Answers

Answers
21. x = − 9 and x = −1
22. x = −1 and x = 8
23. x = − 4 and x = 2
24. x = − 8 and x = − 5
25. x = − 6 and x = −1
26. x = − 2 and x = 8
27. x = − 8 and x = 7
28. x = − 4 and x = 7
29. x = 43
30. x = 102
31. x = 112
32. x = 18
33. x = 26
34. x = 84
35. x = 133
36. x = 26
37. x = 19
38. x = 1275
39. 34
40. 25
41. 82
42. 34
43. 8
44. 12
45. 9
46. 6
11.1 Start Thinking
10π ≈ 31.4 cm
1. 5π ≈ 15.7 cm
3.
2.
5π
≈ 7.9 cm
2
25π
≈ 13.1 cm
6
2. 90°, 15π
3. 135°, 16π
11.1 Cumulative Review Warm Up
1. 130°
2. 54
164 = 2 41
3.
b. 6
11.1 Practice A
c. 11
d. 17
49. ( − 3, − 3)
 15 7 
, 
 2 2
11 

50.  − 2, 
2

 13 7 
,− 
2
 2
1

51.  −1, − 
2

1. 21 m
2. about 169.6 ft
3. about 47.1 in.
4. 12.4 cm
5. Divide the circumference of the tree by π to find
the diameter of the tree. Because the diameter
is 50 ÷ π ≈ 15.9 inches, which is less than
18 inches, the tree is not suitable for tapping.
 11

, − 6
2

52.  −
53.  −
 5 9
55.  − , 
 2 2
 5 1
56.  − , − 
 2 2
58. 2
29
59.
233
60. 2 145
365
62.
109
63.
54. 
57.
745
569
64. 8
6. about 6.28 cm
7. about 47.1 in.
8. about 7.33 ft
9. about 36.57 mm
π
10. about 86.85 in.
11.
12. 225°
13. 1257 ft
3
11.1 Practice B
65. Sample answer: ∠ABC , ∠ABD, and ∠CBD
66. Sample answer: ∠GFJ , ∠GFH , and ∠JFH
67. 107°
218
73. 8
Chapter 11
1. 120°, 8π
48. a. 2 x + 11
70.
b. 17.6 in.
11.1 Warm Up
47. 18
61.
76. a. 7.6 in.
2
68. 113°
69. 2 130
71. 3
72.
74. 2 102
75. a. 18.4 in.
b. 44.4 in.
A108 Geometry
Answers
65
466
1.
36
π
m
2. 10.8π ft
3. about 44.0 cm
4. 160°
5. 200°
6. about 19.54 m
7. about 24.43 m
8. 280°
9. about 34.21 m
11. about 58.03 ft
10. about 114°
12. about 20.53 cm
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Answers
13.
7π
12
14. 150°
15. a. 35.6 in.
b. about 71 teeth
12. about 12.57 m2
13. about 85.13 ft 2
14. about 879.65 mm 2
15. about 21.46 cm2
11.1 Enrichment and Extension
1. a. 18π ≈ 56.55 cm
11. 804.24 yd 2
b. 6π ≈ 18.85 cm
2. 8
3. about 30.16 cm
4. about 36.85 cm
5. about 28.57 in
16. coverage area is 4 times greater;
new coverage area
7260.57 ft 2
≈
≈ 4
old coverage area
1815.14 ft 2
11.2 Practice B
11.1 Puzzle Time
1. about 143.14 in.2
2. about 0.07 mi 2
3. about 4.5 km
4. about 36.4 yd
11.2 Start Thinking
5. about 5.5 mi
6. about 10.5 mi
49π ≈ 153.9 in.
7. about 0.18 m 2 , about 0.33 m 2
SO THAT THE AUDIENCE WOULD BE GLUED TO
THEIR SEATS
2
1.
49π
≈ 77.0 in.2
2
2.
147π
≈ 115.5 in.2
4
9. about 0.39 in.2 , about 0.84 in.2
245π
3.
≈ 42.8 in.2
18
10. about 1.87 yd 2 , about 6.86 yd 2
11.2 Warm Up
1. 81π ≈ 254.5 in.2
3.
100
π
≈ 5.6 mi
2. 4π ≈ 12.6 ft 2
4. 2
42
π
≈ 7.3 m
12. about 0.6 ft
13. about 1099.56 m2
14. about 230.91 ft 2
15. about 32.99 in.2
16. about 26.18 in.3
1.
11.2 Cumulative Review Warm Up
3, y = 4
11. about 2.5 cm
11.2 Enrichment and Extension
5. 36π ≈ 113.1 cm2
1. x = 4
8. about 2.86 cm 2 , about 23.56 cm2
2. x = 7, y = 7
2
3.
3. x = 22, y = 11
11.2 Practice A
3
7π 2
m
3
5. a.
1. 145.27 ft 2
2. 289.53 cm 2
3. 18 m
4. 30 in.
5. about 10,610 people/mi2
6. about 883,573 people
7. about 21.38 ft 2 , about 132.56 ft 2
8. about 184.35 in.2 , about 346.58 in.2
9. about 1.22 m 2 , about 1.92 m 2
2. 36°
π
6. 36
7. a.
1
8
4. 8π square units
b.
3
8
c.
1
2
3 ≈ 62.4 cm 2
Measure
30° 60° 90° 120° 150° 180°
of arc, x
Area of
sector, y
b. y =
π
40
3π
4
3π
2
9π
4
3π
15π
4
9π
2
x
10. about 25.13 cm 2 , about 175.93 cm2
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Geometry A109
Answers
Answers
8. 13.0
11.2 Puzzle Time
A SCREWDRIVER
11.3 Start Thinking
1. Sample answer: To find the area of the triangle,
draw an altitude and create two congruent
30°-60°-90° triangles. Using the special right
triangle, the altitude would have a measure of
2 3 units. Using the formula for the area of a
triangle, you have
1
1
A = bh = ( 4) 2 3 = 4 3 square units.
2
2
(
)
2. Sample answer: To find the area of the regular
hexagon, connect two pairs of non-consecutive
vertices and create four congruent 30°-60°-90°
triangles and a rectangle as shown. Using the
special right triangle, you can find the measures
needed to calculate the area. The area of the hexagon
would be the sum of the areas of the four congruent
triangles and the rectangle in the center, or
1

A = 4  2 3 ( 2) + 4 4 3
2


(
)
(
)
= 24 3 square units.
9. 60°
11. 60°
12. 120°
13. 31.2 square units
14. 554.4 square units
15. 1119.6 square units
16. 178 cm2
17. 1175.6 m2
18. a. about 73.9 ft
b. 3 containers; The area of the floor is about
416.5 square feet. Because 416.5 ÷ 200 ≈ 2.08
and you cannot buy part of a container, you will
need 3 containers of wood sealer.
11.3 Practice B
1. 285.25 square units
2. 110.36 square units
3. 252.5 square units
4. 384 square units
5. 51.5°
6. 25.8°
9. 27.5 square units
8. 154.4°
10. 8.7 square units
11. 3.9 square units
12. 16 ft, 48 ft
13. 22 mm, 44 mm
4
2 3
60°
15 ft
2
8 ft
11.3 Warm Up
3.
b. 1481 ft2
5 ≈ 8.9
2. 15 sin 48° ≈ 11.1
4.6
≈ 5.4
cos 32°
11.3 Cumulative Review Warm Up
1. sometimes
2. always
3. always
4. never
5. always
6. sometimes
1. 202.5 square units
2. 54 square units
3. 126 square units
4. 120 square units
6. 22.6
A110 Geometry
Answers
15. yes; One side length of the 11-gon is
16.5
= 1.5 meters. The length of the apothem a is
11
0.75
a =
. So,
tan 16.4°
A =
1  0.75 

 (16.5) ≈ 21 square meters.
2  tan 16.4° 
11.3 Enrichment and Extension
11.3 Practice A
5. 40°
7. 64.4°
14. a. Sample answer:
120°
1. 4
10. 30°
7. 18°
1. 5.9 ft 2
3. A0 =
2. 3 +
3 2
s
4
4.
π
2
3 2
s
36
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Answers
8.
3 2
3 2
s +
s
4
12
5. 3; A1 =
2
2
6. A2 =
3 2
3 2
3 2
s +
s +
s
4
12
27
7. A3 =
3 2
3 2
3 2
4 3 2
s +
s +
s +
s
4
12
27
243
8. S =
9. An =
sphere with radius 2
9.
9
5
6
2 3 2
s square units
5
4
cone with height 6 and base radius 4
11.3 Puzzle Time
A WATCH DOG
10. yes; It is bounded by polygons only.
11.4 Start Thinking
11.
triangular prism: F = 5, V = 6, E = 9, rectangular
prism: F = 6, V = 8, E = 12, pentagonal prism:
F = 7, V = 10, E = 15, hexagonal prism: F = 8,
V = 12, E = 18; F + V − E = 2 for each of the
prisms; As you increase the number of faces by 1, the
numbers of vertices increases by 2, and the number of
edges increases by 3.
11.4 Warm Up
yes; Sample answer: parallel to a horizontal face
and parallel to a vertical face
12. a. 4 + 4
1. yes
11.4 Practice B
3. no; It is not a plane figure.
1. triangle
11.4 Cumulative Review Warm Up
1. 8
2.
29
5 ≈ 8.94 in.2
b. 4
2. no; The sides are not segments.
5 ≈ 12.94 in.
3.
313
6
4.
2. trapezoid
3. semicircle
6
6
11.4 Practice A
1. yes; rectangular prism
2. yes; triangular pyramid
3. no
4. circle
5. triangle
6. rectangle
cylinder with height 6 and base radius 6
5.
2
7.
5
3
9
cylinder with height 9 and base radius 5
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4
Sample answer: part of a cone with height 3 and
two different circular bases with radii 2 and 4
Geometry A111
Answers
Answers
6.
2.
z
(0, 0, 2)
9
(−3, 0, 0)
9
y
x
hemisphere with radius 9
7. none of the parts are polyhedrons; Sample answer:
Two parts contain some faces that are polygons, but
all three parts also contain faces that are not
polygons.
8.
3. circle
4. no; Sample answer: An oblique cylinder does not
have rotational symmetry, so you cannot draw a
shape and an axis of revolution that could form the
cylinder.
8
5.
6.
2
3
6
Sample answer: two adjacent cylinders that share
the same axis of symmetry
7.
9. a. trapezoid, triangle
b. trapezoid: perimeter = 12 + 2 13 ≈ 19.21 in.,
area = 18 in.2
triangle: perimeter = 4 + 2 13 ≈ 11.21 in.,
11.4 Puzzle Time
area = 6 in.2
c. no; Sample answer: It is flatter than a circle,
more like an oval.
BECAUSE THEY HAD JUST FINISHED A MARCH
OF THIRTY-ONE DAYS
11.5 Start Thinking
9
π
64
11.4 Enrichment and Extension
1.
cubic units;
9
π
64
cubic units; no; no; The volume
of a right solid and an oblique solid will be the same if
the height and cross-sectional area are the same.
11.5 Warm Up
z
1. 168 cm3
(0, 0, 5)
2. 480 in.3
3. 30π ≈ 94.2 cm3
11.5 Cumulative Review Warm Up
y
(0, 4, 0)
x
1. 0.6561
2. 0.9903
3. − 0.2679
4. − 0.5299
5. − 0.9925
6. − 4.3315
11.5 Practice A
A112 Geometry
Answers
1. 84 in.3
2. 864 cm3
3. about 314.16 ft 3
4. about 1130.97 yd 3
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Answers
5. about 2497 g
11.6 Start Thinking
6. 6 m
7. about 11.00 in.
8. 12 ft 2 ; Sample answer: length: 4 ft, width: 3 ft
9. 24 cm 2 ; Sample answer: length: 4 cm, width 6 cm
10. 337.5 m3
11. 323 cm3
12. about 3383.87 gallons
Sample answer: The volume of the cube is clearly
greater than the volume of the pyramid. You can see
that the pyramid would fit inside the cube. The volume
of the cube is 125 cubic inches and the volume of the
pyramid will be one third of that value.
11.6 Warm Up
1. 210 in.2
3. 45 tan 54° ≈ 61.9 cm 2
11.5 Practice B
1. 288 ft 3
2. 17.28 m3
3. about 10,433.62 in.3
4. about 1334.55 cm3
5. 3700 g
6. 4 ft
7. about 3.00 m
8. 18 yd2; Sample answer: length: 6 yd, width: 3 yd
2
11.6 Cumulative Review Warm Up
1.
(x
− 2) + ( y − 5) = 49
2.
(x
+ 3) + ( y − 9) = 9
3.
(x
− 8) + ( y + 4) = 64
4.
(x
+ 11) + ( y + 3) = 169
9. 32 in. ; Sample answer: length: 8 in., width: 4 in.
10. 378π in.3
2. 17.5 m 2
11. about 174 ft3 12. 144 in.3
11.5 Enrichment and Extension
2
2
2
2
2
2
2
2
11.6 Practice A
1. V = x3 + x 2 − 2 x
1. 240 m3
2. 845 in.3
3. 84 ft 3
2. V = π x3 + 2π x 2 + π x
4. 8 cm
5. 7.5 ft
6. 12 m
7. 32 in.3
8. 1024 mm3
9. 704 yd3
3. 7
5
4. Sample answer: The volume is equal to the surface
area, so π r 2 h = 2π r 2 + 2π rh. Solve for r to
2h
obtain r =
. If 0 < h < 2, then r < 0,
h − 2
so h must be greater than 2. Similarly, if you solve
2r
. If 0 < r < 2, then
for h, you get h =
r − 2
h < 0, so r must be greater than 2.
5. 1280
3 ≈ 2217.0 cm3
6. 125π
3 ≈ 680.2 in.3
11. 960 cm3
12. 98
, or 32 23 m
3
11.6 Practice B
1. about 367.04 cm3
2. about 96.99 ft3
3. about 643.79 yd3
4. 6.3 m
5. 12.5 in.
6. 2211.84 cm3
7. 32.175 yd3
8. 290 cm3
9. 3840 in.3
10. 750 m3
11. 8
11.6 Enrichment and Extension
1. 96 cm3
7. 736π ≈ 2312.2 cm3
8. 5 in.
10. 1155 m3
9. 5 ft
1000 2
cm3
3
3. 320 cubic units
11.5 Puzzle Time
POST OFFICE
2.
4. Sample answer: V =
1
B ( h1 + h2 );
3
V ≈ 433.3 units3
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Geometry A113
Answers
Answers
5. V =
10.91
576 + 196 +
3
(
576 • 196
)
9.
= 4029.43 cm 3
11.6 Puzzle Time
2
1
; Sample answer: The original volume is
h; r
2
2
1
1
V = π r 2 h and the new volume is V = π r 2 h.
3
6
A CAR POOL
10. about 10.05 sec
11.7 Start Thinking
11.7 Enrichment and Extension
1. a. 15π ≈ 47.1 m
50
1
250
π ≈ 52.4 in.; 8 in.;
π ≈ 261.8 in.2
3
3
3
b.
11.7 Warm Up
1. about 28.3 cm2
c. about 389.6 m2
2. about 153.3 in.2
3. about 548.2 m2
2.
11.7 Cumulative Review Warm Up
1. y = 2 x − 8
3. y =
2π 3
ft
9
3.
4. 2π ft 3
2. y = x + 9
2π 3
ft
3
5. about 191.45π m3
6. Sample answer: You are increasing the area of the
1
14
x −
5
5
base by a factor of 32 = 9, so the height of the
1
cone must be the original size.
9
11.7 Practice A
1. about 377 ft 2
2. 537.2 m2
3. about 209.44 cm3
4. about 6.76 in.3
5. 512
π mm 3
3
6. 72 π in.3
7. about 871.27 in.3
8. about 47.24 cm3
9. 3h; r
225
π ≈ 176.7 m 2
4
7. 282π ≈ 885.9 ft 2
11.7 Puzzle Time
BECAUSE THEY TAKE TOO LONG TO CHANGE
11.8 Start Thinking
SAcube = 6(9.55) ≈ 547.2 in.2 ,
2
3; Sample answer: The original volume is
V = 13 π r 2 h and the new volume is V = π r 2 h.
10. cylindrical container; The cost of the cylindrical
container is $4.75 per 96π cubic inches (or about
$0.02 per cubic inch), whereas the cost per cubic
inch for the cone-shaped container is $3.25 per
32π cubic inches (or about $0.03 per cubic inch).
11.7 Practice B
2
8. about 634.3 m2
 9.55 
 9.55 
SAcyl = 2π 
(9.55) + 2π 

 2 
 2 
2
≈ 429.8 in.2 ; no; cylinder
11.8 Warm Up
1. 15 in.
2. 5 in.
3. 3 in.
11.8 Cumulative Review Warm Up
1. radius
2. tangent
3. diameter
4. chord
5. secant
6. radius
2
1. about 593.8 yd
2. about 379.3 mm
3. about 1773.95 in.3
4. about 25.13 cm3
5. 756
π ft 3
5
6. 2744
π m3
9
1. about 50.27 in.2
2. about 1256.64 mm2
7. about 2814.87 in.3
8. about 16.20 cm3
3. about 201.06 ft2
4. 3 m
5. 9 yd
6. about 113.10 ft3
7. about 14,137.17 cm3
8. about 659.58 m3
A114 Geometry
Answers
11.8 Practice A
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Answers
9. about 4.19 in.3
10. about 9202.8 km3
3
11. about 3053.63 cm
3
12. about 791.68 ft
13. about 100.53 m2, about 134.04 m3
b. V =
14. about 12.57 yd2; The edge length of the cube,
c.
2 yards, is the diameter of the sphere.
2
6
2
4
V=
2
4
5.
r
πr
πr
3. about 254.47 in.2
4. 5 cm
11.8 Puzzle Time
5. 2.5 in.
6. about 3053.63 ft3
EGG SHELLS
7. about 575.17 cm3
8. about 4188.79 m3
Cumulative Review
12. about 83.38 in.3
13. surface area shrinks to one-fourth original size,
volume shrinks to one-eight original size
14. a. no; You also need to know the radius of one of
the cannonballs.
b. 9 lb
11.8 Enrichment and Extension
1. greater than; The volume of the box is 32 cubic
inches. The volume of the four balls is less than
17 cubic inches. So, there is more than 15 cubic
inches left over inside the box, and the volume of a
fifth ball is less than 6 cubic inches, so the volume
left over inside the box is greater than a fifth ball.
3. 1.0 in.
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4π 2 + 1
7. S = r 2 + 2π r 2 + π r 2
2. about 153.94 yd2
10. about 0.52 mi3
S
r
1. about 113.10 m2
2. 1,876,578 L
8
6. S = 4π r 2 + r 2 + r 2
11.8 Practice B
11. about 5089.38 yd3
6
The shape of the graph is similar to half a
parabola.
volume of new ball
32,000π 3
8000
=
=
volume of old ball
27,436π 3
6859
9. 288π ≈ 904.78 ft3
S Sπ
6π
2
3
8000
 40 
 20 
≈ 1.17 times the
  =   =
38
19
6859
 
 
volume of the old ball, (a 17% increase); Use
the properties of similar solids;
surface area of new ball
1600π
400
=
=
,
surface area of old ball
1444π
361
Sπ
6π
V
 40 
 20 
 400 
  =   =
 ≈ 1.11 times the
38
19
 
 
 361 
surface area of the old ball (an 11% increase),
volume of new ball is
3
S
8
15. surface area of new ball is
2
Sπ
2π
4. a. r =
2
1. 9 ft
2. 16 cm
3. 5 in.
4. 18 km
5. 11 m
6. 18 mm
7. M (1, 2)
8. M ( −1, 1)
9. M ( − 3, 4)
10. M (8, − 2)
11. H (3, 0)
12. G ( − 5, 3)
13. G ( −10, 5)
14. H ( − 6, 8)
15. 1
16.
386
19.
181
17.
21. ∠XYZ , ∠ZYX , ∠Y
157
18.
74
20. 6
22. ∠JKL, ∠LKJ , ∠K
23. Sample answer: consecutive even integers
beginning with 2; 12, 14
24. Sample answer: consecutive odd integers beginning
with −1, alternatively negative and positive; 11,
−13
Geometry A115
Answers
Answers
25. Sample answer: skip two letters in the alphabet,
beginning with A; P, S
26. Sample answer: skip one letter in the alphabet,
beginning with B; L, N
67. A′( 27, − 21), B′( 21, 0), C ′(0, −12)
68. X ′( − 4, − 20), Y ′(0, 22), Z ′( 22, − 22)
69. M ′( − 2, 0), N ′( − 4, 3), P′(3, 2)
27. x = −1
28. x = 1
29. x = − 6
30. x = 1
31. x = 2
32. x = 3
71. D′(32, 8), E ′(32, − 40), F ′( 28, 16), G ′( 24, 4)
33. x = − 7
34. x = 1
72. a. 50.3 in.
35. y = − 4 x − 7
36. y = 5 x + 2
37. y = x + 8
38. y = −10 x + 6
39. y = 35 x − 50
40. y = 9 x − 6
41. x = 3, y = 4
42. x = 15, y = 18
43. x = 15, y = 10
44. x = 21, y = 26
45. 54
46. 47
47. 26
48. 7
49. y = 3x − 39
50. y = − 8 x − 41
51. y = 4 x + 36
52. y = − 2 x − 1
53. y =
1
x −3
4
2
3
70. Q′( − 5, − 4), R′(0, − 6), S ′( 4, 5), T ′(3, − 6)
b. 56 in.
c. 54 in.
73. a. 50.3 cm2
b. 49 cm2
c. 48 cm2
d. the circle
54. y = x − 11
55. y = − x − 7
56. y = − 8 x + 26
57. A′( − 9, 2)
58. B′(6, − 3)
59. C ′( − 6, 7)
60. D (9, 1)
61. E ( −11, − 3)
62. P′( −10, 11), Q′( − 6, 13), R′( − 2, 9)
63. P′(5, − 4), Q′(9, − 2), R′(13, − 6)
64. P′( −13, −1), Q′( − 9, 1), R′( − 5, − 3)
65. P′( −1, − 6), Q′(3, − 4), R′(7, − 8)
66. P′( − 9, −1), Q′( − 5, 1), R′( −1, − 3)
A116 Geometry
Answers
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