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Lesson 6.4
Chapter 6, continued Lesson 6.4 6.4 Guided Practice (pp. 382–383) 1. Because nFGH and nQRS are equiangular, all 6.4 Activity (p. 381) angles measure 608. So, all angles are congruent and nFGH , nQRS by the AA Similarity Postulate. Sample answer: F 2. Because they are both right angles, DFC and EFD T 40° 50° E are congruent. By the Triangle Sum Theorem, 328 1 908 1 m CDF 5 1808, so m CDF 5 588. Therefore, CDF and DEF are congruent. G 40° R 50° S So, nCDF , nDEF by the AA Similarity Postulate. 1808 5 m F 1 m F 1 m G 3. Yes; if S > T (or R > U ), then the triangles are 1808 5 408 1 m F 1 508 similar by the AA Similarity Postulate. 908 5 m F 4. EF ø 15 mm FG ø 13 mm 72 in. 58 in. GE ø 20 mm 1808 5 m R 1 m S 1 m T x in. 1808 5 408 1 m S 1 508 72 in. 58 in. 72x 5 3132 RS ø 22.5 mm x 5 43.5 ST ø 19.5 mm The child’s shadow is 43.5 inches long. TR ø 30 mm Tree height length of tree shadow 5. } 5 }} Your height length of your shadow 1. Sample answer: 15 22.5 FG ST 2 3 }ø}5} 13 19.5 2 3 }ø}5} GE TR 20 30 2 3 }ø}5} Corresponding angles are congruent and corresponding side lengths are proportional, so the triangles are similar. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 2. No; the corresponding sides of two similar triangles are S proportional, so they are not necessarily congruent. G 3. n ABC , nFED by the AA Similarity Postulate. 40° R 70° 40° AC CB BA 4. } 5 } 5 } because the ratios of corresponding side EF FD DE T 1808 5 m E 1 m F 1 m G lengths in similar triangles are equal. 1808 5 708 1 m F 1 408 y AC 25 BA 5. } 5 } because } 5 } . FD EF 15 12 708 5 m F EF ø 18 mm 15 18 FD DE 6. } 5 } because } 5 }. 25 x AC CB FG ø 27 mm y 25 7. y 5 20; } 5 } 15 12 GE ø 27 mm 1808 5 m R 1 m S 1 m T 1808 5 708 1 m S 1 408 708 5 m S TR ø 45 mm 18 30 3 5 15y 5 300 15x 5 450 y 5 20 x 5 30 congruent. ST ø 45 mm EF RS 15 18 8. x 5 30; } 5 } 25 x 9. Because they are both right angles, H and S are RS ø 30 mm }ø}5} Skill Practice of another triangle, then the triangles are similar. F 70° 6.4 Exercises (pp. 384–387) 1. If two angles of one triangle are congruent to two angles 2. Sample answer: E 54 in. x in. }5} 908 5 m S EF RS 54 in. FG ST 27 45 3 5 }ø}5} GE TR 27 45 3 5 }ø}5} So, the triangles are similar. Conjecture: Two triangles with two pairs of congruent corresponding angles are similar. By the Triangle Sum Theorem, 488 1 908 1 m F 5 1808, so m F 5 428. Therefore, F and K are congruent. So, nFGH , nKLJ by the AA Similarity Postulate 10. Because m YNM and m YZX both equal 458, YNM > YZX. By the Vertical Angles Congruence Theorem, NYM > ZYX. So, nYNM ,nYZX by the AA Similarity Postulate. Geometry Worked-Out Solution Key 167 Chapter 6, continued 11. By the Triangle Sum Theorem, 358 1 858 1 m R 5 1808 and 358 1 658 1 m V 5 1808. So, mR 5 608 and mV 5 808. Find BD: BD 5 BE 1 DE 20 5} 15 3 Corresponding angles are not congruent, so the triangles are not similar. 12. Because m EAC and m DBC both equal 658, EAC > DBC. By the Reflexive Property, C > C. So, nACE , nBCD by the AA Similarity Postulate. 13. By the Reflexive Property, Y > Y. By the Triangle Sum Theorem, 458 1 858 1 m YZX 5 1808, so m YZX 5 508. Therefore, YZX and YWU are congruent. So, nYZX , nYWU by the AA Similarity Postulate. 35 5} 3 35 , the correct answer is A. Because BD 5 } 3 } 21. length of AB 5 4 2 0 5 4 } length of AD 5 5 2 0 5 5 } length of AC 5 8 2 0 5 8 AD AB AE AC 5 4 AE 8 }5} }5} 14. By the Reflexive Property, N > N. By the Corresponding Angles Postulate, NMP > NLQ. So, nNMP ,nNLQ by the AA Similarity Postulate. 15. The AA Similarity Postulate is for triangles, not other polygons. p 24 16. B; } 5 } 12 10 240 5 12p AD AB AE AC 7 3 AE 4 }5} 20 5 p The length of p is 20, so the correct answer is B. 17. The proportion is incorrect because 5 is not the length of the corresponding side of the larger triangle. 4 9 Sample answer: A correct proportion is }6 5 }x . 18. Sample answer: 28 AE 5 } 3 28 } 28 , so the coordinates are E 1 } ,0 . The length of AE is } 3 3 2 } 3 cm The sketch shows that corresponding side lengths are not proportional. 19. Sample answer: 9cm 4cm 4cm 6cm 3 cm The sketch shows that corresponding side lengths are not proportional. AD AB AE AC 4 1 AE 6 }5} }5} 24 5 AE } The length of AE is 24, so the coordinates are E(24, 0). } 24. length of AB 5 6 2 0 5 6 } length of AD 5 9 2 0 5 9 } length of AC 5 3 2 0 5 3 AD AB AE AC 9 6 AE 3 }5} CE DE 20. A; Find x: } 5 } BE AE 3 4 }5} 9 2 } 5 AE 5 x }5} 3x 5 20 20 x5} 3 9 } 9 The length of AE is }2, so the coordinates are E 1 }2 , 0 2. 25. a. A 8 B 6 E 0 1 D 168 Geometry Worked-Out Solution Key 5 1 C Copyright © by McDougal Littell, a division of Houghton Mifflin Company. } length of AD 5 4 2 0 5 4 } length of AC 5 6 2 0 5 6 2 cm 3 cm }5} 23. length of AB 5 1 2 0 5 1 2 cm 2 cm 10 5 AE } The length of AE is 10, so the coordinates are E(10, 0). } 22. length of AB 5 3 2 0 5 3 } length of AD 5 7 2 0 5 7 } length of AC 5 4 2 0 5 4 Chapter 6, continued Problem Solving b. Sample answer: AEB > CED by the Vertical Angles Congruence Theorem. 31. The triangles shown in the diagram are similar by the AA Similarity Postulate, so you can write the following proportion. ABE > CDE by the Alternate Interior Angles Theorem. 20 in. d in. c. n AEB is similar to nCED. n AEB , nCED by the 800 5 26d AA Similarity Postulate. BE AE d. } 5 } DE CE 6 15 BE 10 30.8 ø d BA AE }5} DC CE 8 DC }5} 6 15 The distance between the puck and the wall when the opponent returns it is about 30.8 inches. }5} BE 5 4 20 5 DC 32. a. You can use the AA Similarity Postulate to show that the triangles are similar because you can show that two angles of nXYZ are congruent to two angles of nXVW. 26. Yes; Because m J and m X both equal 718, J > X. By the Triangle Sum Theorem, 718 1 528 1 m L 5 1808, so m L 5 578. Therefore, L and Z are congruent. So, nJKL , nXYZ by the AA Similarity Postulate. b. WX ZX WV ZY xm 6m 104 m 8m }5} }5} 27. Yes; If m X 5 908, m Y 5 608, and nJKL contains 8x 5 624 a 608 angle, then the triangles are similar by the AA Similarity Postulate. x 5 78 28. No; Because m J 5 878, nXYZ needs to have an 878 The width of the lake is 78 meters. angle in order for it to be possible that nJKL and nXYZ are similar. This is not possible because m Y 5 948, and the sum of 948 and 878 is 1818, which contradicts the Triangle Sum Theorem. c. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. XY VX ZY WV }5} 10 m 8m 104 m } xm 5} 29. No; By the Triangle Sum Theorem, 858 1 m L 5 1808, and m X 1 808 5 1808, so m L 5 958 and m X 5 1008. So, nXYZ needs to have a 958 angle for it to be possible that nJKL and nXYZ are similar. This is not possible because m X 5 1008, and the sum of 958 and 1008 is 1958, which contradicts the Triangle Sum Theorem. 26 in. (66 2 26) in. }5} 1040 5 8x 130 5 x So, VX is 130 meters. 33. All equilateral triangles have the same angle measurements, 608. So, all equilateral triangles are similar by the AA Similarity Postulate. 30. Because P > P by the Reflexive Property and E PST > R by the Corresponding Angles Postulate. nPST , nPRQ by the AA Similarity Postulate. B Because the triangles are similar, you can set up the following proportion: PT PQ PS PR PT PQ PS PS 1 SR A }5} }5} x a }5} 8 3x a 1 }3x x1 a 1 }3 x 2 5 3ax 8 8 ax 1 }3 x 2 5 3ax 8 } x 2 5 2ax 3 C D F m A 5 m B 5 m C 5 m D 5 m E 5 m F5 608, so A > B > C > D > E > F. 34. f h n g }5} 8 cm hm 3 cm 50 m }5} 400 5 3h 1 133}3 5 h 1 The blimp should fly at a height of 133}3 meters to take the photo. 4 3 }x 5 a 4 So, PS 5 }3 x. Geometry Worked-Out Solution Key 169 Chapter 6, continued 35. Sample answer: 39. Sample answer: U B E V S T A R C N D N Q P } } Angle bisectors SV and PN are corresponding lengths in SV ST 5} by the Corresponding similar triangles. So, } PN PQ Lengths Property on page 375. F M } Let n ABC , nDEF, let BN bisect ABC and let } EM bisect DEF. Because n ABC , nDEF, } } ABC > DEF and A > D. Also, BN and EM bisect congruent angles, so ABN > CBN > DEM > FEM. By the AA Similarity Postulate, n ABN , nDEM. BN AB AB Therefore, } 5} , where } is the scale factor. EM DE DE 36. Sample answer: 40. Sample answer: E A B A E b b A C D F Because they are both right angles, A and D are congruent. The acute angles C and F are also congruent, so n ABC , nDEF by the AA Similarity Postulate. B D C B a C a Because ADC > BEC and C > C, n ADC , nBEC by the AA Similarity Postulate. b 37. a. Sample answer: The ratio of the hypotenuses is }a, so the ratio of the A b corresponding side lengths is also }a. The altitudes are corresponding sides, so their lengths are in the b ratio }a. E C Mixed Review for TAKS B b. Sample answer: m ADE ø 478, m ACB ø 478; m AED ø 298, m ABC ø 298; So, m ADE 5 m ACB and m AED 5 m ABC. c. By the AA Similarity Postulate, n ADE , n ACB. d. Sample answer: AB 5 3 cm, BC 5 4 cm, AC 5 2 cm, AD 5 1 cm, DE 5 2 cm, AE 5 1.5 cm; AD AC AE AB DE CB 1 2 }5}5}5} e. The measures of the angles change, but the equalities remain the same. Yes; the triangles remain similar by the AA Similarity Postulate. 38. Sample answer: Given any two points on a line, you can draw similar triangles as shown in the diagram. Because the triangles are similar, the ratios of corresponding side lengths are the same. So, the ratio of the rise to the run is the same. Therefore, the slope is the same for any two points chosen on a line. 41. C; Length of postcard Width of postcard 5.5 in. 3.5 in. 11 7 }} 5 } 5 } 11 in. 7 in. 11 6.6 in. 4.2 in. 11 Choice A: } 5 } 7 Choice B: } 5 } 7 5 in. 3 in. 5 Choice C: } 5 }3 4.4 in. 2.8 in. 11 Choice D: } 5 } 7 The dimensions 5 inches by 3 inches are not proportional to the dimensions of the postcard. 42. F; 120, 138, 142, 142, 156, 158 856 2 142 1 142 5 142}3 Mean 5 } 6 Median 5 } 5 142 2 Mode 5 142 Range 5 158 2 120 5 38 The mean gives Janice the highest final score. 170 Geometry Worked-Out Solution Key Copyright © by McDougal Littell, a division of Houghton Mifflin Company. D