chapter 12€¦ · 62.83 cm2. lesson 12.1 investigating geometry activity 12.1 ... the base is a...

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399Geometry

Worked-Out Solution Key

Prerequisite Skills (p. 790)

1. The area of a regular polygon is given by the formula

A 5 1 } 2 a p P, where a is the apothem and P is the

perimeter.

2. Two polygons are similar if their corresponding angles are congruent and the lengths of corresponding sides are proportional.

3. tan 258 5 x } 30

x 5 30 tan 258

x ø 13.99

4. cos 708 5 5 } x

x 5 5 }

cos 708

x ø 14.62

5. sin 508 5 x } 30

x 5 30 sin 508

x ø 22.98

6. r 5 2 m

C 5 2πr A 5 πr2

5 2π(2) 5 π(2)2

5 4π 5 4π

ø 12.57 ø 12.57

The circumference is about 12.57 m and the area is about 12.57 m2.

7. d 5 3 in., r 5 3 } 2 in.

C 5 2πr 5 2π 1 3 } 2 2 5 3π ø 9.42

A 5 πr2 5 π 1 3 } 2 2 2 5

9 } 4 π ø 7.07

The circumference is about 9.42 in. and the area is about 7.07 in.2.

8. r 5 2 Ï}

5 cm

C 5 2πr 5 2π(2 Ï}

5 ) 5 4π Ï}

5 ø 28.10

A 5 πr2 5 π(2 Ï}

5 )2 5 20π ø 62.83

The circumference is about 28.1 cm and the area is about 62.83 cm2.

Lesson 12.1

Investigating Geometry Activity 12.1 (p. 793)

1. a. 3 faces meet at each vertex.

b. 5 faces meet at each vertex.

2. The angle measures of all the angles in an equilateral triangle are 608, so for 6 of these angles to meet means the total angle measure is 6 p 608 5 3608. A solid cannot have a vertex with 6 equilateral triangles because they would form a plane.

3. Three congruent regular hexagons meeting at a vertex would result in a plane being formed. The sum of the angles (3 p 1208) is 3608 which will not form a convex vertex.

4. F 1 V 5 E 1 2

12.1 Guided Practice (pp. 795–797)

1. The solid is formed by polygons, so it is a polyhedron. The base is a square, so it is a square pyramid. It has 5 faces, 5 vertices, and 8 edges.

2. The solid has a curved surface, so it is not a polyhedron.

3. The solid is formed by polygons, so it is a polyhedron. The two bases are congruent triangles, so it is a triangular prism. It has 5 faces, 6 vertices, and 9 edges.

4. The dodecahedron has 12 faces, 20 vertices, and 30 edges.

F 1 V 5 E 1 2

12 1 20 5 30 1 2

32 5 32 ✓

5. The cross section is a triangle.

6. The cross section is a circle.

7. The cross section is a hexagon.

12.1 Exercises (pp. 798–801)

Skill Practice

1. The fi ve Platonic solids and their faces are: regular tetrahedron, 4 faces; cube, 6 faces; regular octahedron, 8 faces; regular dodecahedron, 12 faces; regular icosahedron, 20 faces.

2. Euler’s Theorem states the sum of the number of faces and vertices of a polyhedron is equal to the number of edges of the polyhedron plus two.

3. The solid is formed by polygons, so it is a polyhedron. The base is a pentagon, so it is a pentagonal pyramid.

4. The solid is formed by polygons, so it is a polyhedron. The two bases are congruent hexagons, so it is a hexagonal prism.

5. The solid has a curved surface, so it is not a polyhedron.

6. The bases of the prism are triangles, not rectangles. The solid is a triangular prism.

7.

8.

Chapter 12

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400GeometryWorked-Out Solution Key

9.

10.

11. n 1 12 5 18 1 2 12. 5 1 n 5 8 1 2

n 1 12 5 20 5 1 n 5 10

n 5 8 n 5 5

There are 8 faces. There are 5 vertices.

13. 10 1 16 5 n 1 2 14. n 1 12 5 30 1 2

26 5 n 1 2 n 1 12 5 32

24 5 n n 5 20

There are 24 edges. There are 20 faces.

15. There are 4 faces, 4 vertices, and 6 edges.

4 1 4 5 6 1 2

8 5 8 ✓


5 1 5 5 8 1 2

10 5 10 ✓


5 1 6 5 9 1 2

11 5 11 ✓


5 1 6 5 9 1 2

11 5 11 ✓


8 1 12 5 18 1 2

20 5 20 ✓


8 1 12 5 18 1 2

20 5 20 ✓

21. A cube is a solid formed by six congruent faces, so by name it is a regular hexahedron.

22. concave 23. concave 24. convex

25. 26. 27.

circle rectangle triangle

28. A; The cross section parallel to the square base is also a square.

29. Euler’s Theorem proves this is incorrect.

4 1 6 0 4 1 2

10 Þ 8

Instead there should be 4 faces, 4 vertices, and 6 edges.

4 1 4 5 6 1 2

8 5 8 ✓

30. C; A triangular prism and a square pyramid both have 5 faces.

31. D; An octagonal prism has 10 faces, 16 vertices, and 24 edges.

32. 32 1 v 5 90 1 2

32 1 v 5 92

v 5 60

Euler’s Theorem can be used. The solid has 60 vertices.

33. The plane should intersect the cube at an angle so that it touches each of the six faces. The fi gure will look like:

Problem Solving

34. a. There are 10 vertices.

b. 7 1 10 5 E 1 2

17 5 E 1 2

15 5 E

There are 15 edges.

35. There are 18 edges and 12 vertices.

8 1 12 5 18 1 2

20 5 20 ✓

36. The cross section is a circle.

37. The cross section is a square.

38. The cross section is a rectangle.

39. A polyhedron with 4 vertices and 6 edges is a triangular pyramid. No, Euler’s Theorem means that all polyhedrons with 4 vertices and 6 edges have the same number of faces.

40. a. The cross section is a rectangle.

b. The length of the rectangle is the hypotenuse of the right triangle formed by two sides of the cube. The

length is Ï}

62 1 62 5 Ï}

72 5 6 Ï}

2 .

P 5 6 1 6 1 6 Ï}

2 1 6 Ï}

2 5 12 1 12 Ï}

2 ø 28.97

The perimeter is approximately 28.97 inches.

c. A 5 l p w 5 6 Ï}

2 p 6 5 36 Ï}

2

The area of the cross section is 36 Ï}

2 inches2 or about 50.9 inches2.

Chapter 12, continued

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401Geometry


41. a. The cross section is a trapezoid.

b. Yes; Sample answer:

c. The cross section is a square.

d. Yes; Sample answer:

42.

Platonic Solid Faces Vertices EdgesF 1 V 5 E 1 2

Tetrahedron 4 4 64 1 4 5 6 1 2

Cube 6 8 126 1 8 5 12 1 2

Octahedron 8 6 128 1 6 5 12 1 2

Dodecahedron 12 20 3012 1 20 5 30 1 2

Icosahedron 20 12 3020 1 12 5 30 1 2

43. No

44. Yes;

45. Yes;

46. Yes;

47. Yes;

48. Yes;

49. a. There would be 7 faces, 10 vertices, and 15 edges.

b. There would be 7 faces, 10 vertices, and 15 edges.

c. The number of faces, vertices, and edges would remain the same.

d. There would be 9 faces, 14 vertices, and 21 edges.

50. a. There would be 5 faces, 6 vertices, and 9 edges.

b. There would be 5 faces, 6 vertices, and 9 edges.

c. The number of faces, vertices, and edges would remain the same.

d. Cutting two edges would make 6 faces, 7 vertices, and 11 edges.

51. The fi rst solid has 8 vertices and each vertex has three 908 angles.

D 5 3608 2 3(908) 5 908

DV 5 90(8) 5 7208

The second solid has 10 vertices and each vertex has two 908 angles and one 1088 angle.

D 5 3608 2 2(908) 2 1088 5 728

DV 5 72(10) 5 7208

The third solid has 12 vertices and each vertex has two 908 angles and one 1208 angle.

D 5 3608 2 2(908) 2 1208 5 608

DV 5 60(12) 5 7208

Mixed Review

52. x8 5 1 } 2 (738 1 818) 5

1 } 2 (1548) 5 778

The value of x is 77.

53. 1808 2 468 5 1 } 2 (1098 1 x8)

134 5 1 } 2 (109 1 x)

268 5 109 1 x

159 5 x

The value of x is 159.


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54. 1808 2 x8 5 1 } 2 (588 1 1978)

180 2 x 5 1 } 2 (255)

180 2 x 5 127.5

x 5 52.5

The value of x is 52.5.

55. r 5 11 cm

C 5 2πr 5 2π(11) 5 22π ø 69.12

A 5 πr2 5 π(11)2 5 121π ø 380.13

The circumference of the circle is about 69.12 cm and the area is about 380.13 cm2.

56. d 5 28 in., r 5 28

} 2 5 14 in.

C 5 2πr 5 2π(14) 5 28π ø 87.96

A 5 πr2 5 π(14)2 5 196π ø 615.75

The circumference of the circle is about 87.96 in. and the area is about 615.75 in.2.

57. d 5 15 ft, r 5 15

} 2 ft 5 7.5 ft

C 5 2πr 5 2π (7.5) 5 15π ø 47.12

A 5 πr2 5 π (7.5)2 5 56.25π ø 176.71

The circumference of the circle is about 47.12 ft and the area is about 176.71 ft2.

58.3608

} 6 5 608

The segment whose length is the apothem bisects the central angle. So, a 308-608-908 triangle is formed by the segment and radius of the hexagon.

17

} r 5 cos 308

17308

r

r 5 17 }

cos 308

r 5 34 Ï

}

3 } 3 units

x }

17 5 tan 308

x 5 17 tan 308

x 5 17 Ï

}

3 } 3 units

s 5 2x 5 2 1 17 Ï}

3 }

3 2 5

34 Ï}

3 } 3

P 5 ns 5 6 1 34 Ï}

3 }

3 2 5 68 Ï

}

3 ø 117.78 units

A 5 1 } 2 aP 5

1 } 2 (17)(68 Ï

}

3 ) 5 578 Ï}

3 ø 1001.13 sq. units

59. 3608

} 8 5 458

A segment whose length is the apothem bisects the central

angle. So, x }

29 5 sin 22.58 → x 5 29(sin 22.58). One side

of the octagon 5 2x 5 58 sin 22.58.

2922.58

x

P 5 ns P 5 8(58 sin 22.58) 5 464 sin 22.58 ø 177.57 units

a }

29 5 cos 22.58

a 5 29(cos 22.58)

A 5 1 } 2 aP

5 1 } 2 (29 cos 22.58)(464 sin 22.58)

ø 2378.71 sq. units

60. P 5 ns

24 P 5 3(24) 5 72 units

A 5 1 } 4 Ï

}

3 s2

A 5 1 } 4 Ï

}

3 (24)2 5 576 Ï

}

3 } 4 ø 249.42 sq. units

Lesson 12.2


STEP 2

The polyhedron is a rectangular prism. The polyhedron is not regular, but it is convex.

STEP 3

142 sq. units

1. A 5 3 p 7 5 21 5 21 sq. units

P 5 2(3) 1 2(7) 5 20 units

h 5 5 units

2. 2A 1 Ph 5 2(21) 1 20(5) 5 142 sq. units

The values are the same.

3. The surface area of a prism can be found by summing the area of all the faces, or by using the formula S 5 2A 1 Ph.

4. Sample answer:

A

DCB

F

E

S 5 2(5) 1 4(5) 1 4(2) 1 4(5) 1 4(2) 1 2(5) 5 76 units2

S 5 2(10) 1 14(4) 5 76 units2


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403Geometry



1. Sample answer:

2. a.

4 in.

4 in.

4 in.

3 in.

3 in.

A

B C

F

D E

Area 5 4(3) 1 4(7) 1 3(7) 1 4(7) 1 3(7) 1 3(4)

5 122 sq. inches.

b.

7 in.

4 in.3 in.

B 5 3(4) 5 12

P 5 2(3) 1 2(4) 5 14

Area 5 2B 1 Ph 5 2(12) 1 14(7) 5 122 in.2

3. S 5 2πr2 1 2πrh

5 2π(10)2 1 2π(10)(18)

5 200π 1 360π

5 560π

ø 1759.29 in.2

4. S 5 2πr2 1 2πrh

208π 5 2πr2 1 2πr(5)

0 5 2πr2 1 10πr 2 208π

0 5 r2 1 5r 2 104

0 5 (r 1 13)(r 2 8)

r 1 13 5 0 or r 2 8 5 0

r 5 213 or r 5 8

A cylinder cannot have a negative radius, so the radius is 8 feet.

12.2 Exercises (pp. 806–809)

Skill Practice

1.

lateral face

lateral face

lateral face

lateral edge

lateral edgelateral edgebase

base

The bases are the two parallel triangular faces. The lateral faces are the 3 rectangular faces that connect the two bases. The lateral edges are the three edges where the lateral faces meet.

2. The formula states the surface area is equal to two times the area of the base plus the perimeter of the base times the height. The formula is true for cylinders as well as right prisms because πr2 is the area of the base and 2πr is the circumference, or perimeter of the base. So, 2πr2 1 2πrh is really 2B 1 Ph.

3. d 5 4 in., r 5 4 } 2 5 2 in.

S 5 2πr2 1 2πrh

5 2π(2)2 1 2π(2)(10)

5 8π 1 40π

5 48π

ø 150.80

The surface area is about 150.80 in.2.

4. S 5 2B 1 Ph

5 2(16 Ï}

3 ) 1 3(8)(20)

5 32 Ï}

3 1 480

ø 535.43

The surface area is about 535.43 cm2.

5. P 5 ns 5 6(40) 5 240 ft

B 5 1 } 2 aP 5

1 } 2 (34.64)(240) 5 4156.8 ft2

S 5 2B 1 Ph

5 2(4156.8) 1 240(80)

5 8313.6 1 19,200

5 27,513.6

The surface area is about 27,513.6 ft2.

6. B 5 3(8) 5 24 P 5 2(8) 1 2(3) 5 22

S 5 2B 1 Ph

5 2(24) 1 22(2)

5 92

The surface area is 92 ft2.

7. S 5 2B 1 Ph

5 2 1 1 } 2 ( Ï}

61.75 )(3) 2 1 19(9.1)

5 3 Ï}

61.75 1 172.9

ø 196.47

The surface area is about 196.47 m2.


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8. S 5 2B 1 Ph

5 2 1 1 } 2 (tan 548)(2)(5) 2 1 2(5)(3.5)

5 10 tan 548 1 35

ø 48.76


9. S 5 2πr2 1 2πrh

5 2π(0.8)2 1 2π(0.8)(2)

5 1.28π 1 3.2π

5 4.48π

ø 14.07


10. S 5 2πr2 1 2πrh

5 2π(12)2 1 2π(12)(40)

5 288π 1 960π

5 1248π

ø 3920.71

The surface area is about 3920.71 mm2.

11. S 5 2πr2 1 2πrh

5 2π(8)2 1 2π(8)(8)

5 128π 1 128π

5 256π

ø 804.25


12. The diameter of the cylinder was used in place of the radius.

S 5 2π(32) 1 2π(3)(8)

5 2π(9) 1 2π(24)

5 66π

ø 207.35


13. S 5 2B 1 Ph

606 5 2(7 p x) 1 (14 1 2x)(15)

606 5 14x 1 210 1 30x

606 5 44x 1 210

396 5 44x

9 5 x

The value of x is 9 yd.

14. S 5 2πr2 1 2πrh

1097 5 2π(8.2)2 1 2π(8.2)(x)

1097 5 134.48π 1 16.4πx

674.52 5 16.4πx

13.09 ø x

The value of x is about 13.09 m.

15. S 5 2B 1 Ph

616 5 2 1 1 } 2 (8)(17) 2 1 (25 1 Ï

}

353 )(x)

616 5 136 1 43.79x

480 5 43.79x

10.96 ø x

The value of x is about 10.96 in.

16.

9

15

8

Use the Pythagorean Theorem to fi nd the height.

h2 1 92 5 152

h 5 12

S 5 2B 1 Ph

S 5 2 1 1 } 2 (9)(12) 2 1 (9 1 15 1 12)(8)

5 108 1 288

5 396

The surface area is 396 square units.

17. C;

3

3

3

S 5 2B 1 Ph S 5 2B 1 Ph

5 2(l2) 1 (4l)(l) 5 2(3l)2 1 (4(3l))(3l)

5 2l2 1 4l2 5 18l2 1 36l2

5 6l2 5 54l2

5 9(6l2) The surface area is 9 times the original surface area.

18. S 5 2πr2 1 2πrh

Sn 5 2π 1 r }

Ï}

5 2

2 1 2π 1 r

} Ï

} 5 2 1 h

} Ï

} 5 2

5 1 } 5 (2πr2) 1

1 } 5 (2πrh)

5 1 } 5 (2πr2 12πrh)

The new surface area is 1 } 5 the original area.

19. S 5 2B 1 Ph

5 2 1 1 } 2 1 5

} tan 308

2 (6 p 10) 2 1 (6 p 10)(10)

ø 519.62 1 600

ø 1119.62

The surface area of the hexagonal prism is about 1119.62 in.2.


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405Geometry


20. S 5 2πr2 1 2πrh

108π 5 2π(2h)2 1 2π(2h)(h)

108π 5 8πh2 1 4πh2

108π 5 12πh2

9 5 h2

3 5 h

The height of the cylinder is 3 meters.

21. l2 1 (l Ï}

2 )2 5 82

8

2

l2 1 2l2 5 64

3l2 5 64

l2 5 64

} 3

S 5 2B 1 Ph

5 2l2 1 4l2

5 2 1 64 }

3 2 1 4 1 64

} 3 2

5 128

The surface area of the cube is 128 sq. units.

Problem Solving

22. S 5 2πr2 1 2πrh

5 2π(10)2 1 2π(10)(8)

5 200π 1 160π

5 360π

ø 1130.97


23. a. S 5 2B 1 Ph

5 2(12 p 6) 1 36(6)

5 144 1 216

5 360

The minimum amount of paper needed is 360 in.2.

b. The net of the box includes areas that are covered by the top, so there is extra material that is not touched by the paper.

c. The amount in part (a) is the exact amount of paper. It is better to have more so that there is some overlap of wrapping paper.

24. a. S 5 2πr2 1 2πrh

S 5 2π(4)2 1 2π(4)(10)

5 32π 1 80π

5 112π

ø 351.86

The surface area is about 351.86 ft2.

b. Doubling the radius will create a greater surface area.

c. Double radius: Double height:

S 5 2πr2 1 2πrh S 5 2πr2 1 2πrh

S 5 2π(8)2 1 2π(8)(10) S 5 2π(4)2 1 2π(4)(20)

5 128π 1 160π 5 32π 1 160π

5 288π 5 192π

ø 904.78 ø 603.19

When the radius is doubled, the surface area is about 904.78 ft2. When the height is doubled, the surface area is about 603.19 ft2. Doubling the radius creates a greater surface area.

25. A; The white rectangle is touching the base that contains the arrow but not the edge where the arrow starts.

26. Surface area of a prism:

S 5 B 1 Ph

5 (14)(12) 1 (52)(36)

5 168 1 1872

5 2040

Surface area of a cylinder:

S 5 πr2 1 2πrh

S 5 π(6)2 1 2π(6)(36)

5 36π 1 432π

5 468π

ø 1470.27

The rectangular prism has a greater surface area, so it will take more material to make the rectangular prism recycle bin.

27. a. Surface area 5 the number of exposed faces

5 54 sq. units

b. Surface area 5 the number of exposed faces

5 52 sq. units

c. When the red cubes are removed, inner faces of the cubes remaining replace the area of the red cubes that are lost. When the blue cubes are removed, there are still 2 faces of the blue cubes whose area is not replaced by inner faces of the remaining cubes. Therefore, the area of the solid after removing blue cubes is 2 square units less than the solid after removing red cubes.

28. a. S 5 2π(r1)2 1 2π(r1)(h) 2 2π(r2)2 1 2π(r2)(h)

5 2π(12)2 1 2π(12)(8) 2 2π(6)2 1 2π(6)(8)

5 288π 1 192π 2 72π 1 96π

5 504π

ø 1583.36


b. S 5 2π(r1)2 1 2π(r1)(h) 2 2π(r2)2 1 2π(r2)(h)


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29.

1 ft

1 ft

1 ft

4 in.

S 5 2B 1 Ph 2 2πr2 1 2πrh

5 2(12 p 12) 1 (12 p 4)(12) 2 2π(2)2 1 2π(2)(12)

5 288 1 576 2 8π 1 48π

5 864 1 40π

ø 989.66


30. a.

b. S 5 6(5 p 5) 1 8 1 1 } 4 Ï

}

3 (5)2 2 5 150 1 50 Ï}

3 ø 236.60

The surface area is about 236.60 mm2.

Mixed Review

31. 12608 5 (n 2 2) p 1808

7 5 n 2 2

9 5 n

The polygon has 9 sides. It is a nonagon.

32. 10808 5 (n 2 2) p 1808

6 5 n 2 2

8 5 n

The polygon has 8 sides. It is an octagon.

33. 7208 5 (n 2 2) p 1808

4 5 n 2 2

6 5 n

The polygon has 6 sides. It is a hexagon.

34. 18008 5 (n 2 2) p 1808

10 5 n 2 2

12 5 n

The polygon has 12 sides. It is a dodecagon.

35. 3608

} 4 5 908

A segment whose length is the apothem bisects the

central angle. So, tan 458 5 x } 6 .

6

A B

C D

E

F

45°

45°

x 5 6 tan 458 5 6

One side of the square 5 2x 5 12.

A 5 (2x)2 5 122 5 144

The area is 144 square units.

36. 3608

} 6 5 608

A segment whose length is the apothem bisects the central angle. So, a 308-608-908 triangle is formed.

K L

P N

J MH

12

a30°

6 }

a 5 tan 308

a 5 6 }

tan 308 5 6 Ï

}

3

P 5 ns 5 6(12) 5 72

A 5 1 } 2 aP 5

1 } 2 (6 Ï

}

3 )(72) 5 216 Ï}

3 ø 374.12

The area is about 374.12 square units.

37. 3608

} 5 5 728

The segment whose length is the apothem bisects the central angle with the radius drawn, so a 368-548-908 triangle is formed. Use trigonometric ratios to fi nd the length of the apothem (a 5 9 cos 368) and a side. The length of a side is 2(9 sin 368) 5 18 sin 368.

9

36°V

U R

T S

Wx

P 5 ns 5 5(18 sin 368) 5 90 sin 368 ø 52.90

A 5 1 } 2 aP

5 1 } 2 (9 cos 368)(18 sin 368)(5)

5 405(cos 368)(sin 368)

ø 192.59

The area is about 192.59 square units.


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407Geometry


Lesson 12.3


1. l2 5 h2 1 a2

l2 5 4.82 1 5.52 5 53.29

l 5 7.3

The area of each lateral face is

A 5 1 } 2 bl 5

1 } 2 (8)(7.3) 5 29.2 square meters.

2. S 5 B 1 1 } 2 Pl

5 1 } 2 (5.5)(8 p 5) 1

1 } 2 (8 p 5)(7.3)

5 110 1 146

5 256

The surface area is 256 m2.

3. l2 5 h2 1 r2 Lateral area 5 πrl

l2 5 202 1 152 5 625 5 π(15)(25)

l 5 25 5 375π

ø 1178.10

The lateral area is about 1178.10 yd2.

4. S 5 πr2 1 πrl

5 π(15)2 1 π(15)(25)

5 225π 1 375π

5 600π

ø 1884.96

The surface area is about 1885 yd2.

12.3 Exercises (pp. 814–817)

Skill Practice

1.

base

heightslant height

2. The height of a right cone is the perpendicular distance from the vertex of the cone to its base. The slant height of the cone is the height of its lateral surface.

3. A 5 1 } 2 bl 5

1 } 2 (8)(10) 5 40

The area of a lateral face is 40 cm2.

4. A 5 1 } 2 bl 5

1 } 2 (10)(15) 5 75

The area of a lateral face is 75 in.2.

5. l2 5 h2 1 1 1 } 2 b 2 2

l2 5 212 1 202 5 841

l 5 29

Area 5 1 } 2 bl 5

1 }

2 (40)(29) 5 580

The area of a lateral face is 580 ft2.

6. S 5 B 1 1 } 2 Pl 5 (2 p 2) 1

1 } 2 (4 p 2)(3) 5 4 1 12 5 16


7. S 5 B 1 1 } 2 Pl

5 5 1 1 } 2 2 (6.9)(10) 1 1 } 2 (5 p 10)(20)

5 172.5 1 500

5 672.5

The surface area is 672.5 mm2.

8. S 5 B 1 1 } 2 Pl

5 1 } 2 (5) 1 5 Ï

}

3 }

2 2 1

1 } 2 (3 p 5)(8)

ø 10.83 1 60

ø 70.83


9. The slant height l is 5 ft, not 4 ft.

S 5 B 1 1 } 2 Pl 5 62 1

1 } 2 (24)(5) 5 96


10. l2 5 r2 1 h2 Lateral area 5 πrl

l2 5 7.52 1 252 ø π(7.5)(26.10)

l2 5 681.25 ø 195.75π

l ø 26.10 ø 614.98

The lateral area is about 614.98 cm2.


l2 5 12 1 42 5 π(1)( Ï}

17 ) l2 5 17 5 Ï

}

17 π

l 5 Ï}

17 ø 12.95

The lateral area is about 12.95 in.2.


l2 5 3.52 1 122 5 π(3.5)(12.5)

l2 5 156.25 5 43.75π

l 5 12.5 ø 137.44

The lateral area is about 137.44 in.2.

13. S 5 πr2 1 πrl

5 π(4)2 1 π(4)(15)

5 16π 1 60π

ø 238.76


14. S 5 πr2 1 πrl

5 π(13)2 1 π(13)(20)

5 169π 1 260π

ø 1347.74



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15. l2 5 h2 1 r2 S 5 πr2 1 πrl

l2 5 82 1 52 5 π(5)2 1 π(5)( Ï}

89 ) l2 5 89 ø 25π 1 47.17π

l 5 Ï}

89 ft ø 226.73


16. The formula for the surface area of a cone is wrong. The r in the second term should not be squared.

S 5 πr2 1 πrl 5 π(6)2 1 π(6)(10) 5 96π

The surface area is 96π cm2.

17. B; S 5 πr2 1 πrl

200π 5 π(8)2 1 π(8)(l)

200π 5 64π 1 8πl

136π 5 8πl

17 5 l

The slant height is 17 ft.

18.

15 ft

20 ft

S 5 πr2 1 πrl

5 π(15)2 1 π(15)(20)

5 225π 1 300π

5 525π

ø 1649.34


19.

16 m

30 m

l2 5 r2 1 h2 S 5 πr2 1 πrl

l2 5 82 1 302 5 π(8)2 1 π(8)( Ï}

964 ) l2 5 964 ø 64π 1 248.39π

l 5 Ï}

964 ø 981.39


20.

10 in.

24 in.

S 5 B 1 1 } 2 Pl

5 1 } 2 (10)(5 Ï

}

3 ) 1 1 } 2 (3 p 10)(24)

5 25 Ï}

3 1 360

ø 403.30


21.

6 cm

9 cm

3

30°

a

B 5 1 } 2 (a)(6)(6) 5 18a 5 18 1 3

} tan 308

2 5 54 Ï}

3

S 5 B 1 1 } 2 Pl

5 54 Ï}

3 1 1 } 2 (6 p 6)(9)

5 54 Ï}

3 1 162

ø 255.53


22. l2 5 r2 1 h2

5 cm

h2 512 cm

h 5 4 cm

l2 5 52 1 42

l2 5 41

l 5 Ï}

41

S 5 πr2 1 2πrh2 1 πrl

5 π(5)2 1 2π(5)(12) 1 π(5)( Ï}

41 ) 5 25π 1 120π 1 (5) Ï

}

41 π

ø 556.11


23. l2 5 h2 1 1 1 } 2 b 2 2

5 in.5 in.

5 in.

3 in.

l2 5 32 1 1 5 } 2 2 2

l2 5 15.25

l 5 Ï}

15.25

S 5 B 1 Ph 1 1 } 2 Pl

5 (5 p 5) 1 (4 p 5)(5) 1 1 } 2 (4 p 5)( Ï

}

15.25 )

5 25 1 100 1 10 Ï}

15.25

5 125 1 10 Ï}

15.25

ø 164.05


24. (l1)2 5 42 1 32

3 yd

8 yd

4 yd

l2

(l1)2 5 25

l1 5 5

(l2)2 5 32 1 82

(l2)2 5 73

l2 5 Ï}

73

S 5 πrl1 1 πrl2

5 π(3)(5) 1 π(3)( Ï}

73 ) 5 15π 1 3π Ï

}

73

ø 127.65

The surface area is about 127.65 yd2.


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409Geometry


25.

4

4

4

S 5 4 p lateral area

5 4 1 1 } 2 (4)(2 Ï

}

3 ) 2 5 4(4 Ï

}

3 ) 5 16 Ï

}

3

ø 27.71


26.5 in.

5 in.

4 in. x

Cone:

S 5 πr2 1 πrl

5 π(4)2 1 π(4)(5) 5 16π 1 20π

5 36π

Pyramid:

S 5 B 1 1 } 2 Pl

5 x2 1 1 } 2 (4x)(5)

5 x2 1 10x

36π 5 x2 1 10x

0 5 x2 1 10x 2 36π

x 5 2b 6 Ï

}

b2 2 4ac }} 2a

5 210 6 Ï

}}

102 2 4(1)(236π) }}}

2(1)

5 210 6 Ï

}}

100 1 144π� }} 2

x ø 6.75 or x ø 216.75

The length of a base edge is about 6.75 inches.

Problem Solving

27. l2 5 h2 1 1 1 } 2 b 2 2

l2 5 42 1 32

l2 5 25

l 5 5

S 5 B 1 1 } 2 Pl

5 62 1 1 } 2 (4 p 6)(5)

5 36 1 60

5 96

The surface area is 96 in.2.

28. a. lateral area 5 4 1 1 } 2 b l�2� 5 4 1 1 }

2 (8)(14) 2 5 224

The lateral area is about 224 in.2.

b. The lateral faces are not triangles because the top of the pyramid has been cut off. The lampshade is approximately the shape of a regular square pyramid, but not exactly.

29. Regular square pyramid;

l2 5 62 2 32

l2 5 27

l 5 Ï}

27

S 5 B 1 1 } 2 Pl

5 62 1 1 } 2 (6 p 4) Ï

}

27

ø 36 1 62.35

ø 98.35


30. Right cone;

S 5 πr2 1 πrl

5 π(3)2 1 π(3)(9)

5 9π 1 27π

5 36π

ø 113.10


31. a. ∠ A > ∠ D by the Right Angle Congruence Theorem and ∠ C > ∠ C by the Refl exive Property of Congruence. So, by the AA Similarity Postulate, n ABC , nDEC.

b. AC2 1 AB2 5 BC2

42 1 32 5 BC2

25 5 BC2

5 5 BC

AB

} AC

5 DE

} DC

3 }

4 5

DE } 2

4DE 5 6

DE 5 6 } 4 5

3 } 2

AC

} BC

5 DC

} EC

4 } 5 5

2 } EC

4EC 5 10

EC 5 10

} 4 5 5 } 2


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c. Smaller cone:

S 5 πr2 1 πrl

5 π 1 3 } 2 2 2 1 π 1 3 }

2 2 1 5 }

2 2

5 9 } 4 π 1

15 } 4 π

5 6π

Larger cone:

S 5 πr2 1 πrl

5 π(3)2 1 π(3)(5)

5 9π 1 15π

5 24π

The surface area of the smaller cone is 1 }

4 5 25% of the

surface area of the larger cone.

32. a. A 5 m C AB

} 3608

p πr2

b. A 5 20

20 5 1508

} 3608

p πl2

20 5 5 } 12 p πl2

48

} π 5 l2

Î}

48

} π 5 l

3.91 ø l

The slant height is approximately 3.91 m.

c. Lateral area 5 πrl

20 5 πr Î}

48

} π 1.63 ø r

h2 1 r2 5 l2

h2 1 1.632 5 3.912

h2 ø 12.6312

h ø 3.55

The radius is about 1.63 m and the height is about 3.55 m.

33. Original radius to height 5 3 }

1.08

3 }

1.08 5

r2 } 1.08 2 0.83 → r2 ø 0.694

Original volcano:

l2 5 r2 1 h2 lateral area 5 πrl

l2 5 32 1 1.082 5 π(3)(3.188)

l ø 3.188 ø 30.0507

Destroyed top:

l2 5 r2 1 h2

l2 5 0.6942 1 0.252

l ø 0.7377

Lateral area 5 πrl 5 π(0.694)(0.7377) ø 1.6084

30.0507 2 1.6084 5 28.4423

The lateral area of the volcano after 1980 is about 28.44 mi2.

34.

16 in.

6 in.

53°

The height of the original cone 5 8 tan 538 and the

slant height of the original cone 5 Ï}}

82 1 (8 tan 538)2 ø 13.29 in.. The original cone’s lateral area is πrl 5 π(8)(13.29) ø 334.01 in.2. The cone that is cut off the top is similar to the original cone by the AA Similarity Postulate. So, the height of the cone cut off the

top is 3 tan 538 and the slant height is Ï}}

32 1 (3 tan 538)2 ø 4.98 in.. The lateral area of the top cone is πrl ø π(3)(4.98) ø 46.94 in.2. The surface area of the dog’s cone is the surface area of the original cone minus the surface area of the cone that was cut off the top.

S ø 334.01 2 46.94 ø 287.07

The surface area of the collar is approximately 287.07 in.2.

Mixed Review

35. 11x 2 3 5 10x 36. 3x8 5 (6x 2 45)8

23 5 2x 23x 5 245

3 5 x x 5 15

37. A 5 s2 5 72 5 49

The area is 49 mi2.

38. A 5 bl 5 (2 Ï}

2 )( Ï}

2 ) 5 4

The area is 4 yd2.

39. A 5 1 } 2 h(b1 1 b2) 5

1 } 2 (8)(9 1 10) 5 76

The area is 76 mm2.

Quiz 12.1–12.3 (p. 817)

1. F 1 V 5 E 1 2

F 1 8 5 12 1 2

F 1 8 5 14

F 5 6

The polyhedron has 6 faces.

2. S 5 2B 1 Ph

366 5 2(11x) 1 (22 1 2x)8

366 5 22x 1 176 1 16x

366 5 38x 1 176

190 5 38x

5 5 x

The value of x is 5 ft.

3. S 5 2πr2 1 2πrh

717 5 2π(6.1)2 1 2π(6.1)x

717 5 74.42π 1 12.2πx

483.20 ø 12.2πx

12.61 ø x

The value of x is about 12.61 in.


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411Geometry


4. S 5 2B 1 Ph

567 5 2 1 1 } 2 (9)(13) 2 1 (22 1 Ï

}

250 )x

567 5 117 1 (22 1 5 Ï}

10 )x 450 5 (22 1 5 Ï

}

10 )x 11.90 ø x

The value of x is about 11.90 m.

5. S 5 B 1 1 } 2 Pl

5 102 1 1 } 2 (4 p 10)(13)

5 100 1 260

5 360

The surface area is 360 cm2.

6. S 5 πr2 1 πrl

5 π(4)2 1 π(4)(9)

5 16π 1 36π

5 52π

ø 163.36


7. l2 5 h2 1 r2 S 5 πr2 1 πrl

l2 5 162 1 102 5 π(10)2 1 π(10)( Ï}

356 ) l2 5 356 5 100π 1 10π Ï

}

356

l 5 Ï}

356 ø 906.91


Mixed Review of Problem Solving (p. 818)

1. F 1 V 5 E 1 2

F 1 6 5 7 1 2

F 1 6 5 9

F 5 3

For a polyhedron to have 6 vertices and 7 edges it would need to have only 3 faces. Because it is not possible for a polyhedron to have only 3 faces, it is not possible for a polyhedron to have 6 vertices and 7 edges.

2. Method 1: Add the areas of all 6 sides of the rectangular solid.

Method 2: Use the formula S 5 2A 1 Ph where A is the areas of the base, P is the perimeter of the base, and h is the height of the solid.

3. a. S represents the total surface area; r represents the radius of the circumscribed circle for the base and the length of one side of the hexagonal base; h represents the length of the pencil.

b. The new length of the pencil is x and only one end is visible, so the formula for the visible portion is

S 5 3 Ï

}

3 } 2 r2 1 6rx.

c. The visible portion of the pencil is the lateral area of the cone-shaped end plus the lateral area of the hexagonal prism shaft.

Lateral area of Lateral area of hexagonal prism: cone end:

L1 5 6r(x 2 1) l2 5 r2 1 12

L1 5 6rx 2 6r l 5 Ï}

r2 1 1

L2 5 πrl

L2 5 πr( Ï}

r2 1 1 ) S 5 L1 1 L2

S 5 6rx 2 6r 1 πr( Ï}

r2 1 1 ) d. Let D be the difference in the two surface areas and r

be the radius of the circumscribed circle for the base and the length of one side of the hexagonal base.

D 5 3 Ï

}

3 } 2 r2 1 6rx 2 (6rx 2 6r 1 πr( Ï

}

r2 1 1 ))

5 3 Ï

}

3 } 2 r2 1 6r 2 πr( Ï

}

r2 1 1 )

4. Lateral area 5 2πrh 5 2π 1 2.8 }

2 2 (4) 5 11.2π ø 35.19

The lateral area is 35.19 in.2.

5. The only variables in the formula for the surface area of a right cone are the radius and the slant height. The radius is half the diameter. If the diameter and slant height are known, the surface area can be found.

S 5 π 1 d } 2 2 2 1 π 1 d } 2 2 l 6. Answers will vary.

7. a. The cross section is a square.

b. P 5 4s 5 4(10) 5 40

The perimeter is 40 feet.

A 5 s2 5 102 5 100

The area is 100 square feet.

c. Isosceles right triangles are formed.

d.

10

10

Area 5 1 } 2 bh 5

1 } 2 (10)(10) 5 50

The area is 50 square feet.

8.

3x

4x

Slant height 5 Ï}}

(4x)2 1 (3x)2

5 Ï}

16x2 1 9x2

5 Ï}

25x2

5 5x

Surface area 5 πr2 1 πrl

1944π 5 π(3x)2 1 π(3x)(5x)

1944π 5 9πx2 1 15πx2

1944π 5 24πx2

81 5 x2

9 5 x

The value of x is 9 units.


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Lesson 12.4


1. V 5 (1)(1)(1) 1 (3)(1)(2) 5 1 1 6 5 7

The volume is 7 units3.

2. V 5 Bh 5 (5)2(12) 5 300

The volume is 300 ft3.

3. V 5 πr2h

684π 5 πr2(18)

38 5 r2

Ï}

38 5 r

The radius is Ï}

38 inches.

4. V 5 Bh 5 1 } 2 (9 p 5)(8) 5 180

The volume is 180 m3.

5. B 5 Area of triangle 2 area of square

5 1 } 2 (10)(5 Ï

}

3 ) 2 32 5 25 Ï}

3 2 9

V 5 Bh 5 (25 Ï}

3 2 9)(6) ø 205.81

The volume is about 205.81 ft3.

12.4 Exercises (pp. 822–825)

Skill Practice

1. The volume of a solid is measured in cubic units.

2. Not necessarily. For example:

102

3

2

4

8

S 5 2B 1 Ph S 5 2B 1 Ph

5 2(20) 1 (24)(3) 5 2(16) 1 (20)(4)

5 112 square units 5 112 square units

V 5 Bh V 5 Bh

5 20(3) 5 16(4)

5 60 cubic units 5 64 cubic units

Two solids can have the same surface area and not have the same volume.

3. A; Volume of box 5 Bh 5 (15 p 9)(3) 5 405 cubic inches

Volume of cube 5 s3 5 33 5 27 cubic inches

Number of cubes 5 405

} 27 5 15

15 three inch cubes can fi t in the box.

4. V 5 2(3)(1) 1 7(2)(1) 5 6 1 14 5 20 unit cubes

The volume is 20 cubic units.

5. V 5 1(4)(2) 1 5(1)(2) 5 8 1 10 5 18 unit cubes


6. V 5 3(1)(4) 1 5(3)(4) 5 12 1 60 5 72 unit cubes


7. V 5 Bh 5 1 1 } 2 (7 p 10) 2 p 5 5 175

The volume is 175 in.3.

8. V 5 Bh 5 (4 p 2)(1.5) 5 12


9. V 5 Bh 5 6 1 1 } 2 (7.5)(3.75 Ï}

3 ) 2 (18) 5 1518.75 Ï}

3 ø 2630.55

The volume is about 2630.55 cm3.

10. V 5 πr2h 5 π(7)2(12) 5 588π ø 1847.26


11. V 5 πr2h 5 π 1 10 }

2 2 2(16) 5 400π ø 1256.64

The volume is about 1256.64 in.3.

12. V 5 πr2h 5 π 1 26.8 } 2 2 2(9.8) 5 1759.688π ø 5528.22


13. The volume of a right cylinder is πr2h not 2πrh.

V 5 πr2h 5 π(4)2(3) 5 48π

The volume is 48π ft3.

14.

7 ft

3 ft11 in.

In cubic inches,

V 5 Bh 5 (7 p 12)(3 p 12)(11) 5 33,264.

The volume is 33,264 in.3 5 19.25 ft3.

15. V 5 Bh

1000 5 (x p x) p x

1000 5 x3

10 5 x

The value of x is 10 inches.

16. B 5 1 } 2 bh 5

1 } 2 (5)( Ï

}

92 2 2.52 ) ø 21.615

V 5 Bh

45 ø 21.615x

x ø 2.08

The value of x is about 2.08 cm.

17. V 5 πr2h

128π 5 π 1 x } 2 2 2(8)

128π 5 2πx2

64 5 x2

8 5 x


18. V 5 πR2h 2 πr2h

5 π(3)2(7) 2 π(1)(7)

5 63π 2 7π 5 56π ø 175.93

The volume is about 175.93 m3.

19. V 5 Bh 5 [(7.8)(12.4) 2 (1.8)(3)] p 9

5 (96.72 2 5.4) p 9

5 (91.32)9 5 821.88

The volume is 821.88 ft3.


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413Geometry


20. V 5 Bh 1 1 } 2 (πr2h)

5 (4 p 4)(4) 1 1 } 2 (π(2)2(4))

5 64 1 8π

ø 89.13

The volume is about 89.13 in.3

21. A; V 5 πr2h

64π 5 π(4)2h

64π 5 16πh

4 5 h

The height of the cylinder is 4 feet.

22.

4 in.

3 in.

5 in.

h

V 5 Bh

96 5 1 } 2 (3 p 4)h

96 5 6h

16 5 h

The height of the prism is 16 inches.

23. V 5 πr2h

1005.5 5 πr2(8)

40 ø r2

6.33 ø r

d 5 2r ø 2(6.33) ø 12.65

The diameter is about 12.65 cm.

24. V 5 Bh 5 (4 p 7)(6) 5 168

The volume is 168 in.3

25. V 5 πr2h 5 π(8)2(14) 5 896π ø 2814.87


26. sin 608 5 h } 18

18 sin 608 5 h

9 Ï}

3 5 h

V 5 πr2h 5 π 1 12 }

2 2 2(9 Ï

}

3 ) 5 324π Ï}

3 ø 1763.01


27.

8 m8 m

8 m

6 m

6 ms

s 5 Ï}

62 1 82 5 Ï}

36 1 64 5 Ï}

100 5 10

Lateral area 5 P p h 5 4(10) p 8 5 320

The lateral area is 320 m2.

B 5 area of a rhombus 5 1 } 2 d1d2 5

1 } 2 (12)(16) 5 96

Surface area 5 2B 1 Ph 5 2(96) 1 320 5 512

The surface area is 512 m2.

Volume 5 Bh 5 96 p 8 5 768


Problem Solving

28. B 5 Area of rectangular face 2 Area of cylindrical end

5 9.5 2 π 1 3 } 2 2 2

5 45 2 9 } 4 π

V 5 Bh 5 1 45 2 9 } 4 π 2 (17) ø 644.83

The volume is about 644.83 mm3.

29. a. V 5 Volume of large prism

2 Volume of two smaller prisms

V 5 15.75(8)(8) 2 (4.5)(4)(8) 2 (4.5)(4)(8)

5 1008 2 144 2 144

5 720


b. B 5 Area of large rectangle

2 2 p Area of small rectangle

5 15.75(8) 2 2(4.5 p 4)

5 90

V 5 Bh 5 90(8) 5 720


c. The answers are the same.

30. a. V 5 πr2h

5 π 1 1000 }

2 2 2(400)

5 100,000,000π

ø 314,159,265

The volume of Blue Hole is about 314,159,265 ft3.

b. Gallons of water

5 314,159,265 ft3 3 7.48

} 1 ft3

ø 2,349,911,302

There are about 2,349,911,302 gallons of water in the Blue Hole.


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31. C 5 2πr

10 5 2πr

5 } π 5 r

V 5 πr2h 5 π 1 5 } π 2 2(20) ø 159.15

The volume of the column is about 159.15 ft3.

32. The fi rst cylinder has volume

V 5 πr2h 5 π(3)2(5) 5 45π

The second cylinder has volume

V 5 πr2h 5 π(5)2(3) 5 75π

The second cylinder has a greater volume because 75π is greater than 45π.

33. a. V 5 3 } 4 Bh 5

3 } 4 (30 p 10)(20) 5 4500

The volume of the water is 4500 in.3.

b. The volume of the rock is equal to the volume the water rose.

V 5 Bh 5 (30 p 10)(0.25) 5 75

The volume of the rock is 75 in.3.

c. The capacity of the tank is (30)(10)(20) 5 6000 in.3.

Let n represent the number of rocks that can be placed in tank.

Then, 6000 5 4500 1 75n

1500 5 75n

20 5 n

So, 20 rocks can be placed in the aquarium before the water spills out.

34. V 5 Bh 5 1 Area of rectangularbottom

1 Area of triangular top

2 h 9072 5 F (8)(18) 1

1 } 2 (18)( Ï

}

x2 2 92 ) G (36)

252 5 144 1 9 Ï}

x2 2 81

108 5 9 Ï}

x2 2 81

12 5 Ï}

x2 2 81

122 5 ( Ï}

x2 2 81 )2

144 5 x2 2 81

225 5 x2

Ï}

225 5 x

15 5 x

Each half of the roof is 15 ft by 36 ft.

Mixed Review

35. tan 22.58 5 2 } x

x 5 2 }

tan 22.58 ø 4.83

The value of x is about 4.83.

36. sin 368 5 7 } x

x 5 7 }

sin 368 ø 11.91


37. cos 758 5 x } 5

x 5 5 cos 758 ø 1.29


38. A 5 πr2 5 π(9.5)2 5 90.25π ø 283.53

The area is about 283.53 in.2.

39. A 5 1 } 2 aP 5

1 } 2 (7.5)(78) 5 292.5

The area is 292.5 m2.

40. 3608

} 5 5 728

10.6 in. a

36°

x

The apothem bisects a central angle, so the radius and apothem form a 368-548-908 triangle with one half of a side length. So, a 5 10.6 cos 368 and x 5 10.6 sin 368. The length of each side 5 2x 5 2(10.6 sin 368).

Area 5 1 } 2 aP 5

1 } 2 (10.6 cos 368)(5)(2)(10.6 sin 368)

ø 267.15


Problem Solving Workshop 12.4 (p. 827)

1. a. Volume V of pencil holder

5 Volume of large prism

2 Volume of 2 cylinders

5 7.5 p 4 p 4 2 2(π(1.5)2(4)) 5 120 2 18π

ø 63.45

The volume of the pencil holder is about 63.45 in.3.

b. B 5 Area of rectangle 2 Area of two circles

5 (7.5)(4) 2 2(π)(1.5)2

5 30 2 4.5π

V 5 Bh 5 (30 2 4.5π)4 ø 63.45

The volume of the pencil holder is about 63.45 in.3.

2. The student was correct that 4 times the base area of the cylinders needs to be subtracted, but they forgot that the lateral area of each cylinder needs to be added.

S 5 2(7.5 p 4) 1 2(7.5 p 4) 1 2(4 p 4) 2 4(π(1.5)2) 1 2(2π(1.5)4) S 5 152 1 15π

S ø 199.12

The surface area is about 199.12 square inches.


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415Geometry


3. Volume of cylinder 5 πR2h

Volume of hole 5 1 } 2 πR2h 5 πr2h

πr2h 5 1 } 2 πR2h

r2 5 1 } 2 R2

r 5 Ï}

1 }

2 R2

r 5 Ï

}

2 } 2 R

For the hole to have half the volume of the cylinder,

r should be Ï

}

2 }

2 R.

4.Volume of solid

5 Volume of large prism

2 Volume of small prism

V 5 (4)(2)(5) 2 (1)(1)(5) 5 40 2 5 5 35

The volume is 35 ft3.

5. V 5 Area of sector p h

5 608

} 3608

p π(2)2 p 3.5

5 1 } 6 (4π) p 3.5

5 7 } 3 π

ø 7.33


6. a. The surface area can be found by fi nding the sum of the areas of each exposed face.

b. Not all of the faces of the beams are exposed, so adding the individual surface areas will result in a total that is greater than the actual surface area.

Lesson 12.5


Step 3

It takes 3 times to fi ll the prism. So, the volume of the prism is 3 times the volume of the pyramid.

1. The areas of the bases are the same.

2. The heights are approximately the same.

3. The volume of the pyramid is one-third the volume of the prism.

4. V 5 1 } 3 Bh, where B is the area of the base and h is the

height.


1. 3608

} 6 5 608

a30°

4 yd

The apothem bisects the central angle, so a 308-608-908 triangle is formed.

a 5 2 }

tan308 5 2 Ï

}

3

B 5 1 } 2 aP V 5

1 } 3 Bh

5 1 } 2 1 2 Ï

}

3 2 (6 p 4) 5 1 } 3 1 24 Ï

}

3 2 (11)

5 24 Ï}

3 5 88 Ï}

3 ø 152.42

The volume is about 152.42 yd3.

2. h2 5 l2 2 r2

h2 5 82 2 52

h2 5 39

h 5 Ï}

39

V 5 1 } 3 Bh

5 1 } 3 1 π(5)2 2 1 Ï

}

39 2

5 25

} 3 π Ï}

39

ø 163.49


3. V 5 1 } 3 Bh

1350π 5 1 } 3 (π)(18)2h

1350π 5 108πh

12.5 5 h

The height is 12.5 meters.

4. tan 408 5 r } 5.8

r 5 5.8 tan 408

V 5 1 } 3 1 πr2 2 h

5 1 } 3 (π)(5.8 tan 408)2(5.8)

ø 143.86



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5. Volume 5 Volume of cylinder 1 Volume of cone

5 πr2h 1 1 } 3 1 πr2 2 h 5 cm

3 cm

10 cm 5 π(3)2(10) 1

1 } 3 1 π p 32 2 5

5 90π 1 15π

5 105π

ø 329.87

The volume of the solid is about 329.87 cubic centimeters.

6. Flow rate 5 volume

} time

5 101mL

} 3.2 s ø 31.56 mL/s

The flow rate of the sand is about 31.56 mL/s.

12.5 Exercises (pp. 832–836)

Skill Practice

1. A triangular prism has two congruent triangular bases that are parallel to each other. The lateral faces are all parallelograms. A triangular pyramid has a single triangular base with lateral faces are triangles with a common vertex.

prism

pyramid

Triangular prism Triangular pyramid

2. The volume of a square pyramid is one third the volume of a square prism with the same base and height.

3. V 5 1 } 3 Bh 5

1 } 3 1 52 2 (6) 5 50

The volume is 50 cm3.

4. V 5 1 } 3 Bh 5

1 } 3 1 πr2 2 h 5

1 } 3 1 π p 102 2 (13) 5

1300 } 3 π ø 1361.36

The volume is about 1361.36 mm3.

5. V 5 1 } 3 Bh 5

1 } 3 (2.5)(4) 5

40 } 3 ø 13.33


6. V 5 1 } 3 Bh 5

1 } 3 1 πr2 2 h 5

1 } 3 1 π p 12 2 (2) 5

2 } 3 π ø 2.09


7. V 5 1 } 3 Bh 5

1 } 3 1 1 }

2 p 3 p 4 2 (3) 5 6


8. V 5 1 } 3 Bh 5

1 } 3 1 6 p 1 }

2 p 6 Ï

}

3 p 12 2 (17) 5 1224 Ï}

3 ø 2120.03


9. The slant height was used instead of the actual height.

h 5 Ï}

152 2 92 5 Ï}

144 5 12

V 5 1 } 3 π 1 92 2 (12) 5 324π ø 1018

The volume is about 1018 ft3.

10. The volume of the pyramid is 1}3

Bh not 1}2

Bh.

V 5 1}31722 (10) 5

490}

3 ø 163.3


11. D; V 5 1}3Bh

45 5 1}3B(9)

45 5 3B

15 5 B

The area of the base is 15 ft2.

12. V 5 1}3Bh

200 5 1}311022x

200 5 100}

3 x

6 5 x

The value of x is 6 cm.

13. V 5 1}3Bh

216π 5 1}31πr22h

216π 5 1}31π p x22 (18)

216π 5 6πx2

36 5 x2

6 5 x


14. V 5 1}3Bh

7 Ï}

3 5 1}311

}212 Ï

}

3 2 (3) 2 (x)

7 Ï}

3 5 Ï}

3 x

7 5 x

The value of x is 7 feet.

15. tan 608 5 opp.}adj.

Ï}

3 5 22}r

r 5 22} Ï

}

3

V 5 1}31πr22h ø 1}

3π1 22

} Ï

}

3 22(22) ø 3716.85


16. tan 328 5 opp.}adj.

tan 328 5 7}h

h 5 7

}tan 328

V 5 1}31πr22h ø 1}

3π(7)2 1 7

}tan 328 2 ø 574.82



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417Geometry


17. cos 548 5 adj.

} hyp.

cos 548 5 r } 15

r 5 15 cos 548

sin 548 5 opp.

} hyp.

sin 548 5 h } 15

h 5 15 sin 548

V 5 1 } 3 1 πr2 2 h

5 1 } 3 π(15 cos 548)2(15 sin 548)

ø 987.86


18. B; tan 298 5 opp.

} adj.

tan 298 5 5 }

h

h 5 5 }

tan 298

V 5 1 } 3 1 πr2 2 h ø 1 }

3 1 π p 52 2 1 5

} tan 298

2 ø 236.15

The approximate volume of the cone is 236.15 ft3.

19. V 5 1 } 3 1 πr2 2 h

143.6 5 1 } 3 1 π p 42 2 h

143.6 5 1 16 π } 3 2 h

8.57 ø h The height of the cone is about 8.57 cm.

20. V 5 Volume of cylinder 1 Volume of cone

5 πr2h 1 1 } 3 1 πr2 2 h 5 π p 32 p 7 1

1 } 3 1 π p 32 2 (3)

5 63π 1 9π 5 72π ø 226.19


21. V 5 Volume of cube 2 Volume of pyramid

5 s3 2 1 } 3 Bh 5 103 2

1 } 3 1 5 Ï

}

2 2 2(10) ø 833.33


22. V 5 Volume of large cone 1 Volume of small cone

5 1 } 3 1 πR2 2 H 1

1 } 3 1 πr2 2 h 5

1 } 3 1 π p 12 2 (2) 1

1 } 3 1 π 1 1 }

2 2 2 2 (1)

5 2 } 3 π 1

1 } 12 π 5

3 } 4 π ø 2.36


23. V 5 2 p Volume of pyramid 5 2 1 1 } 3 Bh 2 5

2 } 3 1 3.32 2 (2.3)

ø 16.70


24. V 5 Volume of cube 2 Volume of cone

5 S3 2 1 } 3 1 πr2 2 h 5 5.13 2

1 } 3 1 π p 2.552 2 (5.1)

ø 97.92


25. V 5 Volume of cylinder 2 Volume of pyramid

5 1 πr2 2 h 2 1 } 3 Bh

B 5 1 } 2 aP

a 5 2 cos 22.58

P 5 8 1 4 sin 22.58 2

V 5 1 π p 22 2 (3) 2 1 } 3 1 1 } 2 1 2 cos 22.58 2 (8) 1 4 sin 22.58 2 2

ø 26.39


26. V 5 1 } 3 1 πr2 2 h 5

1 } 3 1 π p 22 2 (3) 5 4π ø 12.57

The volume of the solid is about 12.57 cm3.

27. Drawing a diagonal in the square creates two 458-458-908 triangles. The diagonal has a length of 5 Ï

}

2 meters, which is also the diameter of the base of the cone.

So, the radius is 5 Ï

}

2 }

2 .

V 5 1 } 3 1 πr2 2 h 5 m

7 m

5 1 } 3 1 π 1 5 Ï

}

2 }

2 2 2 2 (7)

ø 91.63


28.

a

3 ft

30°

B

C

35°

A

C B

sin 308 5 opp.

} hyp.

tan 358 5 opp.

} adj.

0.5 5 1.5

} BC tan 358 5 AC

} 3

BC 5 3 AC 5 3 tan 358 ø 2.10

tan 308 5 opp.

} adj.

Ï

}

3 }

3 5

1.5 } a

a 5 1.5 Ï}

3

V 5 1 } 3 Bh

ø 1 } 3 1 6 p 1 }

2 p 1.5 Ï

}

3 p 3 2 (2.10)

ø 16.37



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Problem Solving

29. a. V 5 1 } 3 1 πr2 2 h 5

1 } 3 1 π p 42 2 (12) 5 64π ø 201

There are about 201 cubic inches of frosting in the bag.

b. 201 in.3

} 15 flowers

5 13.4 in.3 per flower

About 13.4 cubic inches of frosting are used per flower.

30. V 5 1 } 3 1 πr2 2 h 5

1 } 3 1 π p 32 2 (8) 5 24π ø 75.40

The volume of the small cup is about 75.4 in.3.

31. The cone and the cylinder have the same base and height,

so the volume of the cone is 1 }

3 p Volume of cylinder.

This means you have to buy 3 small cups of popcorn to have the same amount as the large cup.

32. The cost of the large cup is twice the cost of the small cup but the volume is three times as great. So, the large cylindrical cup gives you more popcorn for your money.

33. h 5 Ï}

62 2 22 5 Ï}

32 5 4 Ï}

2

V 5 1 } 3 1 πr2 2 h

2 in.

6 in.

5 1 } 3 1 π p 22 2 1 4 Ï

}

2 2

ø 23.70


34.

5 ft

5 ft

5 ft

5 ft

5 ft

2.5 ft

2.5 ft3h

l 5 Ï}

52 2 2.52 5 2.5 Ï}

3

h 5 Ï}}

1 2.5 Ï}

3 2 2 2 2.52 5 2.5 Ï}

2

V 5 1 } 3 Bh 5

1 } 3 1 52 2 1 2.5 Ï

}

2 2 ø 29.46


35. a. Original volume 5 1 } 3 Bh 5

1 } 3 1 72 2 (10) 5

490 } 3

Let h 5 2(10): V 5 1 } 3 Bh 5

1 } 3 1 72 2 (20) 5 2 1 490

} 3 2

The volume is doubled.

b. Let S 5 side length 5 2(7)

V 5 1 } 3 Bh 5

1 } 3 (2(7))2(10) 5 4 1 490

} 3 2

The original volume is multiplied by four.

c. If the height of a pyramid is multiplied by x, the volume is multiplied by a factor of x. If the side length is multiplied by x, the volume is multiplied by a factor of x2.

36. a. Volume 5 Volume of cylinder 1 Volume of cone

5 πr2h 1 1 } 3 1 πr2 2 h

5 π 1 2.52 2 (7.5) 1 1 } 3 1 π p 2.52 2 (4)

ø 173.44 in.3

173.44 in.3

} 14.4 in.3

ø 12.04

The container can hold about 12 cups.

b. 2 p 1 } 3 5

2 } 3 cups a day.

days 5 2 }

2 }

3 5 12 p 3 }

2 5 18

The feeder can go 18 days before it needs to be refi lled.

37. Volume of big prism 5 6 1 Ï}

3 }

4 S2 2 h

5 6 1 Ï}

3 }

4 (3.5)2 2 (1.5)

ø 47.74

Volume of small prism 5 6 1 Ï}

3 }

4 S2 2 h

5 6 1 Ï}

3 }

4 (3.25)2 2 (0.25)

ø 6.86

Volume of pyramid 5 1 } 3 p 6 1 Ï

}

3 }

4 S2 2 h

5 1 } 3 p 6 1 Ï

}

3 }

4 1 32 2 2 (3)

ø 23.38

Volume of solid 5 47.74 1 6.86 1 23.38 5 77.98

The volume of the deck prism is about 77.98 in.3.

38. a. Average area 5 π p 1 average value of R2 2 5 π 1 R2 }

3 2

b. Vcone 5 1 } 3 Bh 5

1 } 3 πR2h 5

πR2

} 3 p h,

where B is the area of the base of the cone, πR2

} 3 is the

average area of a circular cross section, and h is the height.


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419Geometry


39. a. h 5 2r → r 5 h } 2

V 5 1 }

3 Bh 5

1 } 3 πr2h 5

1 } 3 π 1 h }

2 2 2(h) 5

1 } 3 π 1 h2

} 4 2 (h) 5

πh3

} 12 ,

where B is the area of the base of the cone, r is the radius, and h is the height.

b. Time

(minutes)Height

(meters)

1 1.90

2 2.40

3 2.74

4 3.02

5 3.25

c.

Time (minutes)

Hei

gh

t (m

eter

s)

0 2 4 60

2

4

6h

t

There is not a linear relationship between the height of the water and time. The data points on the graph do not form a straight line, so there cannot be a linear relationship.

40. V 5 [area of base1 1 area of base2 1 geometric mean] 1 }

3 h

5 F π(3)2 1 π(9)2 1 Ï}}

π(3)2 p π(9)2 G 1 }

3 (10)

ø F 9π 1 81π 1 84.82 G 10 }

3

ø [367.57] 10

} 3

ø 1225.22

The volume of the frustum is about 1225.22 cm3.

41. a. h1 1 h2

} r2

5 h1

} r1

h1r2 5 r1(h1 1 h2)

h1r2 5 h1r1 1 h2r1

h1r2 2 h1r1 5 h2r1

h1(r2 2 r1) 5 h2r1

h1 5 h2r1

} r2 2 r1

b. V 5 πr2

2(h1 1 h2) } 3 2 πr1

2h1 } 3

5 πr22 1 r1h2

} r2 2 r1

1 h2 2 }

3 2

πr12 1 r1h2

} r2 2 r1

2 } 3

c. V 5 πr22 1 r1h2

} r2 2 r1

1 h2 2 }

3 2

πr12 1 r1h2

} r2 2 r1

2 } 3

5 1 } 3 1 πr2

2 1 r1h2 1 r2h2 2 r1h2 }}

r2 2 r1 2 2 π 1 r1

3h2 } r2 2 r1 2 2

5 π

} 3 1 r23h2 }

r2 2 r1 2

r13h2 } r2 2 r1 2

5 π h2

} 3 1 r23 2 r1

3

} r2 2 r1

2

5 πh2

} 3 1 (r2 2 r1) 1 r22 1 r1r2 1 r1

2 2 }}

r2 2 r1 2

5 h2

} 3 1 πr22 1 πr1r2 1 πr1

2 2

The volume is equal to one-third of the height times the area of the lower base plus the geometric mean of the two bases plus the area of the upper base.

42. V 5 1 } 3 Bh

r

h 5

1 } 3 1 6 Ï

}

2 2 2(12)

5 1 } 3 (72)(12)

5 288

The volume of the pyramid is 288 cubic feet.

Mixed Review

43. 3.5

} 6 5

x } 12 44.

x }

15 5

8 } 20

6x 5 12(3.5) 20x 5 8(15)

6x 5 42 20x 5 120

x 5 7 x 5 6

45. 6 }

9 5

8 } 9 1 x or

6 }

2 5

9 } x

6(9 1 x) 5 72 6x 5 18

9 1 x 5 12 x 5 3

x 5 3

46. A

F

E

B

D

C

}

AF is a radius, } BE is a diameter, and }

CD is a chord.

47. C BC is a minor arc. 48. C ABE is a major arc.

49. A 5 π(3)2 5 9π ø 28.27

The area is about 28.27 m2.

50. A 5 π(3.5)2 5 12.25π ø 38.48

The area is about 38.48 mi2.

51. A 5 π(0.4)2 5 0.16π ø 0.50

The area is about 0.50 cm2.


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52. 2πr 5 8π

r 5 4

A 5 π(4)2 5 16π ø 50.27


Quiz 12.4–12.5 (p. 836)

1. V 5 Bh 5 1 } 2 (10)(15)(7) 5 525


2. V 5 πr2h 5 π(6)2(10) 5 360π ø 1130.97


3. V 5 πr2h 5 π(4.5)2(16) 5 324π ø 1017.88


4. V 5 1 } 3 Bh 5

1 } 3 1 32 2 (2) 5 6


5. V 5 1 } 3 (πr2)h 5

1 } 3 1 π p 302 2 (40) ø 37,699.11

The volume is about 37,699.11 ft3.

6. V 5 1 } 3 (πr2)h 5

1 } 3 1 π p 82 2 (15) 5 320π ø 1005.31


7. The cups have the same radius and height, so the volume of the cone is one-third the volume of the cylinder. When one conical cup is poured into the cylindrical cup, one third of the cylinder will be full. The full level will be at a height that is one-third the cylinder’s height

or 1 }

3 (6) 5 2 inches.

Spreadsheet Activity 12.5 (p. 837)

1. The radius should be 2.25 cm and the height should be approximately 4.53 cm. The value in the surface area column decreases as you approach 95.81 and increases after.

2. The radius should be about 4.64 cm and the height should be about 9.29 cm.

Lesson 12.6


1. S 5 4πr2 5 4π(20)2 5 1600π ø 5026.55

The surface area is about 5026.55 ft 2.

2. S 5 4πr2

30π 5 4πr2

7.5 5 r2

2.74 ø r The radius is about 2.74 m.

3. C 5 2πr

2πr 5 6π

r 5 3

S 5 4πr2 5 4π(3)2 5 36π

The surface area of the inner ball is 36π, or about 113.10 ft2.

4. V 5 4 } 3 πr3 5

4 } 3 π(5)3 ø 523.60


5. V 5 Volume of cone 1 Volume of hemisphere

5 1 } 3 πr2h 1

1 } 2 1 4 }

3 πr3 2

5 1 } 3 π(1)2(5) 1

2 } 3 π(1)3

1 m

5 m 5 5 } 3 π 1

2 } 3 π

5 7 } 3 π

ø 7.33


12.6 Exercises (pp. 842–845)

Skill Practice

1. The formula for the surface area of a sphere is 4πr2 and

the formula for the volume of a sphere is 4 }

3 πr3, where r

is the radius of the sphere.

2. The center of the sphere must be contained for the intersection to be a great circle.

3. S 5 4πr2 5 4π(4)2 5 64π ø 201.06


4. S 5 4πr2 5 4π(7.5)2 5 225π ø 706.86


5. S 5 4πr2 5 4π(9.15)2 5 334.89π ø 1052.09


6. B; S 5 4πr2

32π 5 4πr2

8 5 r2

Ï}

8 5 r

The approximate radius is 2.83 m.

7. C 5 2πr

9.6π 5 2πr

4.8 5 r

The radius is 4.8 inches.

8. d 5 2r 5 2(4.8) 5 9.6

The diameter is 9.6 inches.

9. S 5 1 } 2 (4πr2) 5

1 } 2 (4π(4.8)2) 5 46.08π ø 144.76

The surface area of the hemisphere is about 144.76 in.2.

10. The surface area of a hemisphere is half the surface area of the sphere.

S 5 1 } 2 (4πr2) 5

1 } 2 (4π (5)2) 5 50π ø 157.08

The surface area of the hemisphere is about 157.08 ft2.

11. C 5 2πr

48.4π 5 2πr

24.2 5 r

S 5 4πr2 5 4π(24.2)2 ø 7359.37



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421Geometry


12. V 5 4 } 3 πr3 5

4 } 3 π(6)3 5 288π ø 904.78


13. V 5 4 } 3 πr3 5

4 } 3 π (40)3 ø 268,082.57

The volume is about 268,082.57 mm3.

14. V 5 4 } 3 πr3 5

4 } 3 π(2.5)3 ø 65.45


15. The formula for the volume of a sphere is 4 }

3 πr3

not 4 }

3 πr2.

V 5 4 } 3 πr3 5

4 } 3 π(8)3 5 682.66π ø 2144.66


16. V 5 4 } 3 πr3 17. V 5

4 } 3 πr3

1436.76 5 4 } 3 πr3 91.95 5

4 } 3 πr 3

343 ø r3 21.95 ø r3

7 ø r 2.80 ø r The radius is about The radius is about

7.00 m. 2.80 cm.

18. V 5 4 } 3 πr3

20,814.37 5 4 } 3 πr3

4969.06 ø r3

17.06 ø r The radius is about 17.06 in.

19. V 5 4 } 3 πr3

36π 5 4 } 3 πr3

27 5 r3

3 5 r

The radius is 3 feet, so the diameter is 2(3) 5 6 feet.

20. A; S 5 4πr2

V 5 4 } 3 πr3 5

r }

3 1 4πr2 2 5

rS } 3

The relationship is V 5 rS

} 3 .

21. S 5 πr2 1 2πrh 1 1 } 2 1 4πr2 2

5 π(3.3)2 1 2π(3.3)(7) 1 1 } 2 (4)(π)(3.3)2

ø 247.78


V 5 πr 2h 2 1 } 2 1 4 }

3 πr3 2

5 π(3.3)2(7) 2 1 } 2 1 4 }

3 2 (π)(3.3)3

ø 164.22


22. S 5 πr2 1 2πrh 1 1 } 2 1 4πr2 2

5 π(5.8)2 1 2π(5.8)(14) 1 1 } 2 (4)(π)(5.8)2

ø 827.24


V 5 πr 2h 1 1 } 2 1 4 }

3 πr3 2

5 π(5.8)2(14) 1 1 } 2 1 4 }

3 2 (π)(5.8)3

ø 1888.21


23. S 5 πrl 1 1 } 2 1 4πr2 2

5 π(4.9)2(13.52) 1 1 } 2 (4)(π)(4.9)2

ø 358.97


V 5 1 } 3 πr 2h 1

1 } 2 1 4 }

3 πr3 2

5 1 } 3 π(4.9)2(12.6) 1

1 } 2 1 4 }

3 2 (π)(4.9)3

ø 563.21


24–27. Radius of

sphereCircumferenceof great circle

24. 10 ft 20π ft

25. 13 in. 26π in.

26. 25 cm 50π cm

27. 21 m 42π m

Surface area

of sphereVolume

of sphere

24. 400π ft2 1 4000 }

3 2 π ft3

25. 676π in.2 1 8788 }

3 2 π in.3

26. 2500π cm2 1 62,500 }

3 2 π cm3

27. 1764π m2 12,348π m3

28. C; Volume of cube 5 s3

s3 5 64

s 5 4

r 5 s } 2 5 2

Surface area of sphere 5 4πr2

5 4π(2)2

5 16π



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-422GeometryWorked-Out Solution Key

29. a. V 5 4 } 3 πr3 5

4 } 3 π(1)35

4 } 3 π in.3

Sample answers: r 5 1 in., h 5 4 } 3 in.

r 5 0.5 in., h 5 16

} 3 in.

r 5 2 in., h 5 1 } 3 in.

b. The surface area of the cylinder will be greater than the surface area of the sphere when 2πrh 1 2πr2 > 4πr2. Solve to get h > r. The surface area of the cylinder will be greater than the surface area of the sphere when the height of the cylinder is greater than the radius.

Problem Solving

30. V 5 πr 2h 1 1 } 2 1 4 }

3 πr3 2

5 π(10)2(60) 1 1 } 2 1 4 }

3 2 (π)(10)3

ø 20,943.95

The volume of the grain silo is about 20,944 ft3.

31. C 5 2πr

24,855 5 2πr

24855

} 2π� 5 r

S 5 1 } 2 (4πr2) 5

1 } 2 (4)(π )

24855 }

2π ø 98,321,312.33

The surface area of the western hemisphere is approximately 98,321,312 square miles.

32. a. V 5 4 } 3 πr3

1427.54 5 4 } 3 πr3

340.8 ø r3

6.99 ø r The radius is about 6.99 cm.

b. S 5 4πr2 5 4π(6.99)2 ø 613.99


33. a. C 5 2πr

8 5 2πr

4 } π 5 r

V 5 4 } 3 πr3 5

4 } 3 π 1 4 }

π 2 3 ø 8.65

The volume of a tennis ball is about 8.65 in.3.

b. V 5 Volume of cylinder 2 3 p volume of one ball

5 πr 2h 2 3 1 4 } 3 πr3 2

5 π(1.43)2(8.625) 2 3 1 4 } 3 π 1 4 }

π 2 3 2

ø 29.47

There is about 29.47 in.3 of space not taken up.

34. a. C 5 2πr

27π 5 2πr

13.5 5 r

V 5 4 } 3 πr3 5

4 }

3 π(13.5)3 ø 10,306

The volume of the balloon is about 10,306 cm3.

b. The radius is cubed, so (2r)3 would give 8r3. This means the volume should be eight times the original volume.

c. V 5 4 } 3 π(2r)3 5

4 } 3 π(2.7)3 ø 82447.96 ø 8(10,306)

The volume is about 82,447.96 cm3. The prediction from part (b) was correct. The ratio of this volume to the original is 8 to 1 or 8 : 1 .

35. a. Surface area of Torrid Zone 5 2πrh

5 2π(3963)(3250)

ø 80,925,856 mi2

Surface area of Earth 5 4πr2

5 4π(3963)2

ø 197,359,488 mi2

b. Probability of Torrid Zone 5 Surface area of Torrid Zone

}}} Surface area of the Earth

5 80,925,856 mi2

}} 197,359,488 mi2

ø 0.41

The probability that a meterorite will land in the Torrid Zone is about 41%.

36. a. Solid I: S 5 4πr2

Solid II: S 5 2πr2 1 2πr(2r) 5 6πr2

Solid III: S 5 2 F πr 1 r Ï}

2 2 1 πr2 G ø 4.83πr2

Solid I, Solid III, Solid II

b. Solid I: V 5 4 } 3 πr3

Solid II: V 5 πr2(2r) 5 2πr3

Solid III: V 5 2 F 1 } 3 1 πr2 2 (r) G 5

2 } 3 πr3

Solid III, Solid I, Solid II

37. A sphere is formed.

S 5 4πr2 5 4π(9)2 ø 1017.88

The surface area is about 1018 in.2.

V 5 4 } 3 πr3 5

4 } 3 π(9)3 ø 3053.63

The volume is about 3054 in.3.

38. a. The diagonal of the cylinder’s height and diameter is equal to the diameter of the sphere.

Using the Pythagorean Theorem: 8

8 x

x

r

r

r2 5 82 2 x2

r2 5 64 2 x2

Volume of the cylinder 5 πr2h

5 π 1 64 2 x2 2 (2x)

5 2πx 1 64 2 x2 2


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423Geometry


b.

22500

2500

The maximum value occurs when x is about 4.62 m.

c. V 5 2πx 1 64 2 x2 2

5 2π(4.62) 1 64 2 4.622 2

ø 1238.22

The maximum volume is about 1238.22 m3.

39. The lateral edge of the cone is tangent to the sphere, so a radius drawn from the sphere is perpendicular to the tangent lateral side. By the AA Similarity Postulate, two similar right triangles are formed by this radius.

} AC has a

length of 4 cm because }

EC is a radius of the sphere and has length 2 cm. By the Pythagorean Theorem,

} AB has

length Ï}

42 2 22 5 Ï}

12 5 2 Ï}

3 cm.

2 cm

6 cm

E

A

D

CB

BC

} DE

5 AB

} EA BC

} DE

5 AC

} AD

2 }

DE 5

2 Ï}

3 } 6

2 }

2 Ï}

3 5

4 } AD

DE 1 2 Ï}

3 2 5 12 2(AD) 5 8 Ï}

3

DE 5 2 Ï}

3 AD 5 4 Ï}

3

The radius of the cone is 2 Ï}

3 cm and the lateral edge is 4 Ï

}

3 cm.

S 5 πrl 1 πr2

5 π 1 2 Ï}

3 2 1 4 Ï}

3 2 1 π 1 2 Ï}

3 2 2

5 36π

ø 113.10


V 5 1 } 3 πr2h 5

1 } 3 π 1 2 Ï

}

3 2 2(6) 5 24π ø 75.40


Mixed Review

40. ratio of sides 5 8 : 5

ratio of areas 5 82 : 52 or 64 : 25

Area of nABC

}} Area of nDEF

5 64

} 25

42 }}

Area of nDEF 5

64 } 25

64(Area of nDEF ) 5 25(42)

Area of nDEF ø 16.41

Area of nDEF is about 16.41 ft2.

41. ratio of sides 5 15 : 21 or 5 : 7

ratio of areas 5 52 : 72 or 25 : 49

Area of JKLM

}} Area of PQRS

5 25

} 49

A }

195 5

25 } 49

49A 5 25(195)

A 5 99.49

The area of PQRS is about 99.49 cm2.

42. P 5 Area of shaded region

}} Total area

5 28(14) 2 2π(7)2

}} 28(14)

ø 0.21

The probability that the point lies in the shaded region is about 21%.

43. P 5 Area of shaded region

}} Total area

5 5 1 1 }

2 2 (7.85)(11.40)

}} π(9.7)2 ø 0.76

The probability that the point lies in the shaded region is about 76%.

44. Volume of cone 5 1 } 3 p Volume of cylinder

5 1 } 3 (330)

5 110 cubic units

Lesson 12.7


Step 1 and Step 2:

Scale factorof Solid A to Solid B

Surface area of

Solid A, SA

Surface area of

Solid B, SB

Pair 1 1 } 2 72 288

Pair 2 2 } 3 28π 63π

Pair 3 3 } 1 675 75

SA } SB

VA VB VA } VB

Pair 1 1 } 4 36 288

1 }

8

Pair 2 4 } 9 20π 67.5π

8 }

27

Pair 3 9 } 1 974.28 36.08

27 }

1

Step 3:

SA

} SB

is the square of the scale factor while VA

} VB is the cube

of the scale factor.

1. The ratio of the surface areas of similar solids is the square of the scale factor of the solids. The ratio of the volumes of similar solids is the cube of the scale factor of the solids.

2. ratio of surface areas 5 1 }

k 2

ratio of volumes 5 1 }

k3


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1. Lengths 5 12

} 9 5 4 } 3

Widths 5 16

} 12 5 4 } 3

Heights 5 4 } 3

The prisms are similar because the ratios of corresponding linear measures are equal.

2. Heights 5 15

} 10 5 3 } 2

Radii 5 10

} 5 5 2

The cones are not similar because the ratios of corresponding linear measures are not equal.

3. C2

} D2 5

54 } 150

C

} D 5 3 Ï

}

6 }

5 Ï}

6

C

} D 5 3 } 5

The scale factor of C to D is 3 } 5 .

The edge length of C is Ï}

54

} 6 5 3 units.

33

} 53 5

C3

} D3

27

} 125

5 27

} D3

D3 5 125

The volume of D is 125 cubic units.

4. 23

} 13 5

Price }

$1.50

Price 5 8($1.50) 5 $12

The price of the large ball should be $12 for neither ball to be a better buy.

12.7 Exercises (pp. 850–854)

Skill Practice

1. Two solids are similar if they are the same type of solid with equal ratios of all corresponding linear measures.

2. If the corresponding linear measures have ratio a :b, then the volumes are in the ratio a3:b3.

3. Radii 5 7 } 4

Heights 5 16

} 10 5 8 } 5

The solids are not similar because the ratios of corresponding linear measures are not equal.

4. Lengths 5 5 } 9

Widths 5 7 } 12.6 5

5 } 9

Heights 5 11

} 14.8 5 55

} 74

The solids are not similar because the ratios of corresponding linear measures are not equal.

5. Lengths 5 6 } 8 5

3 } 4

Widths 5 13.5

} 18 5 3 } 4

Heights 5 4.5

} 6 5 3 } 4

The solids are similar because the ratios of corresponding linear measures are equal.

6. Radii 5 8 } 12 5

2 } 3

Heights 5 18

} 27 5 2 } 3

The solids are similar because the ratios of corresponding linear measures are equal.

7. D; Lengths 5 6 } 15 5

2 } 5

Widths 5 4 } 10 5

2 } 5

Heights 5 10

} 25 5 2 } 5

15 feet by 10 feet by 25 feet

8. Surface area of A

}} Surface area of B

5 a2

} b2

150π }}

Surface area of B 5

12

} 22

Surface area of B 5 600π in.2

Volume of A

} Volume of B

5 a3

} b3

250π }

Volume of B 5

13

} 23

Volume of B 5 2000π in.3



5 a2

} b2

1500 }}

Surface area of B 5

32

} 12

Surface area of B ø 166.67 m2

Volume of A

} Volume of B

5 a3

} b3

3434.6

} Volume of B

5 33

} 13

Volume of B 5 127.21 m3


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425Geometry




5 a2

} b2

2356.2 }}

Surface area of B 5

52

} 22

Surface area of B ø 377 cm2

Volume of A

} Volume of B

5 a3

} b3

7450.9

} Volume of B

5 53

} 23

Volume of B ø 476.86 cm3

11. The ratio of the volumes is a3

} b3 , not

a2 }

b2 .

500π }

Volume of B 5

13

} 43

12. a3

} b3 5

8π } 125 π

a3

} b3 5

8 } 125

a }

b 5

2 } 5

The scale factor of Solid I to Solid II is 2 : 5 .

13. a3

} b3 5

27 } 729

a }

b 5

3 } 9

a }

b 5

1 } 3


14. a2

} b2 5

288 } 128

a2

} b2 5

144 } 64

a }

b 5

12 } 8

a }

b 5

3 } 2


15. a2

} b2 5

192 } 108

a2

} b2 5

16 } 9

a }

b 5

4 } 3


16. C; a3

} b3 5

8π } 27π

a3

} b3 5

8 } 27

a }

b 5

2 } 3

a2

} b2 5

22

} 32

5 4 } 9

17. a3

} b3 5

2π } 16π

a3

} b3 5

1 } 8

a }

b 5

1 } 2

a2

} b2 5

1 } 4

The ratio of the surface area of the smaller sphere to the surface area of the larger sphere is 1 : 4 .



5 a2

} b2

78π }}

Surface area of B 5

22

} 32

The surface area of B or the large cylinder is 175.5π ø 551.35 m2.

19. Scale factor 5 4 } 8 5

1 } 2

Surface area of Solid I 5 πr2 1 2πrh 1 πrl

5 π(2)2 1 2π(2)(4) 1 π(2) 1 Ï}

13 2

ø 85.49 ft2

Volume of Solid I 5 πr2h 1 1 } 3 πr2h

5 π(2)2(4) 1 1 } 3 π(2)2(3)

ø 62.83 ft3

Surface area of Solid I

}} Surface area of Solid II

5 a2

} b2

85.49 }}

Surface area of Solid II 5

12

} 22

Surface area of Solid II ø 341.96 ft2

Volume of Solid I

}} Volume of Solid II

5 a3

} b3

62.83 }}

Volume of Solid II 5

13

} 23

Volume of Solid II ø 502.64 ft3

20. Scale factor 5 8 } 3

Surface area of Solid I 5 B 1 Pl 1 4 1 1 } 4 p Ï

}

3 p 32 2 5 32 1 12(8) 1 4 1 1 }

4 p Ï

}

3 p 32 2 ø 120.59 cm2

Volume of Solid II 5 Bh 1 1 } 3 Bh

5 1 32 2 (8) 1 1 } 3 1 32 2 1 1.5 Ï

}

2 2

ø 78.36 cm3



5 a2

} b2

120.59


5 82

} 32

Surface area of Solid II ø 16.96 cm2

Volume of Solid I


5 a3

} b3

78.36 }}


83

} 33

Volume of Solid II ø 4.13 cm3


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Surface area of Solid I 5 2lh 1 2lw 2 2πr2 1 2πrh

5 2(4)(4) 1 2(4)(4) 2 2π 1 22 2

1 2π(2)(4)

ø 89.13 in.2

Volume of Solid I 5 Bh

5 (42 2 π(22))(4)

ø 13.7345 in.3



5 a2

} b2

89.13 }}

Surface area of Solid II 5

42

} 72

Surface area of Solid II ø 272.96 in.2

Volume of Solid I


5 a3

} b3

13.7345 }}


43

} 73

Volume of Solid II ø 73.61 in.3


Surface area of Solid I 5 5(8) 1 26(5) 1 2π 1 1 } 2 2 (8)3

1 3(1 p 8)

ø 219.13 m2

Volume of Solid I 5 1 52 2 (8)

5 200 m3



5 a2

} b2

219.13


5 82

} 52

Surface area of Solid II ø 85.60 m2

Volume of Solid I


5 a3

} b3

200 }}


83

} 53

Volume of Solid II ø 48.83 m3

23. h 5 2r

S 5 2πr2 1 2πrh

54π 5 2πr2 1 2πr (2r)

54π 5 6πr2

9 5 r2

r 5 3

h 5 2r 5 6

(radius of small cylinder)2

}} (radius of large cylinder)2 5

54π } 384 π

32 }}

(radius of large cylinder)2 5 9 } 64

radius of large cylinder 5 8

height of large cylinder 5 8(2) 5 16

The radius and height of the small cylinder are 3 ft and 6 ft.

The radius and height of the large cylinder are 8 ft and 16 ft.

24. a. The ratio from the small cone to big cone is 8 : 10 or 4 : 5.

The ratio of the area is then 42 : 52 or 16 : 25.

b. 4:5

c. 42 : 52 or 16 : 25

d. 43 : 53 or 64 : 125

e. The volume of the top part of the cone is 64

} 125

.

If 1 represents the volume of the entire cone,

1 2 64

} 125 represents the volume of the bottom part of

the cone. 1 2 64

} 125 5 61

} 125 . The ratio of the pieces is

64

} 125

: 61

} 125

or 64 : 61.

Problem Solving

25. Volume of smaller mug

}} Volume of larger mug

5 a3

} b3

Volume of smaller mug

}} 12

5 3.53

} 43

Volume of smaller mug ø 8.04. So, the capacity of the smaller mug is about 8.04 fluid ounces.

26. The units should be the same for each part of the ratio.

Volume of small binoculars

}}} Volume of large binoculars

5 a3

} b3

5 (0.5 feet)3

} (45 feet)3

5 0.125

} 91125

5 1 } 729,000

The ratio of the volumes is 1 : 729,000.

27. Volume of small bowl

}} Volume of large bowl

5 a3

} b3

Volume of small bowl

}} 64

5 33

} 43

Volume of small bowl 5 27. So, the small bowl requires 27 fluid ounces of lemonade.

28. Sample answer:

8 cm

20 cm

5 cm 2 cm

29. a. Volume of smaller orange 5 4 } 3 π (1.6)3 ø 17.16 in.3

Volume of larger orange 5 4 } 3 π (2)3 ø 33.51 in.3


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427Geometry


b. Ratio of diameters 5 3.2

} 4 5 0.8

Ratio of volumes ø 17.16 }

33.51 ø 0.51 ø (0.8)3

The ratio of the volumes is the ratio of the diameters cubed.

c. Small orange: diameter 5 3.2 2 2 1 1 } 8 2 5 2.95 in.

Large orange: diameter 5 4 2 2 1 1 } 8 2 5 3.75 in.

d. S 5 4π (1.475)2 ø 27.34 in.2

V 5 4 } 3 π (1.475)3 ø 13.44 in.3

S 5 4π (1.875)2 ø 44.18 in.2

V 5 4 } 3 π (1.875)3 ø 27.61 in.3

ratio of surface areas ø 27.34 }

44.18 ø 0.62

ratio of volumes ø 13.44 }

27.61 ø 0.49 ø 0.62(0.8)

The ratio of the volumes is about 0.8 times the ratio of

the surface areas. The original ratio was 3.2

} 4 5 0.8.

30. a. a }

b ;

a3 }

b3

b. Cone I: V 5 1 } 3 1 πa2 2 (2a) 5

2 } 3 πa3

Cone II: V 5 1 } 3 1 πb2 2 (2b) 5

2 } 3 πb3

c. Volume of Cone I

}} Volume of Cone II

5 2 } 3 πa3

} 2 }

3 πb3

5 a3

} b3

The answer is equivalent to the answer from part (a).

31. a. model length

} actual length

5 1 } 18

8 in. }

actual length 5

1 } 18

actual length 5 144 in.

The length of the actual car is 144 in. or 12 ft.

b. Surface area of model tire

}} Surface area of actual tire

5 a2

} b2

121 square inches

}} Surface area of actual tire

5 12

} 182

Surface area 5 3920.4

The surface area of each tire of the actual car is 3920.4 in.2

c. Volume of model engine

}} Volume of actual engine

5 a3

} b3

Volume of model engine

}} 8748

5 13

} 183

Volume 5 1.5

The volume of the model car’s engine is 1.5 cubic inches.

32. Volume of small cylinder

}} Volume of large cylinder

5 a3

} b3

16π

} 432 π

5 1 } 27 5

13

} 33

The common ratio between cylinders is 1 }

3 or 1 : 3.

Lateral area of small cylinder

}}} Lateral area of large cylinder

5 a2

} b2

Lateral area of small cylinder

}}} 72 π

5 12

} 32

Lateral area 5 8π

The lateral area of the small cylinder is 8π square units.

33. The ratio of the small snowball to the medium snowball is 5 : 7. So, the ratio of their volumes is 53 : 73.

53

} 73 5

1.2 } x → x 5 3.2928

The weight of the medium snowball is 3.2928 kilograms.

The ratio of the small snowball to the large snowball is 5 : 9 . So, the ratio of their volumes is 53 : 93.

53

} 93 5

1.2 } x → x 5 6.9984

The weight of the large snowball is 6.9984 kilograms.

1.2 1 3.2928 1 6.9984 5 11.4912

The total weight of the snow used is about 11.5 kilograms.

34. a. P 5 k }

s3 , where k is the constant of variation and s is

the side length.

b. Side length s (cm)

1 }

4

1 }

2 1 2 3

Pressure P (atm) 64k 8k k k }

8

k }

64

c.

Side length (cm)

Pre

ssu

re (

atm

)

0 1 2 3 4 s

P

0

8

16

24

32

40

48

56

64

72

The graph does not show a linear relationship. The points on the graph cannot be connected with a straight line, so there is not a linear relationship.


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35. The top of the pyramid is similar to the entire pyramid.

Let V 5 Volume of top pyramid

h 5 height of top pyramid

V

} 2V

5 h3

} 123

1 }

2 5

h3

} 1728

2h3 5 1728

h3 5 864

h 5 (864)1/3

5 6 p 22/3

ø 9.52

The height of the top of the pyramid is about 9.52 feet.

Mixed Review

36. nABC , nFDE because corresponding angles are congruent.

37. Not similar because corresponding angles are not congruent.

38. nQRT , nPST because corresponding angles are congruent.

39. (n 2 2) p 1808 5 9008

n 2 2 5 5

n 5 7

The polygon has 7 sides. It is a heptagon.

40. By the Triangle Sum Theorem, the polygon is a triangle.

41. (n 2 2) p 1808 5 5408

n 2 2 5 3

n 5 5

The polygon has 5 sides. It is a pentagon.

42. (n 2 2) p 1808 5 10808

n 2 2 5 6

n 5 8

The polygon has 8 sides. It is an octagon.

43. (x 2 2)2 1 (y 2 5)2 5 16

44. (x 1 3)2 1 (y 2 2)2 5 36

45. S 5 2B 1 Ph

6 ft

8 ft

3 ft 5 2(8 p 6) 1 (28)(3)

5 180

The surface area is 180 square feet.

46. S 5 2B 1 Ph

5 2 1 1 } 2 p 12 p 8.26 p 5 2 1 (60)(12)

5 1215.6

12 mm12 mm

The surface area is 1215.6 square millimeters.

47. S 5 2πr2 1 2πrh

4 in.

4 in.

5 2π(4)2 1 2π(4)(4)

ø 201.06


48. S 5 2πr2 1 2πrh

7 cm

9 cm

5 2π(4.5)2 1 2π(4.5)(7)

ø 325.15

The surface area is about 325.15 square centimeters.

Quiz 12.6–12.7 (p. 854)

1. S 5 4πr2 5 4π(7)2 ø 615.75


V 5 4 } 3 πr3 5

4 } 3 π(7)3 ø 1436.76


2. S 5 4πr2 5 4π(11.5)2 ø 1661.90


V 5 4 } 3 πr3 5

4 } 3 π(11.5)3 ø 6370.63


3. S 5 4πr2 5 4π(10.7)2 ø 1438.72


V 5 4 } 3 πr3 5

4 } 3 π(10.7)3 ø 5131.45




5 a2

} b2

114 }}

Surface area of B 5

12

} 32

S 5 1026

Volume of A

} Volume of B

5 a3

} b3

72 }

Volume of B 5

13

} 33

V 5 1944

The surface area of B is 1026 in.2 and the volume is 1944 in.3.



5 a2

} b2

170π }}

Surface area of B 5

22

} 32

S 5 382.5π

Volume of A

} Volume of B

5 a3

} b3

300π }

Volume of B 5

23

} 33

V5 1012.5π

The surface area of B is 382.5π m2 and the volume is 1012.5π m3.


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5 a2

} b2

383 }}

Surface area of B 5

52

} 42

S 5 245.12

Volume of A

} Volume of B

5 a3

} b3

440 }

Volume of B 5

53

} 43

V 5 225.28

The surface area of B is 245.12 cm2 and the volume is 225.28 cm3.

7. Volume of larger cone

}} Volume of smaller cone

5 a3

} b3

729 π

} 343 π

5 729

} 343 5 93

} 73

The scale factor is 9 : 7 or 9 } 7 .

Mixed Review of Problem Solving (p. 855)

1. a. V 5 Bh 5 (24 p 16)(20) 5 7680


b. V 5 Bh 5 (8 p 2)(3) 5 48


c. Number of boxes 5 Volume of container

}} Volume of one box

5 7680

} 48 5 160

So, 160 boxes of cookies can fi t in the container.

2. a. V 5 πr2h 5 π(1.25)2(7) ø 10.94π ø 34.36


b. 18 fluid ounces p 1 1 in.3 }}

0.554 fluid ounces 2 ø 32.49 in.3

The volume of the 18 fluid ounces is about 32.49 cubic inches. Since this volume is less than the volume of the cylinder, all of the juice will fit in the cup.

3. a. V 5 1 } 3 πr2h 5

1 } 3 π(6)2(10) ø 376.99

The volume of the funnel is about 377 cm3.

b. 377 cm3 5 377 mL

45 mL

} 1 sec

5 377 mL

} time

time ø 8.38

It takes about 8.38 seconds to empty the funnel.

c. V 5 1 } 3 π(10)2(6) ø 628.32

The volume of the new funnel is about 628.32 cm3, which means it can hold about 628.32 mL.

45 mL

} 1 sec

5 628.32 mL

} time

time ø 13.96

It would take about 13.96 seconds to empty the funnel.

d. The volume of the funnel in part (c) is greater then the volume of the funnel in part (a). A greater volume means the funnel can hold more oil. If the rate at which the oil flows stays constant, it would have to take the funnel in part (c) longer because it has a greater volume.

4. a. C 5 2πr

29.5 5 2πr

29.5

} 2 π

5 r

S 5 4πr2 5 4π 1 29.5 }

2 π 2 2 ø 277

The surface area of a men’s basketball is about 277 in.2.

V 5 4 } 3 πr3 5

4 } 3 π 1 29.5

} 2 π

2 3 ø 433.53

The volume of a men’s basketball is about 433.53 in.3.

b. C 5 2πr

28.5 5 2πr

28.5

} 2 π

5 r

S 5 4πr2 5 4π 1 28.5 }

2 π 2 2 ø 258.55

The surface area of a women’s basketball is about 258.55 in.2.

V 5 4 } 3 πr3 5

4 } 3 π 1 28.5

} 2 π

2 3 ø 390.92

The volume of a women’s basketball is about 390.92 in.3.

c. Surface area of women’s ball

}}} Surface area of men’s ball

5 a2

} b2

Surface area of women’s ball

}}} 277

5 28.52

} 29.52

Surface area of women’s ball ø 258.54 in.2

Volume of women’s ball

}} Volume of men’s ball

5 a3

} b3

Volume of women’s ball

}} 433.53

5 28.53

} 29.53

Volume of women’s ball ø 390.92 in.3

The answers are approximately equivalent. So, the answers agree with the answers in part (b).

5. Volume of rock 5 volume the water rises

4 }

3 πr3 5 π(2)2 1 9 }

64 2

4 }

3 πr3 5

9 } 16 π

r3 5 27

} 64

r 5 3 } 4

The radius of the rock is 3 }

4 inch.


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6. Sample answer:

14 ft

2 ft

1 ft

2 ft3 ft

6 ft

Original S 5 94 cm2 New S 5 100 cm2

Original V 5 60 cm3 New V 5 56 cm3

7. Volume of cone 5 1 } 3 πr2h 5

1 } 3 π(r)2(r) 5

1 } 3 πr3

Volume of hemisphere 5 1 } 2 1 4 } 3 πr3 2 5 2 1 1 }

3 πr3 2

The volume of the hemisphere is twice the volume of the cone.

8. Sample answer:

r

lh

The slant height of the cone is the hypotenuse of the triangle formed by the height, radius, and slant height. The hypotenuse of a triangle must be longer than each of the legs, so the height will always be less than the slant height.

Chapter 12 Review (pp. 857–860)

1. A sphere is the set of all points in space equidistant from a given point.

2. Sample answer:

w

h

wh

Right rectangular prism

Oblique rectangular prism

An oblique rectangular prism has the same volume as a right rectangular prism if the area of their bases and perpendicular heights are equivalent.

3. 20 1 n 5 30 1 2 4. n 1 6 5 12 1 2

n 5 12 n 5 8

5. 14 1 24 5 n 1 2

36 5 n

6. S 5 2B 1 Ph 5 2 1 1 } 2 2 (6)(7) 1 22.22(10) 5 264.20

The surface area is 264.20 square miles.

7. S 5 2πr2 1 2πrh 5 2π(12)2 1 2π(12)(15) ø 2035.75

The surface area is about 2035.75 square feet.

8. S 5 2B 1 Ph 5 2( Ï}

2 )2 1 (4 Ï

}

2 (5)) ø 32.28

The surface area is about 32.28 square meters.

9. S 5 2πr2 1 2πrh

44π 5 2π(2)2 1 2π(2)h

36π 5 4πh

9 5 h

The height of the cylinder is 9 meters.

10. S 5 B 1 1 } 2 Pl 5 4 1

1 } 2 (8) 1 Ï

}

26 2 ø 24.40

The surface area of the pyramid is about 24.4 square feet.

11. S 5 πr2 1 πrl

500π 5 πr2 1 πr 1 Ï}

152 1 r2 2 500 5 r2 1 r 1 Ï

}

152 1 r2 2 500 2 r2 5 r 1 Ï

}

152 1 r2 2 250,000 2 1000r2 1 r4 5 r2 1 225 1 r2 2

250,000 2 1000r2 1 r4 5 225r2 1 r4

250,000 5 1225r2

10,000

} 49

5 r2

14.29 ø r

The radius is about 14.29 centimeters.

12. S 5 B 1 1 } 2 Pl ø 8 1 1 }

2 2 (1.5)(1.24) 1

1 } 2 (9.94)(2.92) ø 21.95

The surface area is about 21.95 square yards.

13. V 5 Bh 5 (1.5)(2.1)(3.6) 5 11.34

The volume is 11.34 cubic meters.

14. V 5 Bh 5 1 πr2 2 h 5 π(2)2(8) 5 32π ø 100.53

The volume is about 100.53 cubic millimeters.

15. V 5 Bh ø 1 5 1 1 } 2 2 (2)(1.37638) 2 (4) ø 27.53

The volume is about 27.53 cubic yards.

16. V 5 1 } 3 πr2h 5

1 } 3 π(8)2(15) 5 320π ø 1005.31

The volume is about 1005.31 cubic centimeters.

17. V 5 1 } 3 Bh

60 5 1 } 3 B(15)

60 5 5B

12 5 B

The area of the base is 12 square inches.

18. S 5 4πr2 5 4π(1195)2 ø 17,945,091.40

V 5 4 } 3 πr3 5

4 } 3 π(1195)3 ø 7,148,128,072.97

The surface area and volume of Pluto are about 17,945,091.4 square kilometers and 7,148,128,072.97 cubic kilometers.


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431Geometry


19. V 5 S3 1 1 } 2 1 4 }

3 πr3 2 5 63 1

2 } 3 π(3)3

5 216 1 18π ø 272.55

The volume is about 272.55 cubic meters.



5 a2

} b2

62 }}

Surface area of B 5

12

} 42

Surface area of B 5 992

Volume of A

} Volume of B

5 a3

} b3

30 }

Volume of B 5

13

} 43

Volume of B 5 1920

The surface area of Solid B is 992 square centimeters and the volume of Solid B is 1920 cubic centimeters.



5 a2

} b2

112π }}

Surface area of B 5

12

} 32

Surface area of B 5 1008π

Volume of A

} Volume of B

5 a3

} b3

160π }

Volume of B 5

13

} 33

Volume of B 5 4320π

The surface area of Solid B is 1008π ø 3166.73 square meters and the volume of Solid B is 4320π ø 13,571.68 cubic meters.



5 a2

} b2

144π }}

Surface area of B 5

22

} 52

Surface area of B 5 900π

Volume of A

} Volume of B

5 a3

} b3

288π }

Volume of B 5

23

} 53

Volume of B 5 4500π

The surface area of Solid B is 900π ø 2827.43 square yards and the volume of Solid B is 4500π ø 14,137.17 cubic yards.

Chapter 12 Test (p. 861)

1. Faces 5 9

Vertices 5 9

Edges 5 16

F 1 V 5 E 1 2

9 1 9 5 16 1 2 ✓

2. Faces 5 8

Vertices 5 12

Edges 5 18

F 1 V 5 E 1 2

8 1 12 5 18 1 2 ✓

3. Faces 5 10

Vertices 5 16

Edges 5 24

F 1 V 5 E 1 2

10 1 16 5 24 1 2 ✓

4. S 5 2B 1 Ph 5 2(4.5) 1 (18)(8) 5 184

The surface area is 184 square feet.

5. S 5 2B 1 Ph 5 2 1 1 } 2 2 (3.2)(5.471) 1 (14.6)(1.6) ø 40.87


6. S 5 2πr2 1 2πrh 5 2π(4.1)2 1 2π(4.1)(10) ø 363.23

The surface area is about 363.23 square meters.

7. S 5 B 1 1 } 2 Pl 5 72 1

1 } 2 (28)(9) 5 175

The surface area is 175 square centimeters.

8. S 5 πr2 1 πrl 5 π(7.3)2 1 π(7.3)(19.70) ø 619.26


9. S 5 πr2 1 2πrh 1 πrl

5 π(7)2 1 2π(7)(8) 1 π(7) 1 Ï}

65 2

ø 683.09

The surface area is about 683.09 square feet.

10. V 5 Bh 5 (12.7)(4) 5 336

The volume is 336 cubic centimeters.

11. V 5 Bh 5 5 1 1 } 2 p 8 p 5.5055 2 (15.5) ø 1706.71

The volume is about 1706.71 cubic meters.

12. V 5 πr2h 5 π(10.95)2(10.3) ø 3879.85

The volume is about 3879.85 cubic feet.

13. V 5 1 } 3 Bh

324 5 1 } 3 (92)x

324 5 27x

12 5 x


14. V 5 4 } 3 πr3

32π

} 3 5

4 } 3 πx3

8 5 x3

2 5 x

The value of x is 2 feet.

15. V 5 1 } 3 πr2h

180π 5 1 } 3 πx2(15)

36 5 x2

6 5 x

The value of x is 6 centimeters.


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16. S 5 4πr2 5 4π(17.5)2 ø 3848.45

V 5 4 } 3 πr3 5

4 } 3 π(17.5)3 ø 22,449.30

The surface area and volume of the marble are about 3848.45 square millimeters and 22,449.30 cubic millimeters.

17. Surface area of smaller can

}} Surface area of larger can

5 a2

} b2

308π

}} Surface area of larger can

5 22

} 32

The surface area of the larger can is 693π square inches.

Volume of smaller can

}} Volume of larger can

5 a3

} b3

735π }}

Volume of larger can 5

23

} 33

The volume of the larger can is about 2480.63π cubic inches.

Standardized Test Preparation (p. 863)

1. Find the volumes of the small and large canisters.

Volume of small canister 5 4 } 3 πr3

5 4 } 3 π(12)3

5 2304π cm3

Volume of large canister 5 4 } 3 πr3

5 4 } 3 π(18.75)3

5 8789.06π cm3

Write a ratio.

Volume of small canister

}} Volume of large canister

5 Amount small canister can hold

}}} Amount large canister can hold

2304π

} 8789.06π

5 Amount small canister can hold

}}} 2

The amount of flour the small canister can hold is about 0.5 kg. So, the correct answer is A.

2. Let x 5 the height of the pyramid and 3x 5 the height of the prism.

Then x 1 3x 5 height of solid

4x 5 42 ft

x 5 10.5 ft

3x 5 31.5 ft

Volume of the solid 5 volume of prism 1 volume of pyramid

5 Bh 1 1 } 3 Bh

5 1 142 2 (31.5) 1 1 } 3 1 142 2 (10.5)

5 6174 1 686

5 6860

The volume of the solid is 6860 ft3. So, the correct answer is C.

Standardized Test Practice (pp. 864–865)

1. D; There is one pair of parallel congruent bases whose shape is a trapezoid.

2. B; S 5 2B 1 Ph

5 2 1 1 } 2 (b1 1 b2)h1 2 1 Ph2

5 (30 1 46)(30) 1 (140)(30)

5 6480

The surface area is 6480 square inches.

3. A; Percentage 5 Volume of sphere

}} Volume of cube

5 4 } 3 πr2

} s3

5 4 } 3 π(2.5)3

} 63

ø 0.303

The sphere takes up about 30% of the volume of the cube.

4. D; A cylinder is formed with radius } LM and height } KL .

V 5 πr2h 5 π(3)2(1) 5 9π

5. D; Surface area of window 5 4 1 1 } 2 1 3 p 1.5 Ï

}

2 2 2 5 9 Ï

}

2

ø 12.73

Total weight 5 (12.73)(3.25) 5 41.37

So, the total weight of the window is about 41 pounds.

6. C; Volume 5 1 } 3 πr2h

16π 5 1 } 3 π(4)2h

3 5 h

Surface area 5 πr2 1 πrl

5 π(4)2 1 π(4)(5)

5 16π 1 20π

5 36π


7. D; Surface area 5 (2r)h 1 1 } 4 (2πrh) 5 2rh 1

1 } 2 πrh

8. Volume of small prism

}} Volume of large prism

5 a3

} b3

Volume of small prism

}} 175

5 33

} 53

The volume of the small prism is 37.8 cubic inches.

9. F 1 V 5 E 1 2

16 1 10 5 E 1 2

26 5 E 1 2

24 5 E

The polyhedron has 24 edges.


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433Geometry


10. Surface area of Sphere A

}} Surface area of Sphere B

5 a2

} b2

27

} 48

5 a2

} b2

9 }

16 5

a2

} b2

Ï

}

9 }

Ï}

16 5

a }

b

3 }

4 5

a }

b

The ratio of the diameter of Sphere A to the diameter of

Sphere B is 3 }

4 .

11. Let x 5 length of a side of the base and 2x 5 height.

Volume 5 1 } 3 Bh

54 5 1 } 3 1 x2 2 (2x)

54 5 2 } 3 x3

81 5 x3

4.3267 ø x so 2x ø 8.65

The height of the pyramid is about 8.65 meters.

12. Surface area 5 πR2 1 2πRh 1 πr2 1 2πrh

5 π 1 52 2 1 2π(5)(3)1 π 1 32 2 1 2π(3)(3)

5 82π

Cans of frosting 5 82π

} 100 ø 2.58

You will need 2.58 cans of frosting, so you will need to buy 3 cans.

13. Cylinder B Height 5 2h

Cylinder B Radius 5 r } 2

Volume of Cylinder A 5 πr2h

Volume of Cylinder B 5 π 1 r }

2 2 2(2h) 5

1 } 2 πr2h

The volume of Cylinder A is twice the volume of Cylinder B, so Cylinder A has a larger volume.

14. a. V 5 πr2h 5 π(1.2)2(5) ø 22.6

The volume of the tank is about 22.6 cubic feet.

b. Gallons 5 22.6(7.48) 5 169.048

So, about 169 gallons are needed to fi ll the tank.

c. Number of times fi lled 5 1000

} 169 5 5.9

Because it will take 5.9 tanks for a year, the tank needs to be fi lled 6 times.

15. a. Prism:

Surface area 5 2B 1 Ph 5 2 1 32 2 1 12(4) 5 66 in.2

Volume 5 Bh 5 1 32 2 (4) 5 36 in.3

Cylinder:

Surface area 5 2πr2 1 2πrh

5 2π 1 1.62 2 1 2π(1.6)(5)

5 21.12π ø 66.4 in.2

Volume 5 πr2h 5 π(1.6)2(5) 5 12.8π ø 40.2 in.3

b. Prism:

Volume

} Surface area

5 36

} 66 5 6 } 11

Cylinder:

Volume

} Surface area

5 12.8 π

} 21.12 π 5 20

} 33

The ratios show that for the same amount of surface area, the cylinder will have a greater volume. This means that for the same amount of material, the cylinder package can hold more of the product. The manufacturer needs to consider how much of the product can be packaged per square inch of the surface area.

Cumulative Review Chs. 1–12 (pp. 866–867)

1. x 5 75

2. 13x8 5 1308

x 5 10

3. (3x 1 4)8 1 1288 5 180

3x 1 132 5 180

3x 5 48

x 5 16

4. 5 }

6 5

9 } 6 1 x

30 1 5x 5 54

5x 5 24

x 5 24

} 5

5. y }

8 5

5 } 10

10y 5 40

y 5 4

6. 10

} z 5

15 } 18

15z 5 180

z 5 12

7. Both pairs of opposite angles are congruent.

8. One pair of opposite sides are parallel and congruent.

9. The diagonals bisect each other.

10. (r 1 18)2 5 r2 1 302

r2 1 36r 1 324 5 r2 1 900

36r 1 324 5 900

36r 5 576

r 5 16

11. x8 5 1 } 2 (908)

x 5 45

12. 2(y 1 2) 5 3(3 1 5)

2y 1 4 5 24

y 5 10


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13. Shaded area 5 1 858 }

3608 2 1 π(7)2 2 ø 36.35

The shaded area is about 36.35 square inches.

14. Shaded area 5 1 1548 }

3608 2 1 π(23)2 2 ø 710.92

The shaded area is about 710.92 square centimeters.

15. Shaded area 5 Area of circle 2 Area of square

5 πr2 2 s2

5 π 1 Ï}

2 2 2 2 22

ø 2.28

The shaded area is about 2.28 square meters.

16. Surface area 5 2B 1 Ph 5 2(6.4) 1 (20)(13) 5 308 ft2

Volume 5 Bh 5 (6.4)(13) 5 312 ft3

17. Surface area 5 B 1 1 } 2 Pl

5 (4.4)2 1 1 } 2 (4.4)(4) 1 Ï

}}

12.52 1 2.22 2

ø 131.05 in.2

Volume 5 1 } 3 Bh 5

1 } 3 1 4.42 2 (12.5) ø 80.67 in.3

18. Surface area 5 πr2 1 πrl

5 π(10.9)2 1 π(10.9) 1 Ï}}

21.8 1 10.92 2 ø 1207.87 m2

Volume 5 1 } 3 πr2h 5

1 } 3 π(10.9)2(21.8) ø 2712.30 m3

19. Choose the median } TM . Find the coordinates of M, the midpoint of

} OS .

1 0 1 8 }

2 ,

0 1 0 }

2 2 5 1 8 } 2 ,

0 }

2 2 5 (4, 0)

The distance from vertex T(4, 6) to midpoint M(4, 0) is 6 2 0, or 6 units. The coordinates of the centroid are

2 }

3 p 6 or 4 units down from vertex T along the median } TM .

The coordinates of the balancing point P are (4, 6 2 4) or (4, 2).

20. Line symmetry and rotational symmetry;

The figure has two perpendicular lines of symmetry that are vertical and horizontal, and rotational symmetry of 1808 about the center.

21. a. (x 1 2)2 1 (y 2 4)2 ≤ 36

b. Substitute the coordinates in the inequality to see if the locations are valid.

(2, 0): Yes; (3, 9): No; (26, 21): No; (26, 8): Yes; (27, 5): Yes

22. Area 5 2(area of front side) 1 2(area of right side)

5 2 1 (99 1 88)(10.5) }}

2 2 1 2 1 (15)(9.5)

} 2 2

5 1963.5 1 142.5

5 2106

So, the area of the covered bridge is 2106 square feet.

23. a. Surface area 5 2πr2 1 2πrh

5 2π(1)2 1 2π(1)(5.5)

5 13π

ø 40.84 in.2

Volume 5 πr2h 5 π(1)2(5.5) 5 5.5π ø 17.28 in.3

b. Volume gone

}} Original volume

5 Time burned

} Burning time

Volume gone

} 5.5π 5 18

} 30

The amount of candle burned is 3.3π. So, the volume of candle that is left is 5.5π 2 3.3π 5 2.2π, or about 6.91 in.3.

24. Surface area 5 4πr2 5 4π(3960)2 5 62,726,400π mi2

70% of area 5 0.7(62,726,400π) ø 137,942,558.2

So, about 137,942,558.20 square miles of Earth’s surface is covered by water.


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chapter 12€¦ · 62.83 cm2. lesson 12.1 investigating geometry activity 12.1 ... the base is a...

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