Homework Helper Answer Key

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1 Lesson 9-1 Translations 1. III 2. (2, 11) 3. a. B 4. a. Point T: ( 7, 9) Point U: ( 4, 9) Point V: ( 5, 3) b. ( 8, 0) 5. B 6. a. A, C, D b. D 7. a. Answers will vary. Lesson 9-2 Reflections 1. (4, 2) 2. C' B ' A ' 3. a. Graph I A D' D B C b. See student work 4. a.par allelogram JKLM is a reflection of parallelogram ABCD. 5. a.the x-coordinates of the vertices are the same distance away from the y-axis but in the opposite direction. b.the y-coordinates of the vertices are unchanged in the image. c. ArBrCr is the image of ABC after a reflection across the y-axis. 6. a. A, B b. ArBrCrDr is the image of ABCD after a reflection across the y-axis. 7. a. ErFrGr is the image of EFG after a reflection across the line y 1. b. Your friend took the reflection across the x-axis instead of the correct line of reflection. 8. a. b. A reflection changes only a figure s position, not its size or shape. The image of the figure faces the opposite direction as the figure ( 2, 3) 11. a. ( 3, 9) ( 2, 8) ( 4, 7) b. ErFrGr is the image of EFG after a reflection across the line y ( 2, 5) 13. a. y B' J' A' R' A A' C J A R B A C C' B B'' C'' A'' digits 28 Grade 8

2 Lesson 9-3 Rotations 1. C 2. ( 6, 2) 3. a. A 4. a. A 5. ( 3, 2) a. ( 6, 4) 8. a. M b. 270 about the origin, (0, 0) c. Answers will vary. R L b. D I N K T O T 9. a. G, I, K K Lesson 9-4 Congruent Figures L I R M N 1. A 2. reflection across the x-axis, translation 9 units left, translation 7 units up 3. Yes, because a reflection across the x-axis, followed by a translation of 6 units left and 3 units up, maps DEF to DrErFr. 4. ANL JEF 5. a. reflection across the x-axis, translation of 5 units right 6. Yes, because a reflection across the y-axis, followed by a translation of 6 units down and 3 units right, maps XYZ to XrYrZr. 7. reflection across the y-axis, translation of 6 units down, translation of 4 units right 8. a. Translation of 5 units left, translation of 6 units down b. Translation of 5 units right, translation of 6 units up 9. a. Using only translations would make it so ArBrCrDr is not labeled correctly. b. Rotation of 180 about the origin, translation 3 units left, translation 7 units up 10. a. Decide whether a sequence of rigid motions maps DEF to DrErFr. b. Yes, because a rotation of 180 about the origin, followed by translations of 4 units up and 3 units right map DEF to DrErFr. Lesson 9-5 Problem Solving 1. C 2. a. C 3. a. a translation 4 units to the left and 5 units down, followed by a reflection across the y-axis b. She reflected the triangle across the wrong axis. 4. a. a translation 6 units to the left and 1 unit down, followed by a reflection across the y-axis and then a rotation of 90 about the origin 5. a. C 6. a. Move the triangle ABC so the point (3, 24) is on the origin. b. A translation 3 units to the left and 4 units up, followed by a reflection across the x-axis digits 29 Grade 8

3 7. a. Follow the same sequence of rigid motions that map point T to point Q and point U to point R. 8. a. a reflection across the x-axis followed by a translation of 4 units to the left and 2 units down Lesson 10-1 Dilations (2, 1) a (20, 8) 9. a.side DrEr looks about 4 times longer than side DE. b a. A b a b. 5 9 b. 13. Q Q F' P R P R 6. a. 2 7 D' E' Lesson 10-2 Similar Figures 1. a. A 2. C 3. a. Yes 4. a. A, 2 digits 30 Grade 8

4 5. a. C b. A, 2, 3, 2 6. a. Y b. Yes Lesson 10-3 Relating Similar Triangles and Slope 1. a. b. The ratio of the rise to the run of the given triangle is 5 3. The ratio of the rise to the run of the image triangle is a. b. The ratio of the rise to the run of the given triangle is 3. The ratio of the rise to the run of the image triangle is 3. c. A 2. (5, 1) 3. a. b. The ratio of the rise to the run of the given triangle is 1 2. The ratio of the rise to the run of the image triangle is 1 2. The rise to run ratios are the same. c. Answers will vary. 4. a. b. 1 2 c. Your friend found the ratio of the run to the rise. 6. Tile T, Tile S 7. a. Slope Triangle Q can be used to find the slope of the line. 8. a. Find the coordinates of any two points on the triangle. b. 1 3 c. d. The rise to run ratios are the same. change in y 9. a. The formula can be used change in x to find the third vertex, the slope of the second triangle is 1 3, it is similar to the given triangle. b. (12, 3) Lesson 10-4 Problem Solving 1. C, D km 3. d 150 ft cm digits 31 Grade 8

5 5. a. B, D b. She did not find all possible coordinates for Y using another sequence of rigid motions ft 7. 9 in. 8. a. 58 ft b ft 9. a. A, C 10. a. Draw a picture of the similar triangles and use the corresponding side lengths to set up an equivalent ratio equation. b. 150 feet 11. a. Find the length of each side of the actual tractor trailer. b. 110,592, B c. Answers will vary. 13. $76.91 Lesson 11-1 Angles, Lines, and Transversals 1. B 2. m u m a. 3 and 6, 2 and 7 6. a. m w 79 b. Jacob thought that w and y are corresponding angles, when actually w and t are corresponding angles and have the same measure. 7. m a. m a. Find the measure of the corresponding angle, 2, by subtracting m 1 from 180. b a. D b a. m 3 86 and m b. Ball A Lesson 11-2 Reasoning and Parallel Lines 1. D a. Yes, because corresponding angles are 27 each. b. Yes, because alternate interior angles are 81 each. 4. D 5. m 7 n 6. a. 72 b. Your friend found the supplement of a. C 8. D 9. a. 2x b. 16 c. Answers will vary. 10. a. 2x b a. The value of x must equal the value of y. b. 73 Lesson 11-3 Interior Angles of Triangles x 36, m Q a a. 51 digits 32 Grade 8

6 9. a. 35 b. D a. D b m A m B a. D b a. C b. m A 19.3 m B 77.2 m C m A 93 m B 31 m C 56 Lesson 11-4 Exterior Angles of Triangles 1. B 2. a. D b. B, D 3. a. C b. A, C 4. m m 1 32 m a. m a. m a. m 1 54 m 2 68 b. B 9. m m m m 2 35 b a. C b. 97 Lesson 11-5 Angle-Angle Triangle Similarity 1. Yes 2. No 3. B, C 4. a. Yes 5. a. A, B, C b. Anchil did not give all the similar triangles. 6. No 7. Yes 8. A, B, C 9. a. If two angles of WSR are congruent to two angles of ZSP b. No 10. a. A, C, D b. A, B, D 11. B, D, E, F Lesson 11-6 Problem Solving 1. m m B 4. a. First use vertical angles to find m 5. Then use corresponding angles to find m 3. Finally, use alternate interior angles to find m 2. b. m 2 55 c. Answers will vary. 5. a. m b. A 6. m 4 136, m m a. Yes b. Each triangle has angle measures of 30, 59, and a. C b a. m 1 m 2 m 3 b. 36 digits 33 Grade 8

7 11. m 1 23 m 2 49 m m A Lesson 12-1 Reasoning and Proof Subtraction Property of Equality 3. Division Property of Equality Given Multiplication Property of Equality 3. Distributive Property Addition Property of Equality Division Property of Equality 4. a. 2. Multiplication Property of Equality 3. Subtraction Property of Equality 5. a. 2. Definition of Midpoint 4. Substitution Property 5. Subtraction Property of Equality 6. Division Property of Equality 6. It is given that Street A intersects Street B and C, B is parallel to C, and m By vertical angles, m 2 m 4. Then m 4 m 6 because of alternate interior angles. Again by vertical angles, m 6 m 8. By the Transitive Property m 2 m a. Commutative Property of Addition 9x 7 Addition Property of Equality Division Property of Equality 8. It is given that E intersects lines F and G, line F is parallel to G, and m Then m 2 m because 2 and 3 are adjacent angles that form a straight line. m 3 m 5 by alternate interior angles. By vertical angles, m 5 m 7. By the Transitive Property, m 3 m 7. By substitution m 2 m a. A b. Sum of the angles of ABC is 180 Substitution Property Combine like terms Subtraction Property of Equality 10. a. A b.g iven 3x 15 Addition Property of Equality Combine like terms 32 Division Property of Equality Lesson 12-2 The Pythagorean Theorem in ft m a. 37 ft 7. a cm b. D ft 9. " a ft ft b a. B b. 34 in. 14. a m b. go through the park digits 34 Grade 8

8 Lesson 12-3 Finding Unknown Leg Lengths ft yd 4. a. D b. 9.8 ft in in ft 8. a. 125 ft 9. a ft 10. a ft 11. a. 3.9 cm b. She did not square the length of the given leg ft cm 14. a. B b cm 15. Biking mm Lesson 12-4 The Converse of the Pythagorean Theorem 1. Yes 2. No 3. II only 4. a. A 5. B 6. a. Student 1 and Student 3 are incorrect. b. They used the equation a b c, not a 2 b 2 c Triangle 2 and Triangle 3 are right triangles. 8. Triangle 1, Triangle 3 9. No 10. a. A 11. D 12. a. Decide which side length to substitute for the hypotenuse length. b. No 13. a. The sides of the triangle satisfy the equation a 2 b 2 c 2. b. Triangle 1, Triangle 2, Triangle a. Triangle 1 and Triangle 2 only Lesson 12-5 Distance in the Coordinate Plane 1. 5 mi scalene 4. a. You are closer to the school. 5. a. The triangle formed by the points H, P, and L is a right triangle. The distance from the library to your home is the length of the hypotenuse, HL. b. 6 miles 6. 5 yd ft a. Find PR by finding the difference of the x-coordinates. Then use the Pythagorean Theorem to find PQ and QR. b. Isosceles 10. B Lesson 12-6 Problem Solving 1. ( 10, 10), ( 10, 8) 2. a. (5, 10), (5, 14) b. (5, 8), (5, 16) 3. 9,680 ft cm digits 35 Grade 8

9 5. a. B, C b cm 6. a. (7, 3), (7, 5) b. D 7. a. 360 in. 2 b. Yes 8. (4, 4), (4, 4) 9. a. A b. 8 mm c. 2,400 mm a. 5,096 in ( 5, 5), (3, 5) b in ,800 cm a in. Lesson 13-1 Surface Areas of Cylinders cm ,758.8 mm ,179.5 ft in cm 2 6. a m 2 7. a in a. A b in a in. 2 b. She forgot to include the lateral area. 10. a ft 2 b. 11 gal 11. a. D b in ,628.2 cm a. 111p in. 2 b in. 2 Lesson 13-2 Volumes of Cylinders 1. a. 48p in. 3 b. 96p in in , in cm 5. a. 4.9 in. 6. a. 800p m 3 7. a in. 3 b. No c. He found the volume of the rectangular container instead of the cylinder. 8. a. 96p ft 3 b. 6,144p ft 3 c cm ,143p in m b. 147p yd a. Use the height of the original can to find the height of the new can. b cm lb in. Lesson 13-3 Surface Areas of Cones cm cm ,140 in a. 1,884 in. 2 b. 7,536 in a. 1,557 m 2 6. a. 141 cm 2 digits 36 Grade 8

10 7. a. 264 cm 2 b. Your friend did not find the area of the base cm in Cone II cm 2 b. 1,910 cm a. D b a cm 2 b a. 35,408 cm 2 b. $32.19 Lesson 13-4 Volumes of Cones 1. 12p cm 3 2. Cone 1, Cone 3, Cone m 3 4. a. B b cm 3 5. $12, yd feet 8. 7 ft 9. a. B, C b yards 10. a. 75p m a a. 4 units 13. a m 3 b. A 14. a in. 3 b. 1, in cm a. Cone I: 3,072p one 2: 6,000p Lesson 13-5 Surface Areas of Spheres cm , cm cm , yd cm cm ft 8. a. 3, in a ft 10. a. 3, in. 2 b. A 11. $111, a. 32 yd a. B b. 8, ft ,845 km cm m 17. a. B b. 1, in a. 164,176 cm 2 b. 49,253 cm a. Planet A: 1,693 km Planet B: 3,386 km Planet C: 5,079 km Planet D: 6,772 km Lesson 13-6 Volumes of Spheres 1. 2, in ft in ,187 ft 3 digits 37 Grade 8

11 5. a in. 3 b. The volume of a sphere with diameter inches is less than that of a sphere with radius inches cm cm 8. r 6.9 mm, d 13.8 mm 9. a. B b. 33 in a. 44, in a. 4.4 in. 12. a m 3 b. B 13. 9, in a. 7, cm 3 b. 24 minutes yd a. 7.6 cm b cm a. B b cm in a p in.3 b p in.3 c. 3 in. Lesson 13-7 Problem Solving in cm 3 3. a. 578 m m 3 5. a. 1,611 in. 3 b. A cm 3 7. a. 122 in a. 54,937 m gal 10. a. B b. 949 cm a. B b. 8,574 in. 3 digits 38 Grade 8

Grade 8 Mathematics Item Specifications Florida Standards Assessments

Grade 8 Mathematics Item Specifications Florida Standards Assessments MAFS.8.G.1 Understand congruence and similarity using physical models, transparencies, or geometry software. MAFS.8.G.1.2 Understand that a two-dimensional figure is congruent to another if the second