10-3 Surface Area of Cylinders and Prisms
#1: Find the lateral area of the cylinder to the nearest tenth with d = 10 cm and h = 2 cm.
L = 2π rh
d 10
=
=5
2 2
L = 2π ( 5 ) ( 2 ) ≈ 62.8 cm 2
r=
Find (a) the lateral area and (b) the surface area of the prism. Round to the nearest whole
number.
#7: L = Ph
Since the base is a right triangle with legs 8
and 15, we can find the third side using
Pythagorean theorem:
8 2 + 15 2 = x 2
#11: L = Ph
Since the base is a right triangle with legs 15
and 20, we can find the third side using
Pythagorean theorem:
15 2 + 20 2 = x 2
64 + 225 = x 2
225 + 400 = x 2
289 = x 2
625 = x 2
x 2 = 289 = 17
P = 8 + 15 + 17 = 40 m
x 2 = 625 = 25
P = 15 + 20 + 25 = 60 m
L = 40 ( 8 ) = 320 m 2
L = 60 ( 25 ) = 1500 m 2
SA = 2B + L
1
1
B = bh = ( 8 ) (15 ) = 60 m 2
2
2
SA = 2 ( 60 ) + 320 = 120 + 320 = 440 m 2
SA = 2B + L
1
1
B = bh = (15 ) ( 20 ) = 150 m 2
2
2
SA = 2 (150 ) + 1500 = 300 + 1500 = 1800 m 2
Find the surface area of the cylinder in terms of π.
#13: r = 1 m; h = 3 m
SA = 2π r 2 + 2π rh
SA = 2π (1) + 2π (1) ( 3)
2
2π + 6π
8π m 2
Homework: Finish the Practice 10-3 Worksheet.
10-4 Surface Area of Cones and Pyramids
#1: Find the lateral area of the cone with
h = 10 m and r = 2 m to the nearest whole
number.
L = π rl
l is the slant height, which we can find using
Pythagorean theorem:
2 2 + 10 2 = l 2
#4: Find the surface area of the cone with
h = 3 cm and r = 4 cm in terms of π.
4 + 100 = l
l = 25 = 5
104 = l
2
2
SA = π rl + π r 2
32 + 4 2 = l 2
9 + 16 = l 2
25 = l 2
SA = π ( 4 ) ( 5 ) + π ( 4 )
l = 104
L = π ( 2 ) 104 ≈ 64 m
2
= 20π + 16π
2
= 36π cm 2
#7: Find the lateral area of the regular
pyramid to the nearest tenth.
1
L = Pl
2
The base is a square with sides of 7m,
#10: Find the surface area of the regular
pyramid to the nearest tenth.
1
SA = B + Pl
2
The base is a square with sides of 5 m,
so P = 4 ( 7 ) = 28 m.
so P = 4 ( 5 ) = 20 m and B = 5 ( 5 ) = 25 m 2 .
h = 7 and half the base = 3.5
h = 12 and half the base = 2.5
l = 7 + 3.5 = 49 + 12.25 = 61.25
l 2 = 12 2 + 2.5 2 = 144 + 6.25 = 150.25
l = 61.25
1
L = ( 28 ) 61.25 ≈ 109.6 m 2
2
l = 150.25
1
SA = 25 + ( 20 ) 150.25 ≈ 147.6 m 2
2
2
2
2
Homework: Finish the Practice 10-4 Worksheet.
10-5 Volume of Cylinders and Prisms
#1: Find the volume of a cylinder with diameter 12 m and height 10 m to the nearest tenth.
V = Bh = π r 2 h
12
r=
=6m
2
2
V = π ( 6 ) (10 ) ≈ 1131.0 m 2
Find the volume of the prism to the nearest whole number.
#7: length = 8, width = 5, height = 3
V = lwh
= 8 ( 5 ) ( 3)
= 120 in 3
#8: V = Bh
The base is an equilateral triangle with side length 3 ft.
s 2 3 32 3 9 3
=
=
4
4
4
⎛ 9 3⎞
3
V =⎜
⎟ ( 9 ) ≈ 35 ft
4
⎝
⎠
B=
Find the volume of the composite figure to the nearest whole number.
#13: This solid is made of two rectangular prisms.
The bottom one’s dimensions are 6x2x3, and the top’s dimensions are 4x2x5.
(You get 5 by doing 8 – 3.)
Vbottom = 6 ( 2 ) ( 3) = 36 ft 3
Vtop = 4 ( 2 ) ( 5 ) = 40 ft 3
Vtotal = 36 + 40 = 76 ft 3
Homework: Finish the Practice 10-5 Worksheet.
10-6 Volume of Pyramids and Cones
Find the volume of the pyramid.
#1: Base sides = 54 cm, l = 45 cm.
To find height, we need to use Pythagorean Thm:
54 ÷ 2 = 27
27 2 + h 2 = 45 2
729 + h 2 = 2025
#4: B = 400 yd2 and h = 36 yd
1
V = Bh
3
1
V = ( 400 ) ( 36 ) = 4800 yd 3
3
h 2 = 2025 − 729 = 1296
h = 1296 = 36 cm
1
V = Bh
3
B = 54 2 = 2916 cm 2
1
V = ( 2916 ) ( 36 ) = 34992 cm 3
3
Find the volume of the cone. Round to the nearest tenth.
#7: d = 10 cm, h = 24 cm
10
r=
= 5 cm
2
1
V = π r 2h
3
1
2
V = π ( 5 ) ( 24 ) ≈ 628.3 cm 3
3
Find the variable.
1
#13: V = Bh
3
1
1500 = (15 ) (15 ) x
3
1500 = 75x
x=
1500
= 20 units
75
Homework: Finish the Practice 10-6 Worksheet.
#9: d = 26 m, h = 28 m
26
r=
= 13 cm
2
1
V = π r 2h
3
1
2
V = π (13) ( 28 ) ≈ 4955.3 m 3
3
10-7 Surface Area and Volume of Spheres
#1: Find the surface area of a sphere with radius 14 in. to the nearest tenth.
SA = 4π r 2
SA = 4π (14 ) ≈ 2463.0 in 2
2
#7: Find the volume of a sphere with a
diameter of 14 mi to the nearest tenth.
4 3
πr
3
14
r=
= 7 mi
2
4
3
V = π ( 7 ) ≈ 1436.8 mi3
3
V=
#11: Find the volume of a sphere with a
surface area of 90,790 cm2 to the nearest
tenth.
SA = 4π r 2
90790 = 4π r 2
90790
r2 =
4π
90790
r=
≈ 84.99905
4π
3
4 ⎛ 90790 ⎞
V = π⎜
≈ 2, 572, 354.6 cm 3
3 ⎝
4π ⎟⎠
#13: Find the surface area of a sphere with
V = 1200 ft3. Round to the nearest whole number.
4
1200 = π r 3
3
1200
r3 =
4
π
3
1200
r=
≈ 6.592208
3 4
π
3
SA = 4π ( 6.592208 ) ≈ 546 ft 2
2
Homework: Finish the Practice 10-7 Worksheet.
#16: Find the volume of a baseball with
C = 24 cm.
C = 2π r
24 = 2π r
24 12
r=
=
2π π
3
4 ⎛ 12 ⎞
V = π ⎜ ⎟ ≈ 233 cm 3
3 ⎝π⎠