Section 7.3
Answers to Classwork
5”
Classwork 1 Find the volume of the right
triangular prism shown at right.
5”
Answer: For the figure at right to be considered a
prism we must use one of the triangular faces as
8”
the base. So
Area of base = 0.5(3)(8) = 12 in2
The height is 12 inches so Volume = (Area of base)(height) = (12in2)(12 in) = 144 in3.
12”
Classwork 2: Draw a net for the right triangular prism shown at the top
of this page, and find its surface area.
Answer: Here is one possible net:
The surface area is the sum of the areas of the three rectangular faces and
the two triangular faces. So
Surface Area = (8)(12) + 2(5)(12) + 2(0.5)(8)(3) in2
= 96 + 120 + 24 in2
= 240 in2.
3”
5”
8”
5”
Classwork 3: Find the volume of the cylinder shown at right.
Answer: The radius of the base is 3’, so the area of the base is
A = π (3)2 ft2 = 9 π ft2
So Volume = (9 π )(10) ft3 = 90 π ft3 ≈ 282.6 ft3.
10’
6’
Classwork 4: Sketch the net for a right cylinder whose circular base has
radius r and which has height h. The net should consist of two circles and
a polygon. What kind of polygon is it? What are the dimensions of the
polygon in terms of r and h? What is the surface area of the cylinder in
terms of r and h?
Answer: The polygon is rectangle of dimensions h by C = 2 π r.
The surface area of a cylinder = (h)( 2 π r) + 2 π r2
Surface Area of Cylinder = 2 π rh + 2(Area of Base) units2
Classwork 5: Find the surface area of the cylinder from Classwork 3.
Answer: Using the formula above,
Surface Area = 2 π (3)(10) + 2 π (3)2 = 78 π ft2 ≈ 244.92 ft2
r
2π r
h
Classwork 6: (a) The diagram shows a right square pyramid. The base
is a square of side 10” and the height h of the pyramid is 12”. The
slant height s of one of the lateral triangular faces is the hypotenuse of
h = 12”
a right triangle that has the altitude of the pyramid as one of its sides.
Sketch that right triangle, and use the Pythagorean Theorem to find the
slant height s.
Answer: Since the altitude meets the base at the center of the base and
10”
perpendicular to the base, the slant height s is the hypotenuse of a right
triangle with sides 12” (the height) and 5” (half the width of the base). So
s
s = 12 2 + 5 2 = 169 = 13 in
(b) This slant height s is the height of each of the triangular lateral faces of the pyramid. Find the
area of one of these triangular faces.
Answer: The base of a lateral face is 10 in, so the area of a lateral face is 0.5(10)(13) = 65 in2.
(c) Since the base of the pyramid is a square and the pyramid is a right pyramid, all the triangular
lateral faces are congruent. Use this to find the total surface area of the pyramid.
Answer: The total surface area of the pyramid = area of base + 4(area of a lateral face)
SA = (10)(10) + 4(65) = 360 in2.
Classwork 7 An ice cream cone has a diameter of 10 cm and a height of 12 cm.
Find its lateral surface area.
Answer: The radius of the base is half the diameter, or 5 cm. The slant height is the
hypotenuse of a right triangle like that in Classwork 6, and
2
12
5
2
s = 12 + 5 = 169 = 13 cm
Using these in Theorem 7.3.6,
Lateral Surface Area = π (5)(13) cm2 = 65 π cm2 ≈ 204.1 cm2.
Classwork 8 Find the volume of the pyramid shown at right.
Answer: Area of Base = (10)2 in2. So
Volume =
1
(10)2 (12) = 400 in 3
3
s
h = 12”
10”
Classwork 9 In Classwork 7 we found the lateral surface are of an ice cream cone with a
diameter of 10 cm and a height of 12 cm. What volume of ice cream will it take to fill up the
cone?
Answer: The radius of the base is 5 cm, so Area of base = π (5)2 = 25 π cm2.
1
So V = (25π )(12) = 100π cm 3 ≈ 314 cm 3
3
Classwork 10: What are the volume and surface area of the smallest right
cylinder surrounding a sphere of radius r, in terms of r? (Hint: What is the
height of the cylinder in terms of r?)
r
Answer: The height of the cylinder is h = 2r, so the volume of the smallest
surrounding cylinder is
V = π r 2 (2r) = 2π r 3 .
The surface area of a cylinder is SA = 2(Area of Base) + 2 π rh so for this cylinder
SA = 2π r 2 + 2π r(2r) = 6π r 2 .
Classwork 11 Suppose an igloo has the shape of a hemisphere. If its inner diameter is 10 m.,
what is the volume of the space inside the igloo? What is the surface area of the inside of the
igloo?
Answer: The volume and surface are of a hemisphere are half those of a sphere.
So in this problem
1⎛ 4
1
⎞ 250
V = ⎜ π 53⎟ =
π m 3 = 83 π m 3 ≈ 262 m 3
⎠
2⎝ 3
3
3
1
SA = 4π 5 2 = 50π m 2 ≈ 157 m 2
2
(
)