2. Prism
Pyramid
Cone
Cylinder
Sphere

3. Prism
Pyramid
Cone
Cylinder
Sphere
Link

5. Check It Out: Example 1B
All faces and bases are congruent
squares.
The figure is a cube.
Identify the bases and faces of the figure.
Then name the figure.
Surface Area and Volume

6. Check It Out: Example 2B
Classify each figure as a polyhedron or not a
polyhedron. Then name the figure.
There are two triangular
bases for the figure.
The faces are all polygons, so
the figure is a polyhedron.
The figure is a triangular
prism.

7. Identify the bases and faces of the figure.
Then name the figure.
There are two octagonal bases.
The figure is an octagonal prism.
There are eight rectangular faces.

8. Lesson Quiz: Part I
Identify the bases and faces of each figure.
Then name each figure.
Two pentagon bases, 5 rectangular
faces; pentagonal prism
One square base, 4 triangular
faces; square pyramid
1.
2.

9. Lesson Quiz: Part II
Classify each figure as a polyhedron or not a
polyhedron. Then name the figure.
polyhedron,
rectangular prism
polyhedron,
triangular prism
3.
4.

10. 2. Identify the bases and faces, and then the
name of the given figure.
A. pentagon; triangles; pentagonal pyramid
B. heptagon; triangles; heptagonal pyramid
C. pentagon; rectangles; pentagonal prism
D. heptagon; rectangles; heptagonal prism
Lesson Quiz for Student Response Systems

11. 3. Classify the figure as a polyhedron or not a
polyhedron, and then name the given figure.
A. polyhedron; pentagonal prism
B. polyhedron; pentagonal pyramid
C. not a polyhedron; cylinder
D. not a polyhedron; cone
Lesson Quiz for Student Response Systems

12. Vertices (points)
Edges (lines)
Faces (planes)
6
9
5
The base has 3 sides.

13. Vertices (points)
Edges (lines)
Faces (planes)
8
12
6
The base has sides.4

14. Vertices (points)
Edges (lines)
Faces (planes)
10
15
7
The base has sides.5

15. Vertices (points)
Edges (lines)
Faces (planes)
12
18
8
The base has sides.6

16. Vertices (points)
Edges (lines)
Faces (planes)
16
24
10
The base has sides.8

17. Name Picture Base Vertices Edges Faces
Triangular
Prism
Rectangular
Prism
Pentagonal
Prism
Hexagonal
Prism
Heptagonal
Prism
Octagonal
Prism
3 6 9 5
4 8 12 6
5 10 15 7
6 12 18 8
7 14 21 9
8 16 24 10
Any Prism n 2n 3n n + 2
Draw it
No
picture

18. Prism
Surface area of wall = Length around the base.  high
= (3+4+5)  6
= 72 Square centimeter
Base = 43
2
1

= 6 Square centimeter
Surface area = Surface area of wall + (2  Area of base )
= 72 + (2 6)
= 72+12
= 84 Square centimeter
= 84 Square centimeter

19. Prism
surface Area of wall = Length around the base.  high
= ( 5+5+5+5 ) 10
= 20  10
= 200 cm2
Area of base = 5  5
= 25 cm2
Surface area = surface Area of wall + (2  Area of base)
= 200 + (2  25)
= 200 + 50
= 250 cm2
= 250 cm2

20. Prism
Surface Area of wall = perimeter of base.  high
= ( 2+5+2+5 ) 10
= 14  10
= 140 cm2
Area of base = 5  2
= 10 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 140 + (2  10)
= 140 + 20
= 160 cm2
= 160 cm2

22. J.Stewart Highcliffe
This is a Pentagonal based Prism
• This shape has 7 faces
• Five faces are
rectangular
• Two faces are
pentagons
• It has 10 vertices and 15
edges

23. Prism
Surface Area of wall = perimeter of base.  high
= ( 6+6+6+6+6) 15
= 30  15
= 450 cm2
Area of base = 62 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 450+ (2  62)
= 450 + 124
= 574 cm2
= 574 cm2
Find the surface area of a regular pentagonal prism if its base area is 62
cm2

24. Pythagoras : C2 = A2 + B2
C2 = 32 + 42
C2 = 9 + 16
C2 = 25
C = 5
Surface Area of wall = perimeter of base.  High
= (3+4+5) 10
= 12  10
= 120 cm2
Area of base = 6 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 120+ (2  6)
= 120 + 12
= 132 cm2
= 132 cm2

25. Surface Area of wall = perimeter of base.  High
= (3+6+3+6) 12
= 18  12
= 216 cm2
Area of base = 18 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 216+ (2  18)
= 216 + 36
= 252 cm2
= 252 cm2

26. Surface Area of wall = perimeter of base.  High
= (5+5+5+5) 11
= 20  11
= 220 cm2
Area of base = 25 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 220+ (2  25)
= 220 + 50
= 270 cm2
= 270 cm2
Find surface area square prism if it has side of base 5 and
high 11 cm.

27. Find Surface area of rectangular prism long side is 2 and 5 cm high 8
cm .
Surface Area of wall = perimeter of base.  High
= (2+5+2+5) 8
= 14  8
= 112 cm2
Area of base = 10 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 112+ (2  10)
= 112 + 20
= 132 cm2
= 132 cm2

28. 6.Find surface area of right triangular prism if adjacent side of base is 6 and 8 and
high 15 cm.
Pythagoras : C2 = A2 + B2
C2 = 62 + 82
C2 = 36 + 64
C2 = 100
C = 10
Surface Area of wall = perimeter of base.  High
= (6+8+10) 15
= 24  15
= 360 cm2
Area of base = 24 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 360+ (2  24)
= 360 + 48
= 408 cm2
= 408 cm2

29. Find surface area of octagonal prism if side of base is 4 cm.
Area of base is 60 and high 20 cm
Surface Area of wall = perimeter of base.  High
= (4+4+4+4+4+4+4+4) 20
= 32  20
= 640 cm2
Area of base = 60 cm2
Surface Area = Surface Area of wall + (2  Area of base)
= 640+ (2  60)
= 640 + 120
= 760 cm2
= 760 cm2

30. J.Stewart Highcliffe
Hexagonal Prism
• This shape has 8 faces
• Six faces are rectangular
• Two faces are hexagons
• It has 12 vertices and 18
edges

31. Find surface area square prism if it
have side of base 5 and high 11 cm
1 2 3
4 5
Find area of rectangular prism long
side is 2 and 5 cm high 8 cm
6
Home work

33. Prism

34. Solution
Area of Base = highBase
2
1
= 34
2
1

= 6
= 6 cm2
Volume =Area of base x High
= 6 x 10
= 60 cm3
= 60 cm3

35. Prism

36. Solution
Area of Base = Width X length.
= 6 x 2
= 12
= 12 cm2
Volume =Area of base x High
= 12 x 11
= 132 cm3
= 132 cm3

37. Volume = 1.72 × (base side) 2 x high
Area of base = 2
4
3
5 SideBase
Pentagonal Prism

38. Solution
Area of Base =
2
4
3
5 SideBase
= 1.7 x Base side 2
= 1.7 x 52
= 41.5 cm2
Volume =Area of base x High
= 41.5 x 12
= 498 cm3
= 498 cm3
Pentagonal Prism

39. Solution
Area of Base =
2
4
3
5 SideBase
= 1.72 x Base side 2
= 1.72 x 52
= 43 cm2
Volume =Area of base x High
= 43 x 12
= 516 cm3
= 516 cm3
Pentagonal Prism

40. Solution
Area of Base =
2
4
3
5 SideBase
= 1.7 x Base side 2
= 1.7 x 62
= 61.2 cm2
Volume =Area of base x High
= 61.2 x 15
= 918 cm3
= 918 cm3

41. Solution
Area of Base =
2
4
3
5 SideBase
= 1.7 x Base side 2
= 1.7 x 92
= 137.7 cm2
Volume =Area of base x High
= 137.7 x 10
= 1377 cm3
= 1377 cm3

42. Solution
Area of Base =
2
4
3
5 SideBase
= 1.7 x Base side 2
= 1.7 x 112
= 205.7 cm2
Volume =Area of base x High
= 205.7 x 8
= 1645.6 cm3

43. Base side High
1 9 17
2 6 23
3 3 19
4 8 77
5 3 6
6 By your self By your self
Volume(cm3 )
2340.9
1407.6
290.7
8377.6
4596.8

44. Volume = 2.5 × (base side) 2 x high
Area of base = 2
4
3
6 SideBase
Hexagonal Prism

45. Volume = 3.6 × (base side) 2 x high
Area of base = 2
4
3
7 SideBase
Heptagonal Prism

46. Volume = 4.8 × (base side) 2 x high
Area of base = 2
4
3
8 SideBase
octagonal Prism

47. Prism Base side High
Triangular 3,4,5 16
Rectangular 10,12 20
Pentagonal 12 30
hexagonal 14 50
heptagonal 21 24
octagonal 33 15
Volume
96
2400
7344
24500
38102.4
78408

48. Prism
Pyramid
Cone
Cylinder
Sphere

49. Cones and Pyramids
Pyramid
Slant
Heigh
t
Heigh
t
h
Slant
Heigh
t

Pyramid

50. Theorem 10-3
• Lateral Area and Surface Area of a Regular Pyramid
PL
2
1

Lateral
Area
Base
Perimeter
Slant Height
BLS 
Surfac
e Area
Lateral
Area
Base
Area
B


51. #1 Finding Surface Area of a Pyramid
• Find the surface area of a square pyramid with base edges 5
m and slant height 3 m.
m3
m5
BLS 
PL
2
1

)5(4P
2530S
20
)3(20
2
1
L 2
m30
2
sB 
2
5B
2
m25B
2
m55S

52. Base edges Slant height
6 9
5 13
7 10
11 12
3 19
Surface Area
144
155
189
385
123

53. #2 Finding Surface Area of a Pyramid
• Find the surface area of a regular hexagonal pyramid with a
slant height 20 in and sides 8 in.
BLS 
pL
2
1

)8(6P
160480 S
2
in48
)20(48
2
1
L 2
in480
2
5.2 BasesideB 
2
85.2 B
2
in160B
2
in640S
in8
in02

54. BLS 
pL
2
1

)9(5P cm45
)10(45
2
1
L 2
cm225
2
97.1 B
2
7.1 basesideB 
2
7.137 cmB 
7.137225S
2
7.362 cmS 
Find the surface area of pentagonal pyramid
with base Edges 9 cm
and Slant height 10 cm

55. Find the surface area of Hexagonal pyramid
with base Edges 11 cm
and Slant height 20 cm
BLS 
pL
2
1

)11(6P cm66
)20(66
2
1
L 2
cm660
2
115.2 B
2
5.2 basesideB 
2
5.302 cmB 
5.302660 S
2
5.962 cmS 

56. Pyramid Base edges Slant Height
Pentagonal 12 16
Hexagonal 21 24
Heptagonal 8 32
Octagonal 31 12
Triangular 13 42
Pentagonal 15 16
Octagonal 6 22
Find the surface area of ………. pyramid
with base Edges ……. cm
and Slant height ……..cm
Surface area
724.8
2614.5
1126.4
6101.8
892
982.5
700.8

58. A Pyramid is a three dimensional figure with a
regular polygon as its base and lateral faces
are identical isosceles triangles meeting at a
point.
Pyramids
base = quadrilateral base = pentagon base = heptagon
Identical isosceles
triangles

60. Volume of a Pyramid:
V = (1/3) Area of the base x height
V = (1/3) Ah
Volume of a Pyramid = 1/3 x Volume of a
Prism
Volume of Pyramids
+ + =

61. Theorem 10-8
• Volume of a Pyramid
BhV
3
1

Volum
e
HeightBase
Area
B

62. Finding Volume of a Pyramid
• Find the volume of a square pyramid with
base edges 5 m and height 3 m.
m3
m5
BhV
3
1

)3)(25(
3
1
V
2
sB 
2
5B
2
m25B
3
m25V

63. • Pyramid – Is a polyhedron in which one face
can be any polygon & the other faces are
triangles.
hVp = ⅓Bh
Area of the Base
A = l•w
A = ½bh
Height of the pyramid,
not to be confused with
the slant height (l)
Volume of a Pyramid

64. • Find the volume of a square pyramid with
base edges of 15cm & a height of 22cm.
22cm
15cm
15cm
V = (⅓)Bh
= (⅓)l•w•h
= (⅓)15•15•22
= (⅓)4950
= 1650cm3
Square
Volume of a right Pyramid

65. • Find the area of a square pyramid w/ base
edges 16ft long & a slant height 17ft.
h
17ft
16ft
8ft
V = (⅓)Bh
= (⅓)l•w•h
= (⅓)16•16•___
= (⅓)3840
= 1280ft3
a2 + b2 = c2
h2 + 82 = 172
h2 = 225
h = 15
15
Another square pyramid

66. Finding the Volume of a Pyramid
The base of a pyramid is a square. The side length of
the square is 24 feet. The height of the pyramid is 9
feet. Find the volume of the pyramid.
Write formula for volume of a pyramid.
Substitute 242 for B and 9 for h.
Simplify.
ANSWER The volume of the pyramid is 1728 cubic feet.
V = Bh
1
3
= 1728
= (242)(9)
1
3
Volumes of Pyramids and Cones

67. Find the volume of a ………… pyramid with base edges
………. m and height …………. m.
Pyramid Base edges Height
Pentagonal 12 16
Hexagonal 21 24
Heptagonal 8 32
Octagonal 31 12
Triangular 13 42
Pentagonal 15 16
Octagonal 6 22
1305.60
8820.00
2457.60
18451.20
1017.38
2040.00
1267.20

68. Surface Area of a Cylinder
h
A cylinder is a prism with a circular cross-section.
2r

69. Surface Area of a Cylinder
A cylinder is a
prism with a
circular cross-
section.
r2
r2

Removing
top and
bottom
Open out2r
h
Surface Area = 2r2 + 2rh
= 2r(r + h)

70. Surface Area of a Cylinder
8cm
Surface area = 2r(r + h)
3cm
Find the surface area of
the cylinder.
SA = 2 x  x 3(3 + 8)
= 6 x 11
= 66
= 207 cm2 (3 sf)

71. 6 cm
Calculate the surface area of the following cylinders.
15 cm
2.1mm
9.2 mm
3 cm
2 cm
1
2
3
SA = 2 x  x 3(3 + 6)
= 6 x 9 = 54 = 169.6 cm2
SA = 2 x  x 2(2 + 15)
= 4 x 17 = 68 = 213.6 m2
SA = 2 x  x 4.6(4.6 + 2.1)
= 9.2 x 6.7 = 193.6 mm2
2r(r + h)

72. 2 ( )SA r r h 
The Surface Area of a Cylinder
Find the surface area of the cylinder.
2 m
6m
2 3(3 2)x x 
2
94 26 5 . mx 

73. The Surface Area of a Cylinder
Find the surface area of the cylinder.
0.7 m
1.5m
2 ( )SA r r h 
2 0.7(0.7 1.5)x x 
2
1.4 2. 9.2 7mx 

74. The Surface Area of a Cylinder
Find the surface area of the cylinder.
1.9 m
8m
2 ( )SA r r h 
2 4(4 1.9)x x 
2
8 5.9 148.3x m 

75. The Surface Area of a Cylinder
Find the surface area of the cylinder.
0.76 m
1.3m
2 ( )SA r r h 
2 0.76(0.76 1.3)x x 
2
1.52 2. 9.06 8x m 

76. Radius
• Volume of a Cylinder
BhV 
Volume Height
2
rB 
Base
Area
Radius
Base
Areah
Base
Area

77. #3 Finding the Volume of a Cylinder
10 cm
5 cm
BhV 
• Find the volume of a cylinder with
height 10 cm and radius 5cm.
2
rB 
2
)5(B
2
cm25B
)10(25V
3
cm250V

78. #4 Finding the Volume of a Cylinder
9 ft
6 ft
BhV 
• Find the volume of the cylinder.
2
rB 
2
)6(B
2
ft36B
)9(36V
3
ft324V

79. Prism
Pyramid
Cone
Cylinder
Sphere

80. SA = LSA + B
or
Lateral Surface Area:
Total Surface Area: l
Cones
SA = rℓ +
2
r
LSA = rℓ

81. • Find the slant height then find the surface area of the
cone.
222
cba 
222
25.1 
2
425.2 
2
25.6 
5.2

82. Find the lateral area and surface area of the cone.
LSA = rℓ
= (8)(17)
= 136 in2
 427.04 in2
SA = rℓ + r2
= 136 + (8)2
= 200 in2
 628 in2
Example

83. Ex: Find the lateral & surface areas of the right cone.
LA = rl
LA = (6)(10)
LA = 60 in2
S = r2 + rl
S = (62) + 60
S = 36 + 60
S = 96 in2
6 in
8in
l
How do you find the slant
height?
Use Pythag. Thm!
82 + 62 = l 2
10 = l
l = 10
S  301.44 in2

84. • Find the surface area of the cone.
Surface Area of Pyramids and Cones
2
rrSA   
2
5135  SA
 2565 SA
2
in90SA
2
in6.282SA

85. • Find the slant height then find the surface area of the
cone.
Surface Area of Pyramids and Cones
2
rrSA   
 25.275.3 SA
2
m6SA
2
5.15.25.1  SA
2
m84.18SA

86. No. high Radius
1 3 4
2 6 8
3 5 12
4 10 24
5 9 12
6 25 60
7 15 20
SA
113.0
452.2
942.0
3768.0
1017.4
23550.0
2826.0
Slant high
5
10
13
26
15
65
25

87. V =
or
Volume:
Cones
Bh
3
1
V = hr2
3
1

h

88. • Volume of a Cone
BhV
3
1

Volume HeightBase
Arear
B

89. Find the volume of the cone.
V = 1/3 Bh or 1/3 (TTr2h)
V  1/3 (3.14  42  16)
V  1/3 (3.14  16  16)
V  1/3 (50.24  16)
V  1/3 (803.84)
V  267.95 cm3

90. Volume of a Cone
• Formula
• Find the volume of a cone with a diameter of
13.5 m and a height of 10m
hrV 2
3
1


91. 91
EXAMPLE 2 Finding the Volume of a Cone
Find the volume of the cone shown.
Round to the nearest cubic millimeter.
Write formula for volume of a cone.
Substitute 6.75 for r and 10 for h.
Evaluate. Use a calculator.
The radius is one half of the diameter, so r = 6.75.
V = πr2h
1
3
= π(6.75)2(10)
1
3
≈ 477.1
ANSWER
The volume of the cone is about 477 cubic millimeters.
Volumes of Pyramids and Cones

92. Finding the Volume of a Cone
• The radius of the base of a cone is 6 m. Its
height is 13 m. Find the volume.
m6
BhV
3
1

)13)(36(
3
1
V
2
rB 
 2
6B
2
m36B
3
m156V
m13
3
m84.489V

93. Find the volume of a cone.
Step 1 Use the Pythagorean
Theorem to find the height.
162 + h2 = 342
h2 = 900
h = 30
Example
Step 2 Use the radius and height to find the volume.
= 2560 cm3
 8050.96 cm3

94. 11in V = ⅓Bh
= (⅓)r2h
= (⅓)(3)2(11)
= (⅓)99
= 33  103.7in3
Circle
3in
Ex.3: Find the volume of the following right cone w/ a
diameter of 6 in.

95. • The following cone has a volume of 110.
What is its radius.
10cm
r
V = ⅓Bh
V = ⅓(r2)h
110 = (⅓)r2(10)
110 = (⅓)r2(10)
11 = (⅓)r2
33 = r2
r = √(33) = 5.7cm
Ex.5: Solve for the missing variable.

96. Find the volume of a cone with base circumference
25 in. and a height 2 in. more than twice the
radius.
Step 1 Use the circumference to find the radius.
Step 2 Use the radius to find the height.
h = 2(12.5) + 2 = 27 in.
2r = 25
r = 12.5
Step 3 Use the radius and height to find the volume.
= 1406.25 in3
= 4415.63 in3

97. Finding Volumes of Cones
Step 2 Use the
radius and height to
find the volume.
Volume of a cone
Substitute 16 for r and 30 for h.
 2560 cm3  8042.5 cm3 Simplify.
Find the volume of a cone.

98. Finding Volumes of Cones
Find the volume of a cone.
Step 1 Use the Pythagorean
Theorem to find the height.
162 + h2 = 342 Pythagorean Theorem
h2 = 900 Subtract 162 from both sides.
h = 30 Take the square root of both sides.

99. No. Radius high
1 9 13
2 6 16
3 8 10
4 2 21
5 5 12
6 8 15
Volume
1102.1
602.9
669.9
87.9
314.0
1004.8

100. 8cm
10cm
4cm
Volume of Cone first!
Vc = ⅓Bh
= (⅓)r2h
= (⅓)(8)2(10)
= (⅓)(640)
= 213.3 = 670.2cm3
Volume of Cylinder NEXT!
Vc = Bh
= r2h
= (8)2(4)
= 256  804.2cm3
VT = Vc + Vc
VT = 670.2cm3 + 804.2cm3
VT = 1474.4cm3
Volume of a Composite Figure

101. Check It Out!
Find the volume of the cone.
Volume of a cone
Substitute 9 for r and 8 for h.
≈ 216 m3 ≈ 678.6 m3 Simplify.

102. Example 4: Exploring Effects of Changing
Dimensions
original dimensions: radius and height divided by 3:
Notice that . If the radius and height are divided by 3, the
volume is divided by 33, or 27.
The diameter and height of the cone are
divided by 3. Describe the effect on the
volume.

103. Check It Out! Example 4
original dimensions: radius and height doubled:
The volume is multiplied by 8.
The radius and height of the cone
are doubled. Describe the effect
on the volume.

104. Example 5: Finding Volumes of Composite
Three-Dimensional Figures
Find the volume of the composite
figure. Round to the nearest
tenth.
The volume of the upper cone is

105. Example 5: Finding Volumes of Composite
Three-Dimensional Figures
The volume of the cylinder is
The volume of the lower cone is
The volume of the figure is the sum of the volumes.
Find the volume of the composite figure.
Round to the nearest tenth.
Vcylinder = r2h = (21)2(35)=15,435 cm3.
V = 5145 + 15,435 + 5,880 = 26,460  83,126.5 cm3

106. The volume of a sphere is two thirds
the volume of its corresponding cylinder.
Spheres and Cylinders
r
Sphere
r
Cylinder
h=2r
CylinderVolume
3
2
=SphereVolume
cylindersphere V
3
2
=V
h)r(
3
2
=V 2
sphere 
2r)r(
3
2
=V 2
sphere 
3
sphere r
3
4
=V 
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107. The Sphere
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108. A sphere is a three dimensional figure
in which every point on the surface is
equally distant from the centre O.
Volume of a
Sphere:
The cross-
section of
a sphere is
a circle.
Surface Area of
a Sphere:
SA = 4r2
r
O
3
r
3
4
=V 
Sphere
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109. The volume of a sphere is twice the
volume of its corresponding cone.
Spheres and Cones
r
Sphere
r
Cone
h=2r
ConeVolume2=SphereVolume
conesphere V2=V
)
3
hr
(2=V
2
sphere

3
sphere r
3
4
=V 
)
3
2rr
(2=V
2
sphere

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110. 2r
2r

r
The Surface Area of a
SphereThe formula for the surface area of a sphere was discovered by Archimedes. In
the diagram below a cylinder just encloses a sphere of radius r. Archimedes was
able to determine the formula by showing that a pair of parallel planes
perpendicular to the vertical axis of the cylinder, would enclose equal areas on
both shapes. 2r
Surface area =
2r x 2r
Surface area =
4r2

111. Surface Area
4r2
Archimedes did not have the advantage of a
sophisticated algebra like we use today. He had to
express relationships in terms of simpler geometric
shapes. For him the surface area of a sphere was equal to
the area of 4 of the greatest circles that it could contain.
r2
r2
r2
r2
Archimedes was intrigued
by this amazing discovery.
Why is the answer exactly
4 and not 4.342?
Painting the surface of a
sphere uses the same
amount of paint as
painting four of its
greatest circles!

112. Example 2
A spherical balloon is blown up from a radius of 7 cm to
one of 21 cm. What is the change in its surface area?
Step 1
Find the surface area
of the small balloon
A = 4  r2
= 4(3.14)(72)
= 615.44 cm2
Step 2
Find the surface area
of the big balloon
A = 4  r2
= 4(3.14)(212)
= 5538.96 cm2
The change in the surface area
5538.96 – 615.44 = 4923.52 cm2
Step 3
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113. 12
cm
7.3
cm
SA = 4r2
SA = 4 x  x 7.32 = 669.7cm2
SA = 4r2
SA = 4 x  x 122 = 1809.6 cm2
Example Questions: Calculate the surface area of the
spheres below. (to 1 dp)
1 2
SA = 4r2

114. radius 3.14 22/7
7
21
28
35
42
radius 3.14 22/7
7 615.44 616
21 5538.96 5544
28 9847.04 9856
35 15386 15400
42 22155.84 22176

115. Questions: Calculate the surface area of the spheres
below. (to 1 dp)
SA = 4r2
SA = 4 x  x 3.22 = 128.7 m2
SA = 4r2
SA = 4 x  x 2.42 = 72.4 m2
3.2
m
2.4
m
1 2
SA = 4r2

116. Example Questions: Calculate the radius of the spheres
shown below. (to 1 dp)
SA = 1500
cm2
1 2
4r2 = 1500
SA = 3500
cm2
4
15002
 r
cmr 910
4
1500
.

4r2 = 3500
4
35002
 r
cmr 716
4
3500
.

SA = 4r2

117. Questions: Calculate the radius of the spheres shown below.
(to 1 dp)
SA = 8.4
m2
4r2 = 8.4
SA = 1200
cm2
4
482 .
 r
mr 820
4
48
.
.


4r2 = 1200
4
12002
 r
cmr 89
4
1200
.

1 2
SA = 4r2

118. = 324 Simplify.
Find the surface area of the sphere below. Leave
your answer in terms of .
S.A. = 4 r 2
Use the formula for the surface area of a sphere.
= 4 • 92
Substitute r = = 9.18
2
The surface area of the sphere is 324 ft2
= 1073.88 ft2

119. S.A. = 4 r 2
Use the formula for the surface area of a sphere.
13 = 2 r Substitute 13 for C.
C = 2 r Use the formula for circumference.
Step 1: Use the circumference to find the radius in terms of .
The circumference of a rubber ball is 13 cm. Calculate its
surface area to the nearest whole number.
= r Solve for r.
13
2
Step 2: Use the radius to find the surface area.
= 4 • ( )2
Substitute for r.13
2
13
2
To the nearest whole number, the surface area of the rubber ball is 54 cm2
.
= Simplify.
169
53.794371 Use a calculator.

120. Find the volume of the sphere. Leave your
answer in terms of .
V = r 3
Use the formula for the volume of a sphere.
4
3
30
2
= • 153
Substitute r = = 15.
4
3
= 4500 Simplify.
The volume of the sphere is 4500 cm3.

121. The volume of a sphere is 1 in.3. Find its surface area to
the nearest tenth.
Step 1: Use the volume to find the radius r.
V = r 3
Use the formula for the volume of a sphere.
4
3
1 = r 3
Substitute.
4
3
= r 3
Solve for r 3.
3
4
= r Find the cube root of each side.
3
4
3
r 0.62035049 Use a calculator.

122. Step 2: Use the radius to find the surface area.
S.A. = 4 r 2
Use the formula for the surface area of a sphere.
4 (0.62035049)2 Substitute.
4.8359744 Use a calculator.
To the nearest tenth, the surface area of the sphere is 4.8 in.2.
(continued)

123. A cylindrical container used to store chocolate is 12 cm high with a
diameter of 6 cm. Each chocolate is a sphere with a diameter of 3
cm. Each chocolate sells for $1.50 What is the cost of one contai-
ner full of chocolates?
Example 1
Step 1
Find the radius of the container
R = 6  2 = 3 cm
12 cm
6 cm
Container
Step 2
Find the radius of the chocolate
r = 3  2 = 1.5 cm
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3 cm

124. Example 1 Cont’d
Step 3 Find the volume of the container
V = r2h
= (3.14)(32)(12)
= 339.12 cm3
12 cm
6 cm
3 cm
Container Chocolate
Step 4 Find the volume of one chocolate
3
r
3
4
=V 
)5.1)(14.3(
3
4
= 3
3
cm13.14=
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125. Example 1 Cont’d
Step 5
Find how many chocolates fit into
the container
V container  V chocolate
= 339.12  14.13
= 2412 cm
6 cm
3 cm
Container Chocolate
Step 6
Find the cost of one container
Cost = 24 ($1.50)
= $36
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