# chapter 12

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9:18 AM
Page 245
CHAPTER 12
Application (p. 717)
1.
17. Not regular because not all faces are congruent polygons.
2402 10,053 ft2
Convex; any two points on its surface can be connected
by a segment that lies entirely inside or on the
polyhedron.
18. Not regular because not all faces are congruent polygons.
Skill Review (p. 718)
1
1
1. 2:1 2. 3:4 3. A 4 342 4. A 2 33 6
6.9 in.2
5. A 1
2 1.28
Not convex because some segments which connect two
points lie outside the polyhedron.
6
93.5 m2
1
19. False; a polyhedron can be convex without being regular.
See Example 2(b) on p. 720.
20. False; a polyhedron must have at least 4 faces.
4.8 ft2
21. True; a cube is composed of six congruent squares.
22. True; a tetrahedron has exactly four faces.
Lesson 12.1
23. False; a cone is not a solid bounded by polygons.
12.1 Guided Practice (p. 723)
24. True; an example would be a square pyramid
1. Sample Answer: A group of polygons put together in
such a way as to enclose a single region of space.
2. Yes; yes; a segment connecting any two points on the
surface of any Platonic solid lies entirely inside or on the
solid.
3. Polyhedron; each face is a polygon and the solid encloses
or a triangular prism.
25. circle
26. circle
27. pentagon
29. circle
30. rectangle or square
28. rectangle
31. rectangle
32. pentagon
33. yes
34. square
a single region of space.
4. Polyhedron; each face is a polygon and the solid encloses
a single region of space.
5. Not a polyhedron because some of its faces are not
polygons.
7. F V E 2
6. F V E 2
8.
F 6 12 2
5V92
F8
V6
FVE2
FVE2
9.
F 10 15 2
20 12 E 2
F7
30 E
35. yes
36. tetrahedron
37. octahedron
38. dodecahedron
12.1 Practice and Applications (pp. 723–726)
39. dodecahedron
10. No; not every face is a polygon.
40. octahedron
11. Yes; each face is a polygon and the solid encloses
42.
41. cube
43. F V E 2
a single region of space.
5692
12. Yes; each face is a polygon and the solid encloses
11 11
a single region of space.
13. F 5
V5
14. F 8
V 12
15. F 10
V 16
E8
E 18
E 24
16. Regular; all faces are congruent regular polygons.
Convex; any two points on its surface can be connected
by a segment that lies entirely inside or on the polyhedron.
regular octahedron
44. F V E 2
45. F V E 2
5582
5692
10 10
11 11
Geometry
Chapter 12 Worked-out Solution Key
245
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Chapter 12 continued
46.
Platonic solid
Tetrahedron
Cube
Octahedron
Dodecahedron
Icosahedron
Faces
4
6
8
12
20
E 1220
52. E 2 12
1
Edges
6
12
12
30
30
3
48.
30 edges
FVE2
20 V 30 2
12 V 30 2
V 20
55. B.
FVE2
F 12 18 2
F8
3 6 4
1
2 24 1
2 48
24
V 12
53. 6 molecules
30
54. C.
E 128
5
1
2 60
FVE2
FVE2
4462
6 8 12 2
8 6 12 2
12 20 30 2
20 12 30 2
Platonic solid
Tetrahedron
Cube
Octahedron
Dodecahedron
Icosahedron
47.
Vertices
4
8
6
20
12
QS2 h2 h2
QS2 h2 h2
4
4
QS2 h2
2
QS 24
56.
57.
58.
59.
2
2
h
2
FVE2
14 V 24 2
V 12
E
49.
1
2 8 6 6
1
2 48 24
1
2 72
4
36
FVE2
14 V 36 2
V 24
E 1218
50.
4 8 3
1
2 72 1
2 96
24
48
FVE2
26 V 48 2
V 24
E 124
51.
6 4 3
1224 12
1236
18
FVE2
8 V 18 2
V 12
246
Geometry
Chapter 12 Worked-out Solution Key
Lesson 12.1
12.1 Mixed Review (p. 726)
61. A 2 hb1 b2
1
60. A bh
128 121621 14
96 in.2
835 280 ft2
1
1
62. A 2 d1d2
63. A 2 aP
124930
124.648
735 m2
110.40 m2
64. A 1
2 aP
1
2 4.2228
59.08 ft2
66. A 6
143s 2
6
A 3s2
4
1
65. A 4 3s2
14382
27.71 cm2
1
2
67. A aP
1
8
16 12
2 tan 15
6
34 ft2
4
4
192
tan 15
41.57 ft2
2866.22 in.2
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Chapter 12 continued
68. A 115
360
72
115
360
49
The lateral area is the area of 3 rectangles, one 3 cm by 2
cm, one 4 cm by 2 cm, and one 5 cm by 2 cm, for a total
of 24 cm2. The surface area is the lateral area plus the
area of the bases, 2 right triangles with a total area of
12 cm2. So the surface area is 36 cm2.
69. A 432
1849
5808.80 ft2
49.17 cm2
3. right cylinder
180 140
70. A 322
360
40
360
4. triangular prism 5. rectangular prism
6. top and bottom bases: 22 cm
front and back bases: 16 cm
1024
left and right bases: 26 cm
7. top and bottom bases: 5 cm
357.44 in.2
front and back bases: 8 cm
12.2 Developing Concepts (p. 727)
left and right bases: 3 cm
1.
S.A. Area A Area B Area C Area D Area E 8. top and bottom bases: 110 cm2
front and back bases: 128 cm2
Area F
3
7 5 7 3 5 7 5 3 5 3 7
left and right bases: 78 cm2
9. top and bottom bases: 24 cm2
front and back bases: 15 cm2
21 35 15 35 15 21
left and right bases: 40 cm2
142 sq units
2. A 3
10. S.A. 2B Ph
7 21 sq units
224 cm2 110 cm2
P 23 27 20 units
158 cm2
h 5 units
3. 2A Ph 221 205
14 m
42 100
5.9 ft
142 sq units
22 m
They are equal.
or 2A Ph
w 2w h 2l h
12.2 Guided Practice (p. 731)
3.4 ft
3.4 ft
12.2 Practice and Applications (p. 732–734)
1. Sample Answer: Both have congruent faces which are in
parallel planes. In both cases, the surface area is the sum
of the area of the bases and the area of the lateral surface(s). A cylinder has bases which are not polygons
whereas a prism does. The lateral area of a cylinder is the
area of its curved surface. For a polyhedron, the lateral
area is the sum of the areas of the lateral faces.
13. right hexagonal prism
17. pentagonal prism
598 in.2
5 cm
4 cm
2 cm
2 cm
21.
S.A. 2B Ph
22
3 cm
3 cm
11 18 2210
198 400
5 cm
4 cm
5 cm
18. cylinder
20. S.A. 2B Ph
29
3 cm
15. rectangle
19. triangular prism
3 cm
14. six
16. VF, TE, SD, RC, QB, PA (any four)
9 18 47
36 154
190 m2
Geometry
Chapter 12 Worked-out Solution Key
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Chapter 12 continued
22. S.A. 2B Ph
S.A. 2 r 2 2 rh
31. 2.4 in.
1
2
3 142 426
4
21.22 21.26.1
2.88 14.64
6.1 in.
983 252
17.52
421.74 ft2
55.04 in.2
23. S.A. 2B Ph
32.
212 42 4 167.2
298 27
162 115.2
242 22x
24. S.A. 2B Ph
11 ft x
2 12 26.4 15.112.9
33.
12.8 43.82
S 2 r 2 2 rh
1202 27.5 2 27.5 z
26. S.A. 2 r 2 2 rh
1202 112.5 15z
26 ft2 26 ft11 ft
848.57 15z
72 132
18.01 in. z
204
35. For h 1 in.
640.88 ft2
S 2B Ph
27. S.A. 2 r 2 2 rh
28 cm2 28 cm8 cm
128 128
22
2 4.21
88
16 in.2
256
For h 2 in.
804.25 cm2
S 2B Ph
28. S.A. 2 r 2 2 rh
22
23.12 23.110
2 4 22
8 16
81.22
24 in.2
255.16 in.2
Doubling the height does not double the surface area.
S.A. 2B Ph
218 18
10
36 180
3 ft
27 m y
34.
93.98 in.2
30.
810 30y
14 3 22 2 66.1
20.78 73.2
6 ft
12 5 5 12 13y
870 60 30y
25. S.A. 2B Ph
10 ft
S 2B Ph
870 2 12
56.62 cm2
29.
4 22x
298 56 22x
137.83 m2
26
S 2B Ph
216 ft2
12 mm
36. For h 1 in.
S 2B Ph
214332 3
31
7.79 9
17 in.2
For h 2 in.
S 2B Ph
S.A. 2B Ph
26
14 3 122 12 612
748.25 864
1612.25 mm2
248
Geometry
Chapter 12 Worked-out Solution Key
214332 3
32
7.79 18
26 in.2
Doubling the height does not double the surface area.
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Chapter 12 continued
37. For h 1 in.
46. S 2 r 2 2 rh
S 2B Ph
26
1
2
4 32
232 238
6 21
18 48
20.78 12
66
33 in.2
207 in.2
For h 2 in.
47. Sample Answer: Let r, h, and S be the radius, height, and
S 2B Ph
26
1
2
4 32
surface area of the original cylinder. Then the surface
area of the larger cylinder is 22r2 22r2h 6 22
20.78 24
45 in.2
Doubling the height does not double the surface area.
8r 2 8rh 42r 2 2rh 4S.
48. S 24
6 25 6 22 6 23 6
220 6
196 cm2
49. S 21.56 20.56 21.52 20.52
18 6 4.5 0.5
28
87.96 in2
50.
mA 61
51.
5
tan 29 BC
BC 41. Sample Answer: At least two sides would be longer or
BC 9.02
wider than the corresponding sides of the folded box.
Several sides have to fold over to enclose the box and
make its structure more rigid.
5
sin 29 AB
42. S 2B Ph
2642 64
AB 4414
43. L.A. 2rh
52.
214
25 in.2
AB 44. A B Ph B Ph
6 12 3 11 6 113
2
314.37 198 64.95 90
667 in.2
667 in.2
6 cans
130 in.2 can
45. S 2 r 2 rh
2
AB 10.5
AB
10.5
sin 32
AB 19.81
tan 32 BC 10.5
BC
10.5
tan 32
BC 16.80
12
AB
12
cos 46
AB 17.27
2
6 14 3 5 6 53
sin 32 mB 44
cos 46 8
5
sin 29
AB 10.31
114,176 m2
5
tan 29
mA 58
tan 46 BC
12
tan 46 BC
12.43 BC
1
53. A 2 aP
1219 cos 36219 sin 365
1805 cos 36 sin 36
858.33 m2
21.52 21.54
4.5 12
16.5
52 in.2
Geometry
Chapter 12 Worked-out Solution Key
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Chapter 12 continued
55. A 64 3 s 2
54. A r 2
1
1
12. S B 2 Pl
14 ft2
614 3 82
9 12122.5
196
166.28 in.2
24 m2
1
13. S B 2 Pl
615.75 ft2
253 123012
8
QS
9
PU
16
41% 57. P 73%
PW
22
PW
22
11
56. P 58. P 7
QU
14
64%
PW
22
11
59. P 4
TW
8
36%
PW
22
11
223.30 in.2
12.3 Practice and Applications (pp. 738–741)
l2 122 42
14.
l2 160
PV
20
10
60. P 91%
PW
22
11
l 410 m
Lateral Area 12 bl
128410
Lesson 12.3
50.6 m2
12.3 Guided Practice (p. 738)
l2 222 112
15.
1. Sample Answer: All the faces of a pyramid are polygons
l2 605
and two lateral faces intersect in a lateral edge. A cone
has no lateral faces. Its base is a and its lateral surface
is curved, so it has no lateral edges. Both are space
figures with a vertex and a base. The altitude of each is
the distance from the vertex to the plane containing the
base.
2. No; all the faces of any pyramid are triangles.
3. C
4. E
5. B
6. A
l 115 in.
Lateral Area 12bl
12 22115
270.6 in.2
l 2 22 7.12
16.
l 2 54.41
7. D
l 54.41 ft
ART FOR EXERCISES 8–11
Lateral Area 12 bl
12454.41
14.8 ft2
l
7 ft
17. S B 1
2 Pl
11.22 124
3 ft
8. A r 2
9. l 2 72 32
32
l 2 58
9
l 7.62 ft
28.27 ft2
10. Lateral Area r l
37.62
22.86
71.82 ft2
506.24
18.
l2
11.217
mm2
132 42
l2 153
l 317
S B 12 Pl
1288 cos 30 123
8317 27.71 148.43
176.14 cm2
11. S.A. r 2 rl
32 37.62
9 22.86
31.86
100.09 ft2
250
Geometry
Chapter 12 Worked-out Solution Key
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Chapter 12 continued
19. 2 92 2.752
26. regular square pyramid
2 7 2 3.5 2
2 73.4375
l2 36.75
73.4375
36.75
1
S B Pl
2
S B 12Pl
72 127
1
1
aP Pl
2
2
1 2.75
5.5
2 tan 30
1
6 25.5
73.4375 49 84.87
6 133.9 ft2
27. right cone
S r 2 rl
78.59 141.40
22 26
219.99 cm2
4 12
20. 2 142 82
16
2 260
50.3 cm2
260
16.1 in.
436.75 28.
21. 2 9.22 5.62
12.2 cm
2 116
12 cm
116
10.78 cm
22.
2
2 22
8 cm
2
2 6
S B 12Pl
6
14382 123
2.4 ft
27.71 146.4
23. S r 2 rl
7.82 7.810
60.84 78
138.84 m2
812.2
174.1 cm2
29.
5.8 m
6.2 m
24. S r 2 rl
5.92 5.910
3m
34.81 59
93.81 mm2
25. 2 4.5 2 112
2 141.25
141.25
S r 2 rl
4.52 4.5141.25
20.25 53.48
73.73 in.2
S B 12Pl
6 14
332 126 36.2
23.38 55.8
79.2 m2
30. S r 2 rl
5.52 5.57.2
30.25 39.6
69.85
219.4 ft2
7.2 ft
11 ft
Geometry
Chapter 12 Worked-out Solution Key
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Chapter 12 continued
31. 2 122 92
S r 2 rl
36.
2 225
75.4 32 3y
15
75.4 9 3 y
47.13 3 y
S r2 rl
9 915
47.13
y
3
2
81 135
5 in. y
216
y 2 32
x2
678.58 in.2
x2 25 9
x2 16
12 in.
x 4 in.
9 in.
333 126.142 1242l
32. 2 62 8.82
333 128.1 21l
2 113.44
204.9 21l
113.44
9.8 m l
S 2 12Pl
2[124
S B 12Pl
37.
h2
12113.44 ]
2 6.12
h2 9.82 6.12
4810.65
h2 58.83 m
511.2 sq. units
h 7.7 m
33. 2 1.52 32
1
38. Lateral Area 2 Pl
2 11.25
1228
11.25
56520 S B Ph 412 b 32 126 4 3 11.25
1
2
81 20.1
101.1 sq. units
34.
2
32 42
2 25
S r2 2rh r l
32 2310 35
9 60 15
84
263.9 sq. units
35. S 72 4 18
p 18 2 9 cm
q2 p2 122
92
122
q2 225
q 15 cm
252
1277 cm2
1
39. S B 2 Pl
707.752 124
471
2
707.75 35378 2
500,910.06 1415.5589.125
1,334,817 ft2
5
q2
4182 142 Geometry
Chapter 12 Worked-out Solution Key
40. Base 704 ft
Height 446 ft
S B 12Pl
7042 124
704446 2 3522 495,616 1408568.1725
1,295,603 ft2
The surface area is about 39,214 ft2 less.
41. Lateral Area rl
812
96
302 in.2
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Chapter 12 continued
4.5
sin 30
r
42.
12 ft
9 ft r
7.79 ft
9 ft
4.5
r
sin 30
1
1
aP P
2
2
1 0.3825
8
2 tan 22.5
l2
225
122
49. Sample Answer: The surface area increases. As the
number of sides increases, the surface area will
approach the surface area of the cone, 1 3 .
l 15
lateral area r l
12.3 Mixed Review (p. 741)
915
1
50. A 4 3 s 2
135
424 ft2
1
51. A 2 aP
1
2
4 3 21
12582
190.96 sq units
43. The surface area of the cup is about
the original circle. mABC 29.
1
4
the surface area of
Column A: Area of base 42 16
52. A 6
1
4
3 3
2
23.38 sq units
46. B
Column A: Lateral Area 12P 121621 36.66
Column B: Lateral Area rl 3.1435 47.12
47. S r 2 rl
12 12 12 2
82.84 sq units
53.
A 12 r 2
190 123.14r 2
11 r ; about 11 in.
Column A: Lateral edge length 42 32 5
Column B: Slant height 42 32 5
12.3 Quiz 1 (p. 742)
1. regular; convex
E
1
2 4
2. not regular, convex
3 6
E 124
E 14
4V62
FVE2
V4
8 V 14 2
V8
3. not regular, nonconvex
E 122
8.58 sq units
6 6 4 18
FVE2
48. Square
8 V 18 2
S B 12Pl
1.4142 124
1.4141.58
6.47 sq units
V 12
4. S 2B Ph
Hexagon
214
1
S B Pl
2
336.44 ft2
1
1
aP P
2
2
3 4 4
FVE2
3
1 3
5 tan 22.5
121 r 2
Column B: Area of base 32 28.27
45. C
1
0.765 28 0.7651.68
7.97 sq units
l2
44. B
1
S B 2Pl
30
92
Octagon
12 tan0.530 6.1 216.11.65
7.55 sq units
3 72 2114
42.44 294
1
5. S B 2 Pl
102 124
1092 52 100 205.91
305.91 m2
6. S r 2 r l
92 992 162
81 165.22
246.22
773.52 mm2
Geometry
Chapter 12 Worked-out Solution Key
253
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Chapter 12 continued
13. V Bh
12.3 Math & History (p. 742)
14. V 4512
888
1. Length 4 16 20 in.
240 cm3
512 in.3
Width 6 12 6 12 2 38 in.
15. V Bh
Area 2038
6
760 in.2
1
2
4 33.5
10 in.
16. V r 2h
12215
318.26 in.
2160
3
2. 12 in. by 6 in.;
6785.84 m3
S.A. B Ph
6
17. V r 2h
12 212 2616
18. V r 2h
72 576
3210.2
3.529.9
648 in.2
91.8
121.275
288.40
381.00 cm3
ft3
V Bh
19.
16 in.
4215
6 in.
12 in.
240 m3
15 m
Lesson 12.4
4m
4m
12.4 Guided Practice (p. 746)
V Bh
20.
1. square units; cubic units
2423
2. The two stacks have the same height and the same cross-
72 ft3
sectional area at every level. Therefore, by Cavalieri’s
Principle, the two stacks have the same volume.
Vl
w
h
3
3
0.01500
3 ft
21.
45 in.3
4.
5.
6.
7. V h
5
10
5.5
3t
Volume
255
160
161.04
54t3
r2h
8c
m
w
3
8
6.1
3t
60
1438211.2
w
h
10810
540
800 in.
310.38 cm3
22.
3
14 in.
840
V r2h
4m
8m
1696.46 in.3
w
428
128
h
6 in.
cm
8 cm
16 15
9. V l
.2
11
V Bh
8. V l
2
m
8c
l
17
2
4.8
6t
3.
30
402.12 m3
10 in.
in.3
12.4 Practice and Applications (pp. 746–749)
10. 120 unit cubes; Sample Answer: 3 layers of 4 rows of 10
cubes each
11. 100 unit cubes; Sample Answer: 4 layers of 5 rows of 5
23.
21.4 ft
V r 2h
21.4t233.7
33.7 ft
15433.252
48,484.99 ft3
cubes each
12. 36 unit cubes; Sample Answer: 6 layers of 3 rows of 2
cubes each
254
Geometry
Chapter 12 Worked-out Solution Key
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Page 255
Chapter 12 continued
24.
37. V 8211 32112
V r 2h
15 in.
7.5 26
704 99
1462.5
605
2
26 in.
4594.58 in.
1900.66 in.3
3
38. First, convert measurements to yards.
25. V Bh
26. V r 2h
V Bh
61114
5 8
924 m3
628.32 ft3
2
27. V Bh
28.
50 6
100
3 3 36 V Bh
V v3
29.
560 56u
10 ft u
2700 v3
3 100 v
3
13.92 yd v
V Bh
80 1285w
80 20w
4 cm w
31.
V Bh
72.66 614322x
72.66 10.39x
6.99 in. x
32.
V r 2h
3000 9.32 y
3000 271.72y
11.04 ft y
33.
6.67 yd
V Bh
cm3
560 78u
30.
4
39. First, convert measurements to yards.
3315 sin 60
116.91
10636 92.59 yd
40. First, convert measurements to yards.
V r 2h
8
8236
44.68 yd
41. No; the circumference of the base of the shorter cylinder
is 11 in., so the radius is about 1.75 in. and the volume is
about 82 in.3. The circumference of the base of the taller
cylinder is 8.5 in., so the radius is about 1.35 in. and the
42. Vblock 10920
1800 cm3
Vcyl 4.5212
763.4 cm3
1800 cm3
2 candles
763.4 cm3
43. Vprism 2 6810
1
240 cm3
V r 2h
1696.5 z215
From Ex. 42, the volume of the block is 1800 cm3, so
1696.5 47.12z2
1800 cm3
7 candles.
240 cm3
36 z2
6mz
34. V 1.310.660.66 20.330.390.66
0.57 0.17
0.40 ft3
44. V 6
1
433212
280.6 cm3
1800 cm3
6 candles
280.6 cm3
They are the same.
35. V 1062 325
120 30
150 ft3
36. V 12.47.89 1.839
870.48 48.6
821.88 m3
Geometry
Chapter 12 Worked-out Solution Key
255
mcrbg-1203-sk.qxd
6-18-2001
9:29 AM
Page 256
Chapter 12 continued
45. Vprism Bh
Vcyl r 2h
433
3.141.5 in.25.1 in.
36
36.05
in.3
s 72
in.3
A
s2
2
66 in.2
Scyl 250.28
8.5 tan 30
in.2
72 3 124
2r 2
128.562
s2 98
Sprism 2B Ph
23
1
56. A 2 aP
55. s2 72 72
98 m2
2rh
57.
21.5 21.55.1
2
13 in.
4.5 15.3
19.8
12 in.
62.20 in.2
3 in.
They both have approximately the same volume, but the
surface area of the cylinder is less so less metal is needed, thus reducing the cost.
46. V r 2h
12.42 12.417
28,902.65 ft3
153.76 210.8
28,902.65 ft3
216,175 gallons
0.1337 ft3 gal
364.56
1145.30 ft2
47. 216,175 gal8.56 lb 1,850,458 lb
49. E
4
1l
x
l
1 4x
3 3013
58. S r 2 rl
9200
V Bh
212
462 in.2
20223
48. A
S 2B Ph
1
V r 2h
6210
360
59. S B 2 Pl
62 124
144
1
60. S B 2 Pl
6
1
x
4
69
cm2
1
2
4 34
126
48
137.57 in.2
50. Vhor r 2h
Vvert r 2h
325
523
45 in.3
75 in.3
12.4 Activity (p. 750)
1. r 2.25 and h 4.53; for values of r less than 2.25, S
decreases and for values greater than 2.25, S increases.
The solid about the vertical line has the greater volume
because volume is proportional to the square of the
Lesson 12.5
Developing Concepts 12.5 (p. 751)
1. The area of the bases are the same: 4 in.2.
Lesson 12.4
2. The heights are the same.
12.4 Mixed Review (p. 749)
51. 2x 5x 5x 180
52. x 2x 3x 180
12x 180
6x 180
x 15
x 30
30, 60, 90
30, 75, 75
53. 3x 4x 5x 180
12x 180
5.062
x 15
25.6036
45, 60, 75
256
54. A r 2
3. The volume of the prism is greater; Sample answer:
4. The pyramid was filled about 3 times. So, the volume of
the prism is 3 times greater.
Extension: The volume of a pyramid with base of area B
and height h is equal to 13 Bh.
80.44 ft2
Geometry
Chapter 12 Worked-out Solution Key
mcrbg-1203-sk.qxd
6-18-2001
9:29 AM
Page 257
Chapter 12 continued
12.5 Guided Practice (p. 755)
22.
53 13143 23 h
2
1. cylinder
2. Yes; according to Cavalieri’s Principle, if two solids have
the same height and the same cross-sectional area at
every level, then they have the same volume.
3. Yes; the volume of the rectangle prism is y2x. The
1
volume of the pyramid is 3y23x y2x.
2
2
4. (a) A 5 25 cm
1
1
100
(b) V 3Bh 3 254 3 cm3 33.3 cm3
53 3h
5 cm h
23. V Vcube Vpyramid
V 63 13626
288 ft3
24. V 2 3 Bh
1
V 2 3.322.3
5. (a) A 22 12.57 ft2
(b)
6. (a)
V 13Bh
1
3
V 13Bh 1312.57 ft24 ft 16.76 ft3
A 143112 52.39 m2
V 13Bh 1352.3914 244.51 m3
(b)
7. The height of the cone would be equal to
slant height2 radius2. Solve by using the
Pythagorean Theorem.
16.70 cm3
25. V Vcube Vcone
V 5.13 133.142.5525.1
132.65 34.73
97.92 m3
26. V Vcyl Vcone
12.5 Practice and Applications (pp. 755–757)
2.527.5 132.524
8. A 910.1 90.9 in.
46.875 8.33
9. A r2 3.146.12 116.9 ft2
55.205
2
10. A 6
11. V 1
2
4 318
841.8 mm2
1
3 Bh
1
2
3 10 12
173.43 in.3
12. V 400 cm3
1
3 Bh
1
2
3 7 5
81.7 m3
1
1 1
2
13. V 3 Bh 3 4 39.2 12.7 155.2 ft3
14. V 15. V 16. V 17. V 1
1 1
2
3
3 Bh 3 4 3 14 18 509.2 in.
1
1
1
2
3
3 Bh 3 6 4 3 10 14.2 1229.8 mm
1
1
1
2
3
3 Bh 3 6 4 3 12 20 2494.2 cm
1
2
3 r h
1
2
2
2
3 3 6 3
173.43 in.3
12 cups
14.4 in.3 cup
27.
The feeder will hold enough food for 3 days.
1
28. V 3 Bh
133.144.152 10.12 4.152 18.04 84.79
166 cm3
93
48.97 ft3
166 cm327 4482.0 cm3
1
29. V 3 Bh
133.145210 cm
1
18. V 3 r 2h
261.8 cm3
135.75215.2
167.517
526.27
19. V 12 cup23 3 cups
80 65 15 mL s in the funnel after 1 second.
261.8 cm3
17.5 seconds
15 cm3 s
cm3
1
2
3 r h
1
2
3 7 13
30.
The solid is a triangular prism with
congruent base and faces that are
equilateral triangles with edge length
5 feet.
212.333
667.06 in.3
20.
V 13Bh
21.
1
270 392h
270 27 h
10 m h
V 13Bh
100 13r212
25 r 2
5 in. r
5 ft
Let x the apothem of the base
triangle and y the radius of the
base triangle. Then half of a base
edge will be the longer leg in a
30 60 90.
5
5
y
30
x
5
2
—CONTINUED—
Geometry
Chapter 12 Worked-out Solution Key
257
mcrbg-1203-sk.qxd
6-18-2001
9:29 AM
Page 258
Chapter 12 continued
38. V area base1 area base2 geometric mean 3 h
1
30. —CONTINUED—
22 62 22
5
y 2x
x3
2
5
53
x
2
23
6
53
53
x
6
3
The height of the prism is one leg
of a right triangle with hypotenuse
an edge of the prism and other leg
a radius of the base triangle.
4 36 37.703
52 h2 53
3
2
1
1
52
3
4
1252
12
56
h
3
163.364 3
490.09 ft3
39. V 5
h
12.5 Mixed Review (p. 758)
5 3
3
3 56
3
14.73 ft3
V 13r 2 h
31.
133.14628
10 cm
301.59 cm3
40.
9 2180
140 m int.; m ext. 40
9
41.
10 2180
144 m int.; m ext. 36
10
42.
19 2180
161 m int.; m ext. 19
19
43.
22 2180
163.6 m int.; m ext. 16.4
22
44.
25 2180
165.6 m int.; m ext. 14.4
25
45.
30 2180
168 m int.; m ext. 12
30
46. A r 2 12.52 490.87 in.2
6 cm
47. A r 2 16.32 834.69 cm2
V 13r 2h
32.
133.1442 43
8m
116.08 m3
48.
C 2r
48 2r
22 26 50.3 m3
V 0.043
0.4 mi
h
1.83 mi
m arc
360
2 36
360
mi3
V 13r 2h
0.043 133.140.42h
0.043 0.17h
3 mi
not drawn to scale
1809.56 ft2
49. Arc length 1
1
2
3 r h 3 3.14
34.
A 242
24 r
4m
33. V 13 hr12 r22 r1r2
1
3hr12 r22 r1r2
1
V Bh
3
25
25 h2 3
50
h2
3
62139
2r
A r2
102
2r
314.16 m2
10 r
50. E 2 12
1
8 20 3
78
FVE2
32 V 78 2
0.25 mi h
1.83 0.25 1.58 miles
1
35. V 3 r 2 h
133.143.923.9
V 48
51. E 1
2 6
4 8 6
36
FVE2
14 V 36 2
V 24
62.12 in.3
36. Sample answer: No. It would take 62.12 min. for the sand
to fall through not 60 minutes.
37. Sample answer: Assume that half the sand has
accumulated after 30 min. Let r be the radius of the pile
containing all the sand and r1 the radius of the pile
containing half. 13r13 1213 r 3, so r13 12r 3 and
r1 258
3 1 r.
2
Geometry
Chapter 12 Worked-out Solution Key
12.5 Quiz 2 (p. 758)
1. V Bh
2. V Bh
1517
10186
128
1080
1020 ft3
in.3
3. V r 2 h
1
4. V 3 r 2 h
5214
134.529
1099.56 cm3
190.85 m3
mcrbg-1206-sk.qxd
5-25-2001
11:34 AM
Page 259
Chapter 12 continued
1
19. S d 2
1
5. V 3 Bh
6. V 3 Bh
143729
1342236
13
21,168 mm3
63.65 in.3
Neptune: S 30,7752 2,975,404,400 mi2
7. V volume prism volume pyramid
1
V Bh Bh
3
1
2
5
tan 22.5
1 1
80 8 3 2
5
tan 22.5
8011
3862.74 1770.42 5633 ft3
Nereid:
S 2112 139,900 mi2
Note: Values for the diameter may vary depending on the
source.
4
4
20. V 3 r3
21. V 3 r3
4
3
3 22
432.53
44,602.24 cm3
22. V 12.6 Guided Practice (p. 762)
1
2
4
3
3 r
3. TR, TS, or QS
4
3
3 r
4
3 9.1
65.45 in.3
mm3
3156.55 mm3
1. equidistant; point
the diameter or 5 mm.
of sphere of great circle
of sphere
r 2.
not
7 mm
14 mm
196 mm2
1372
mm3
3
43.1432
24.
6 in.
12 in.
144 in.2
288 in.3
113.1 sq. units
25.
5 cm
10 cm
100 cm2
500
cm3
3
26.
10 m
20 m
400 m2
4000 3
m
3
5. QS
7. S 4r2
23
18.85 units
4
3
3 r
9. V Volume
of sphere
23.
4. PQ, PS, or PR
6. C 2r
8. V S 16802 8,866,800 mi2
Lesson 12.6
Volume Triton:
4
3
3 r
4
3.140.5 3
433.1433
108 cm3
113.1 cu. units
5.24 1025 cm3
27. a. S r2 2rh 2 4r2
1
12.6 Practice and Applications (p. 762–765)
10. S 4r2
4.82 24.89 1244.82
11. S 4r2
43.146.52
43.14182
530.93
4071.50
m2
12. S 4r2
488.58 in.2
13. hemisphere
28. a. S r2 2rh 2 4r2
1
ft2
C 2r
14.
15.
29. a. S rl 2 4r2
1
17. S 4r2
18. SEarth 4r2
1 4
3
2 3 r
10218 23103
b. V r2h 7749.26 cm3
86.02 in.2
24,903
2
2073.45 cm2
7.4 in.
1
2
2 4r 1
2
2 43.143.7
4
102 21018 2102
d 2r
23.7 in.
7.4 2r
7.4
r
2
3.7 in. r
16. S 4.83
419.82 in.3
43.143.852
186.27
1 4
3
2 3 r
2
4.8 9 12 43
b. V r2h cm2
43.141.92
5.113.22 25.12
45.4 in.2
375.24 ft2
1
b. V 3 r2h Smoon 4r2
2
197,402,900
mi2
41077.52
14,589,600
1 4
3
2 3 r
1
2
3 5.1 12.2
235.13
610.12 ft3
mi2
The surface area of Earth is about 13.5 times greater than
the surface area of the moon.
Geometry
Chapter 12 Worked-out Solution Key
259
mcrbg-1206-sk.qxd
5-25-2001
11:34 AM
Page 260
Chapter 12 continued
30.
1 meter
V
2 meter
4
3
3 r
4
3
3 1
4
3
V
S 4r2
3 meter
4
3
3 r
4
3
3 2
32
3
V 43r3
S 4r2
422
432
4
16
36
4 meter
5 meter
V
4
3
3 r
4
3
3 4
256
3 V
S 4r2
64
100
1m
z 12
A 122
452.39 sq. units
No. You would need to know the
distance from the plane to the center of the sphere. Only then could
the radius of the sphere be determined and from that, the surface
area.
38. S 4r2
4
2m
4
3
V 36
S
36
V
S
16
1
3
4m
39. 267.31000 267,300 ft2
4
40. V 3 r3
433.14
1
2
3
No; the surface of the building
does not appear to be smooth.
165 2
2
85,529.86 ft2
3m
32
3
V
S
4
41. Vslug r2h
165 3
2
326
54
2,352,071 ft3
5m
256
V
3
S
64
2
S 4r2
452
6 63
36. z 2
4
3
3 r
4
3
3 5
500
3 442
31.
3.1422
12.57 sq units
S 4r2
412
A r2
2
4
3
3 3
36
22 2 22
35. y VBB 43r3
500
V
3
S
100
4
3
54 43r3
40.5 r3
3.43 cm r
5
3
42. Vcyl 4
2.5724.8
It is a straight line.
32.
43. VBB 3 53
r2h
500
3 31.70
V
is a function of r and
S
directly proportional to r.
VBB 31.70 23.78 Vcyl 2h
4
3
3 r
4
3
3 r
500
3 500
3 500
48
r3
2.88 cm r
33. No; let r be the radius of a sphere and S its surface area.
If the radius is tripled, then the surface area is
43r2 49r2 94r2 9S. The surface area is
multiplied by 9. Another way to consider this is to note
that surface area is a function of r2, not of r, so tripling
the value of r multiplies the value of S by 32 9.
757.12
100 80%
946.34
45.
47.
r
10.42 cm h
757.12 cm3
r22r
2r3
h
4
0.5 10231.446 109
3
V r h
2r
16h
4
44. V r3number of air bubbles
3
2
34.
42h
4
3
46.
3 r
1
2
2 3 r h
r2 2r 2r3
2 13r2 r
23r3
48. Sample answer: The volume of the cylinder is greater
than the volumes of both other solids; the volume of the
cylinder is equal to the sum of the volumes of the sphere
and the double-cone shaped solid.
260
Geometry
Chapter 12 Worked-out Solution Key
mcrbg-1207-sk.qxd
5-25-2001
11:33 AM
Page 261
Chapter 12 continued
52 x2 r2
49.
12. Yes; height :
25 x2 r2
4.8 1.2
6 1.2
; base : 4
1
5
1
25 x2 r
13. always
V 13r2h
17. S 282 112
14. sometimes
2
2
1
2 x 3 25 x
1
2
3 25 x x 5
15. always
V 2023 160 cm3
5
18. S 125.532 1129.5 m2
V 8733 2349 m3
12.6 Mixed Review (p. 765)
19. S 2442 384 ft2
V 1243 768 ft3
50. translation
51. translation, 180 rotation, vertical line reflection,
horizontal glide reflection 52. translation
53. translation, 180 rotation
54. Similar by AA. CAB CDE and ACB DCE
by vertical .
Area of ABC 7 59.5 19.43 sq. units
4 2
55. Similar by AA. ABC EDC and ACB ECD
by vertical .
Area of ABC 1289 36 sq. units
56. Similar by AA. C C and ABC EDC
Area of ABC 13 2
9 37.35
77.93 sq. units
57. C 2r
20. S 3602 2250 in.2
5 2
V 40052 6250 in.3
3
21. Scale factor 3 27 3 1 1
216
8
2
Scale factor is 1:2.
22. Scale factor 3 27 3
125
5
23. Scale factor 121.5 3
24. Scale factor 24
384
14
3
2
36
25. h 5.516 88 in.
26. S 12.9162 3302.4 in.2
27. V 2163 8192 in.3
28. True; all spheres are similar; the scale factor of A to B is
23.1413.25
x:y, so the ratio of their volumes is x3:y3.
83.25 inches
29. True; all cubes are similar; the scale factor of A to B is
100 ft12 in.ft
14.4 revolutions
83.25 in.
x:y, so the ratio of their surface areas is x2:y2.
30.
0.125
1
or 1:96
12
96
31. S Lesson 12.7
32. S 3
31 height: 12
4 1 The solids are similar
because the radii and height are in the same ratio, 3:1.
6
2
2
1
6
3
4
2
2
1
4. B The scale factor is 3:2.
6. A The scale factor is 4:3.
7.
Vmodel 34.
9
2
1 3 3
2 2
34,051 144 12
66.5 in.3
963
1
11
27
16 116 cup
Vlarge 24144 1344
14
7
2 2
1
2
, , and
3 3
1.5 3
3
35. E Vsmall 1272 168
4
9. No; radius: 2 ; height: 3
10. Yes; radius: 4.5 1 ; height:
34,051 ft3
8. 1:32 1:9 so S 369 324 m2
3
144
63 in.2
1
1 4
r3
2 3
6
11
12.7 Practice and Applications (p. 769–771)
2
1
25
3
3
5. C The scale factor is 2:1.
3 216
1331
4032
962
V
33.
4
4
Side: Side: 1 No. The
solids are not similar because one pair of sides is not in
the ratio 2:1.
11. Yes; sides:
4r2
4032 ft2
1. p2: q2; p3:q3
3. Height:
1
2
S 1242513 2
12.7 Guided Practice (p. 769)
16. never
cm2
21
36.
3 8 2
125
5
1344
168
8
4
S.A. 25 25
2
D
Geometry
Chapter 12 Worked-out Solution Key
261
mcrbg-1207-sk.qxd
5-25-2001
11:33 AM
Page 262
Chapter 12 continued
3. S 4r2
V 12 4
3
3 r
4
3
3 4.775
12 1741.97
4
3
3 r
4
3
3 4.135
94.27
96 2r2
96 6r2
2r2r 150 2r2
4r
5r
42. LKJ
S
43. BAC
416.42
2228.02 m2
3379.85 in.2
3
40,447.07 544.38h
74.3 in. h
r2
Small
rl
S r2 rl
244.86
108.80 ft2
ft2
V 13r2h
1
2
3 4.5 12
13328
4r2
75.40 ft3
8.
2
314.16 ft2
1
V 4,188,790.21 20
3
523.60 ft3
4,188,790.21 ft3
Chapter 12 Review (p. 774)
FVE2
2.
32 V 90 2
12.7 Quiz 3 (p. 772)
r 100
20 5 ft
1
S 125,663.71 20
V 43r3
433.141003
1.
FVE2
F 6 10 2
V 60
2.
46.5
78 cm3
cm3
125,663.71 ft2
40,447.07 1322.82h
d 24 m
S 4r2
43.14102
43.141.882
1256.64 cm2
44.41 in.2
4188.79 cm3
813
43.141002
V 13r2h
49.
12 m r
V Bh
32 38.54
7. S 182 1821.4
144 m2 r2
4 146.5
115 cm2
V 13r2h
47. S 4r2
14,476.46 m3 r232
23
254.47 ft3
959.86 ft2
V r2h
6 2813
4.52 4.512.82
2143 152 4517
V 43r3
433.14103
S 2B Ph
Large
45. S 2B Ph
1. S 4r2
S 2B Ph
6. y 3 ft
44. JLK
48.
Small
624
12.7 Mixed Review (p. 772)
46. S r2 rl
158,058.33 m3
Large
6
2r2r
10 h
41. CA
433.14 155 3
V Bh
150 6r
25 r2
40. JL
V 43r3
460 cm2
2
16 r2
39. LK
14,137.17 m2
28
S 2r2 2rh
8h
366.4 ft2
5. x 6.5 cm
Because of space between the volleyballs and a lack of
knowledge of the dimensions of the crate, one can only
estimate how many volleyballs will fill the crate.
S 2r2 2rh
43.14 155 2
659.58 ft3
in.3
1741.97
18 volleyballs
94.27
38.
S 4r2
43.145.42
V 43r3
433.145.43
in.3
Volume volleyball 4.
V 43r3
433.141.883
27.83 in.3
3. F V E 2
55E2
8E
F6
4.
S 2B Ph
24 12 329
96 288
384 m2
5. S 2r2 2rh
23.1462 23.1465
414.69 ft2
6. S 2r2 2rh
23.145.52 23.145.518
812.10 in.2
262
Geometry
Chapter 12 Worked-out Solution Key
mcrbg-120R-sk.qxd
6-18-2001
9:34 AM
Page 263
Chapter 12 continued
1
8. S r2 rl
7. S B 2 Pl
S B 12Pl
6.
62 12245
3.1462 3.1468
340 100 1240l
96 cm2
263.89 in.2
240 20l
9. S B 6
1
2 Pl
10. V Bh
1
2
4 34
124.71
43 1
2 24
648
512 cm3
in.2
6 14321237.2
3.143.528
42,621.96 m3
307.88 in.3
1
1
13. V 3 Bh
14. V 3 Bh
17 yd
10,500 in.3
1198.43 cm3
1
S 4r2
43.14342
34 cm
14,526.72 cm2
16. S 4r2
133.148215
43.14142
1005.31 ft3
2463.01 m2
433.14143
11,494.04 m3
17. S 4r2
18.
1 2
2
40
20
2
1
12
6
3.14 in.2
12201518
3.146221
2700 ft 3
2375.04 cm3
11. V 2
1
The solids are similar with
a scale factor of 2:1.
10. V r2h
9. V Bh
V 43r3
1
3 Bh
1
3 756
70 ft3
12.
20 ft
20 ft
4
3.140.53
3
0.52 in.3
19.
1093.27 yd2
12 yd
14319223
13
15. V 3 r2h
S 3.14122 3.141217
8.
1330235
V
S r2 rl
7.
12. V r2h
11. V Bh
43.14
12 ft l
100 ft2
25 ft
15 ft
18 ft
5
15
4
8
13; 12
13; 16
12 They are not similar because the
ratios are not the same.
20 ft
Chapter 12 Test (p. 777)
12 cm
1. F 8; V 12; E 18
2. F 5; V 6; E 9
3. F 9; V 14; E 21
S 2B Ph
4.
244 307
7m
21 cm
12 cm
298 m2
44
m2
6.5 ft
6.95 ft
6.95 ft
5 ft
5.
7 ft
S 2r2 2rh
8.6 in.
784 28.62 28.6h
6.5 ft
784 147.92 17.2h
636.08 17.2h
36.98 in. h
13. r 5
3 cm 15 cm
4
V 3
3
4
153
3
14,137.17 cm3
Geometry
Chapter 12 Worked-out Solution Key
263
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Chapter 12 continued
14. A plane may intersect a sphere in a point or in a circle. If
the plane contains a diameter of the sphere, the intersection
is a great circle.
15.
3 8
2
3
3
27
23.1422 23.1427.64
128 in.
121.14
in.2
3.14227.64
96 in.3
96.01 in.3
The surface area of the cylinder is less than that of the
prism, but the volumes are the same.
15459
3540 ft2
V Bh
28259
16,638 ft3
19. A:
1
1.25
2
5531.25
1
1.25
3
11. D Vsmall r2h
Vlarge 4380
425
5120 ft3
80
12. Side Roof 152 82 17 ft
Area roof 25217 1768 ft2 not including ends
Area ends 22 308 240 ft2
1
1768
56 sheets
32
301052 1230852
15,600 6240
1
1.95
21,840 ft3
V 1.25 16,638
14. S 2B Ph
32,496.09 ft3
230812 6452
Chapter 12 Standardized Test (p. 778)
1. B
240 3328
3568 ft2
15. V r2h
S 2B Ph
3.14929
2608 26.828 69.6h
2290.22 cm3
2227.2 69.6h
16. Vbowl 32 m h
3. D L.A. 2rh
361.73
10. C
13. V Volume right prism Volume triangular prism
ft2
3
2. E
9. A Column A: S 4r2 43.147.62 725.83 cm2
1768 240
63 sheets including ends
32
1
1.5625
Lateral Area 1.56253540
V:
Column B: P 215.2 6 36.4 cm
18. L.A. Ph
282 ft2
13
1
3
1
Column B: S B 2Pl 2362 23614.9
166
1
2 aP
1
2 9.4154
18
54
8. A Column A: C 2r 23.147.6 47.75 cm
V r2h
V Bh
13
1
V 3 27
S 2r2 2rh
2
12
36
1 3
216 166
40.79 ft3
Cylinder
S 2B Ph
A
643322.5 3.141.522.5
7. E
16. Prism
17.
6. D V Bh r2h
4. C
V Bh
21.28
1650 1215x22
19.2 cm2
1650 165x
3.14103
1 4
2 3
2094.40 cm3
Find the volume of the bowl. If it is less than the volume
of the souffle dish, then the peaches will fit in the souffle
dish.
10 m x
5. B
V
972 4
3
3 r
4
3
3 r
729 r3
9 yd r
264
Geometry
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Chapter 12 continued
17. volume of custard cups 2
3.52 3.5
269.39 cm3
volume of bowl 12
43103
2094.40 cm3
x
12
2094.40 269.39
2094.40x 12269.39
x 1.5 peaches or 112 peaches
18. Sample answer: Mark is not correct. Volume is a cubic
quantity. If the volume is to be reduced by 12, each
dimension should be reduced by
12 or
3
19.
1
3 2
0.79.
0.72
0.57
20. h 1.14 1.26 1.44 cm
21. S 2r2 2rh
23.140.572 23.140.571.14
6.12 cm2
Large cylinder:
S 6.12 cm21.262
9.72 cm2
22. V r2h
3.140.5721.14
1.16 cm3 for small cylinder
V 1.16 cm31.263
2.32 cm3
23. S 9.72 cm232
87.48 cm2
V 2.32 cm333
62.64 cm3
Geometry
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Chapter 12 continued
Cumulative Practice, Chs. 1–12 (p. 780)
6. Statements
1. 4x 2 6x 3
4x 2 6x 18
Reasons
1. j k, m1 73
1. Given
2. m1 m2 180
2. If two lines are cut by a
3. 73 m2 180
3. Substitution prop. of
trans. the cons. int. are
supplementary
16 2x
8x
So 4x 2 30.
equality
4. m2 107
The measures of the four are 30, 30, 150, 150.
4. Subtraction prop. of
equality
2.
B
A
7. Statements
Reasons
1. ABDE and CDEF are
P
C
D
Consecutive interior are supplementary. If each interior is
bisected, then two of the interior of the triangle are complementary. The measure of the third must be 90 in order for the
sum of the measures of the to be 180.
3.
2. 4 D
2. Opp. of a are .
3. m4 mD
3. Def. of .
4. m6 mD 180
4. Consecutive of a 5. m6 m4 180
5. Substitution Prop. of
are supp.
Equality
6. 4 and 6 are supp.
8. always
y
T (0, 2k)
1. Given
parallelograms.
6. Def. of supplementary
9. never
10. 180 rotation about the origin
y
Y(5, 8)
R (0, 0)
X(3, 3)
S(2h, 0) x
X (3, 3)
RST is a rt. with vertices R0, 0, S2h, 0 and
T0, 2k. The length of the hypotenuse is
0 2h2 2k 02 4h2 4k2 2h2 k2. The midpoint of the
Y (5, 8)
The length of the median is h 02 k 02 1
h2 k2. Therefore the median to the hypotenuse 2
the length of the hypotenuse.
5. XY2 100; YZ2 64; XZ2 36. Since
X (3, 3)
Y (5, 8)
lines are cut by a transversal then corr. are so
ABC ADE. By the AA Similarity Postulate
12.
x
3
4 5
5x 12
x2
4. AB2 144; BC2 64; AC2 225; since
AC2 > AB2 BC2 ABC is obtuse. B is the largest
. A is the smallest .
x
2
11. Sample answer: A A by the reflexive prop. If two
hypotenuse ST is
0 2 2h, 0 2 2k or h, k.
2
13.
2
5
Perimeter of ABC
5
Area of ABC
5
8
2
25
64
XY2 YZ2 XZ2, XYZ is a rt. . Z is the
largest , and Y is the smallest.
266
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9:34 AM
Page 267
Chapter 12 continued
h
7
24 25
25h 168
14. Hyp2 72 242
Hyp2 625
Hyp 25
31. C 2r
23.145
31.4 units
h 6.72
33. the bisectors of the rt. formed by the lines
Y
24
h
Z
7
34. 25 2180 4140
X
35. A 2 aP
W
1
25
1215 tan 67.5240
7
mA sin1 25
16.3
25
B
16.
4345.6 cm2
7
sin A 25
15. C
7
mDB
36. Let s side of the square.
P
mB 90 16.3 73.7
A
24
32. center of the hexagon
mAB
1
2 50
90
s2
360
s2
25
180 50
1 2
s
4
2 s
4
130
18. mACB 2 50 25 19. mEDB 2 180 90
1
1
1
29%
4
2
130 65
2
1
21.
1
40 12180 mCD
80
22. mBC
80 180 mCD
1
37. V 3 r2h
100
mCD
76.97 ft
100 in.3
1562.5
15.625
100
1
3
15.625 2.5
25. BAC and BPC are supplementary. ABPC is a
Small:large 1:2.5
quadrilateral with two right angles, so the sum of the
measures of the other two angles is 180.
26. It is a kite. 2 pairs of sides are congruent, but opposite
2:5
A r
2
39.
3.14 22 2
sides are not congruent.
ED AE EB
4 6 AE
28. DF BF
2
25.13 ft2
2 2
36 16 4x
20 4x
5x
EB ED CE
10 7 3.5 CE
29. AE
4
2
40. L.A. Ph
0.2560.75
1.125 in.2
20 CE
2 6 1 5
30. Center 2, 2
,
2
2
FC
62 4 4 x
12 AE
455
3
24. mBGC 2 80 130 105
18
V Bh
38.
1
2
3 3.5 6
23. mABD 50 180 230
27. EC
Area of 14 circle
Area square
41. S 4r2
43.140.752
7.07 in.2
42 32
25
5 units
x 2 y 22 25
2
Geometry
Chapter 12 Worked-out Solution Key
267
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