Chapter 11 Review ex. answers

Chapter 11, Review
Set I (pages 476–478)
Greek Cross.
•1. 2. 2. – 1
3
. 3. 2. 4.
. •5. – 4
2 4 3
10. Their product is –1. (Or, one slope is the opposite of the reciprocal of the other.)
Doubled Square.
•11. d = s 2 . 12. s 2 13. d 2 14. 2s 2 [d 2 = (s 2 ) 2 = 2s 2 .]
Escalator Design.
. 6. AC ^ CD. 7. AC|| DE. 8. HF ^ BG. 9. They are equal.
•15. 28 ft. [c = 2a = 2(14).] •16. About 24.2 ft. (b = a 3 = 14 3 ≈ 24.2.)
17. 20 ft. (40 = 2a.) 18. About 34.6 ft. (b = a 3 = 20 3 ≈ 34.6.)
•19. About 195 ft. (b = 338; so a = 338
Basketball Angles.
•21. About 5.7°. (tan B = h 10 - 7.5
=
D 25
22. About 14°. (tan B = 2.5 = 0.25, Ð B ≈ 14°.)
10
3
≈ 195.) 20. About 390 ft. [c = 2a ≈ 2(195).]
= 2.5 = 0.1, Ð B ≈ 5.7°.)
25
Equal Parts.
23. Three. 24. The altitude to the hypotenuse of a right triangle forms two triangles similar to it and to each other.
25. Three. •26. Two. 27. Five. 28. No. The longest sides of the two triangles are not equal.
Pentagon.
29. Ð BOC = 36°. ( 360 ° .)
1, 000
•30. BC = 100 ft. (
10
10
32. OC ≈ 170.1 ft. (sin 36° = 100 100
OC sin 36°
33. OB 2 + BC 2 ≈ 137.6 2 + 100 2 ≈ 28,934 and OC 2 ≈ 170.1 2 ≈ 28,934.
•34. About 408 feet. ( rD OBC ≈ 137.6 + 100 + 170.1 ≈ 408.)
.) •31. OB ≈ 137.6 ft. (tan 36° = 100
OB , OB = 100
tan 36° ≈ 137.6.)
Three Ratios.
•35. 0.87. (Letting the sides of the triangle and square be 1 unit, CF =
36. 0.75.( rD ABC = 3 and r ABDE = 4; so
r D ABC
r ABDE
= 3
4 = 0.75.)
3
37. 0.43. (aD ABC = (1) 2 and a ABDE = 1 2 aD
ABC 3
; so = ≈ 0.43.)
4 a ABDE 4
Set II (pages 478–480)
Isosceles Right Triangles.
38.
39. 2 , 2, 2 2 , 4, 3 2 , and 7 units.
2
3
CF
and EA = 1; so = 3
EA 2
≈ 0.87.)
•40. 7 units.
41. 1 + 1 + 2 + 4 + 8 + 9 + 24 1 = 49.
2 2

Two Birds.
42. •43. AF = BF. 44. 50 – x.
45. 30 2 + x 2 = 40 2 + (50 – x) 2 , 900 + x 2 = 1,600 + 2,500 – 100x + x 2 , 100x = 3,200, x = 32.
46. 32 ft and 18 ft.
•47. About 44 ft. (AF 2 = 30 2 + 32 2 = 1,924, AF ≈ 43.9, BF 2 = 40 2 +18 2 = 1,924, BF ≈ 43.9
Road Systems.
48. 3 mi.
49. Approx. 2.7 mi. In 30°­60° right D AGE, the longer leg AG = 1 ; so the shorter leg
2
Sun and Moon.
GE =
1
2 1
= and the hypotenuse AE = 2GE = 1 1
. EF =1 – 2( )= 1 –
3 2 3 3
2 3
The sum of the five segments is (1 –
50. The second solution is better.
1
) + 4( 1 ) = 1 +
3 3
1
.
3
3
= 1 + 3 ≈ 2.732.
3
51. In right D OMS, cos Ð O = m
1
. Because cos 87° ≈ 0.052 ≈ , m = 1 .
s 19 s 19
52. Because the product of the means is equal to the product of the extremes (or multiplication), s = 19m.
•53. About 380 times as far. (cos 89.85° = m ; so s = 1
s m cos 89.85° ≈ 382.)
Eye Chart.
54. 20 ft = 240 in. tan P = 3.75 = 0.015625, Ð P ≈ 0.895°; so Ð P ≈ 0.895° × 60 minutes per degree ≈ 54 minutes.
240
•55. Approximately 1
4.8
in. [4.8 minutes = ( )° = 0.08°. tan 0.08° = x
, x = 240 tan 0.08° ≈ 0.34 in.]
3 60 240
UFO Altitude.
56. 75°. (180° – 60° – 45° = 75°.) •57. About 600 ft. (
AC
=
820 820 sin 45 °
, AC = ≈ 600.)
sin 45° sin 75° sin 75 °
58. About 520 ft. (sin 60° ≈ CD ,CD ≈ 600 sin 60°≈ 520.) 59. About 735 ft. (
BC
=
820 820 sin 60 °
, BC = ≈ 735.)
600
sin 60° sin 75° sin 75 °
60. About 520 ft; so it checks. (sin 45° ≈ CD , CD ≈ 735 sin 45° ≈ 520.)
735
Outdoor Lighting.
61.
•62. tan Ð ABE = h - x .
d
63. x = h – (tan Ð ABE)d. (d tan Ð ABE = h – x.)
Number Mystery.
65.
2 2
75 - 45 = 60;
2 2
97 - 65 = 72;
64. They came from the equation for exercise 63
(and the fact that Ð ABE = 90° – A.)
2 2
481 - 319 = 360;
2 2
769 - 541 = 546.516 . . .;
2 2
2929 - 1679 = 240. With one exception, each pair of numbers is part of a Pythagorean triple;
(the lengths of one leg and the hypotenuse.)
66. As the list for exercise 65 shows, the mistake is in the pair of numbers 541, 769. (541 should be 481.)
2 2
1249 - 799 = 960;