VWX. line segment line segment (spell out the words) Line segment AB bisects line segment CD. line line (spell out the word) Line AB is parallel to line CD. ray ray (spell out the word) Ray AB goes through point P. angle symbol ∆ ABC triangle symbol Geometry EOC Sample Test Items—October 2013 45 Geometry Reference Sheet 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 Geometry EOC Sample Test Items—October 2013 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 Geometry EOC Sample Test Items—October 2013 47 Louisiana Department of Education Office of Assessments

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