Answers to Geometry Unit 3 Practice
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1 Lesson a. (4, 9) b. (8, 0) c. (, 1) d. 9, e. (0.10,.). a. b. Aʹ(4., 4.), Bʹ(6, 7.), Cʹ(9, 0), Dʹ(6, 6). C 4. a. b. enlargement c. (1, ) d. Pʹ(7, 10), Qʹ(9, 4), Rʹ(, 6). Aʹ(, 7.), Bʹ(, 7.), Cʹ(, 1.), Dʹ(, 1.) Lesson D x y 7. (x, y), x y, 8. Pʹ(6, 1), Qʹ(, 1), Rʹ(, 9) 9. D o, No. Sample answer. A dilation can change the size of a preimage. Since a rigid transformation does not affect the size of a figure, a dilation is not a rigid transformation. Lesson y 7., z 1 1. a. no, not congruent; yes, similar b. Aʹ(, 17), Bʹ(10, 0), Cʹ(16, 1) Answers to Geometry Unit Practice 14. B 1. a. Yes. In ABC, C B because AB AC. Dilations preserve angle measure, so Cʹ Bʹ and AʹBʹCʹ is isosceles. b. Lesson Triangle II is similar to Triangle III. Sample answer. The third angle in Triangle II is 70, so Triangles II and III are similar by the AA Similarity Postulate. 17. a. AB 1 4 AC XY 1 ; 16 4 ; A X. Since XZ 0 two sides are proportional and the angles formed by the sides are congruent, the triangles satisfy the SAS similarity criterion. 4 BC b. Use the scale factor : 4 YZ ; 18 4, YZ YZ (18)().. 4 c. There is a sequence of transformations, including a dilation, that maps ABC to XYZ. d. YZX 18. C 19. a. Sample answer. PST PQR or PTS PRQ b. PS PT PQ PR 0. a. m D 107, m E 8 b Lesson a. b. ZSR c. 1 1 ; 1.1 d..67 e. RZ 1, XZ Sample answer. Using 8, 9, and 6 in the numerator and 8, 10, and 1 in the denominator, we can form the ratios 6 8, 8 10, and 9, and show that 1 each ratio is equivalent to. That means 4 corresponding sides are proportional, so the two triangles are similar by the SSS similarity criterion. 01 College Board. All rights reserved. A1
2 4. a. 0 units, 16 units. D b. 7 1 units, 9 units c. 9 units, 1 units d. 1 4 units, units Lesson C 7. a d b c or b c a d 8. a. AD DB or ED CB b. AD 40.9, DB a. 7. b.. c. 9 d a. 7 b. 118 Lesson a. PMT, SPT b. PTM c. TPM d.. a. BE b. CJ, GJ c. EF d. AGJ, AGC. a C b. 8.8 c..8 d. 1.. Sample answer. KJL MJK MKL Lesson C 7. a. 4 b. 4. c. 8 d a. 8 b. c. 18. d a. 0 b. 18 c. d. 100 e. ab 40. a. f 8 b. area 1 bh 1 (4 1 16)(8) 1 (0)(8) 80 units c. a 4, b 8 d. area 1 bh 1 (4 )(8 ) 1 ()() 80 units e. Sample answer. You get the same value for the area of triangle ABC whether you use AB and CD as the base and height or whether you use BC and CA as the base and height. Lesson C 4. 4 ft 4. a. 1 ft b. 61 ft c. 8 ft d. ft 01 College Board. All rights reserved. A
3 44. a. 14. units b. 8 units c. 7.7 units d. 9.9 units 4. a. 4.1 cm b. 6.6 cm Lesson a. acute b. right c. right d. obtuse e. acute 47. a. Yes b. No fi a. 1.9, s, 8.6 cm 49. D b. 10., l, 19.1 cm c. 6.0 cm d. 1.6 cm 0. a. b. c. 1 or 1 d. 6 e. Lesson a. 1 in.; in. b. cm;.6 cm c. 7a ft; 9.90a ft d. a b 1.41a units; units b. a. 11 in.; 1.6 in.. B b. 9. cm or 19 cm; 1.44 cm c..a ft;.4a ft d. c d 4. a. 6 units b. 6 cm units; c. ( 6 1 ) cm d. 0. unit 0.71c units d. a. leg: 7 units; hypotenuse: 7 units b. leg: 10 units; hypotenuse: 10 units c. leg: units; hypotenuse: units d. leg: m units; hypotenuse: m units Lesson 1-6. a. longer leg: 1 in.; hypotenuse: 0 in. b. longer leg: 4 cm; hypotenuse: 16 cm c. longer leg: a ft; hypotenuse: a ft d. longer leg: 1 units; hypotenuse: 6 units 7. a. shorter leg: 1. cm; longer leg: 1. cm b. shorter leg: 4 in.; hypotenuse 8 in. c. shorter leg: 10 ft; hypotenuse: 0 ft d. shorter leg: 1 or units; longer leg: 1 unit 8. D 9. a. legs: cm, cm; hypotenuse: 10 cm b. legs; 6, 6 ; hypotenuse: 1 c. legs: 1, 1 ; hypotenuse: 0 d. legs: a, a ; hypotenuse: a 60. a, b, c, d College Board. All rights reserved. A
4 Lesson a. MT b. MT c. NT d. NT 6. a. 1 b. 8.1 c. 8 X 61.9 Y d. Sample answer. I used the same angle measures as in QRS. I multiplied each side length of QRS by. to find the side lengths of XYZ cm, 6.1 cm 64. C 6. Scale Factor 0.4, m A 8.1, m E 61.9, AC 1, EF., DE 6.8 Lesson - p 66. a. r b. q p p c. r d. q r p e. q 67. a. 48 or b. 48 c d. 7 e. 7 Z 68. a b C 70. B c d. 0 e. 1 Lesson - a 71. a. sin ; 0.97 a 10 ; a (10)(0.97) 19.1 b b. cos ; b 10 ; b (0.746)(10) a. sin 6 ; m m sin perimeter: units area: 1 (7)(7.6) 96.6 units b. tan 4 ; p p tan sin 4 ; q q sin perimeter: units area: 1 (17.8)(19.1) units 7. B 74. a. cos AD, AD (BD)(cos ) BD (4.)(0.6018). sin AB, AB (BD)(sin ) BD (4.)(0.7986).8 b. The area of ABCD is (.)(.8) 861.9, so the area of ABD is (0.)(861.9) Using AT as height and BD as base in ABD, A 1 bh; 40.9 (4.)(AT); so AT (40.9)() College Board. All rights reserved. A4
5 c. sin AT, AT (AD)(sin ) AD (.)(0.7986) 0.4. d. Sample answer. The results are the same. I prefer the method in Part c because it is faster. 7. a m b. 78. m c m d m Lesson a. b. 44 c. 78. d. 76 e a b AB 1., CB 1.9, m B DF 9.8, m F 66.9, m E D Lesson a. sin Q h r b. sin R h q c. h r sin Q, h q sin R d. r sin Q q sin R e. sinq q sin R r sin M sin N 8. m n 8. a b D 8. C sint t Lesson A sin8 sin Q 87. a. 1 1 ; sin Q (1)(sin 8 ) 1 (1)(0.617) 0.77; m Q 0. 1 b. sin8 1 sin T 1 ; sin T (1)(sin 8 ) 1 (1)(0.617) 0.77; m T 0. 1 c. Sample answer. In PQR, m Q 0., but in STV, m T fi 0.. d. Sample answer. The supplement of 0. is 19.7, and sin The actual measure of angle T is the supplement of a., 1 b. 8, Sample answer. Y X W X W 90. a. 7, 17 b. 10.8,. Lesson C D cm 9. a. 4 b. 78 Y 01 College Board. All rights reserved. A
6 Lesson B 97. a. side, side, side b. Law of Cosines c d. 6.4 e a. angle, angle, side b. Law of Sines c. 18. d. 8.6 e a. side, angle, side b. You can use the Law of Cosines to find HK and then either the Law of Sines or the Law of Cosines to find m K or m H. c. 1.1 d. 7 e m T 180 ( 1 9) sin 16 sin sin TB TA TB TA 100 sin sin sin 9 sin 16 (100)(0.46) (100)(0.4848) The surveyor at point B is closer to T, by m.. m 9.9 m 01 College Board. All rights reserved. A6
Answers. Chapter 9 A92. Angles Theorem (Thm. 5.6) then XZY. Base Angles Theorem (Thm. 5.6) 5, 2. then WV WZ;
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