Surface
Area
of
10-4
10-4Surface Area of Prisms and Cylinders
Prisms and Cylinders
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Geometry
Holt
Geometry
10-4 Surface Area of Prisms and Cylinders
Warm Up
Find the perimeter and area of
each polygon.
1. a rectangle with base 14 cm and height
9 cm P = 46 cm; A = 126 cm2
2. a right triangle with 9 cm and 12 cm
legs P = 36 cm; A = 54 cm2
3. an equilateral triangle with side length
6 cm
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Objectives
Learn and apply the formula for the
surface area of a prism.
Learn and apply the formula for the
surface area of a cylinder.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Vocabulary
lateral face
lateral edge
right prism
oblique prism
altitude
surface area
lateral surface
axis of a cylinder
right cylinder
oblique cylinder
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Prisms and cylinders have 2 congruent parallel bases.
A lateral face is not a base. The edges of the base
are called base edges. A lateral edge is not an edge
of a base. The lateral faces of a right prism are all
rectangles. An oblique prism has at least one
nonrectangular lateral face.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
An altitude of a prism or cylinder is a perpendicular
segment joining the planes of the bases. The height of
a three-dimensional figure is the length of an altitude.
Surface area is the total area of all faces and curved
surfaces of a three-dimensional figure. The lateral
area of a prism is the sum of the areas of the lateral
faces.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
The net of a right prism can be drawn so that the
lateral faces form a rectangle with the same height as
the prism. The base of the rectangle is equal to the
perimeter of the base of the prism.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
The surface area of a right rectangular prism with
length ℓ, width w, and height h can be written as
S = 2ℓw + 2wh + 2ℓh.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Caution!
The surface area formula is only true for right
prisms. To find the surface area of an oblique
prism, add the areas of the faces.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 1A: Finding Lateral Areas and Surface Areas
of Prisms
Find the lateral area and surface area of the
right rectangular prism. Round to the nearest
tenth, if necessary.
L = Ph
P = 2(9) + 2(7) = 32 ft
= 32(14) = 448 ft2
S = Ph + 2B
= 448 + 2(7)(9) = 574 ft2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 1B: Finding Lateral Areas and Surface Areas
of Prisms
Find the lateral area and surface area of a
right regular triangular prism with height 20
cm and base edges of length 10 cm. Round to
the nearest tenth, if necessary.
L = Ph
= 30(20) = 600 ft2
P = 3(10) = 30 cm
S = Ph + 2B
The base area is
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 1
Find the lateral area and surface area of a cube
with edge length 8 cm.
L = Ph
= 32(8) = 256 cm2
P = 4(8) = 32 cm
S = Ph + 2B
= 256 + 2(8)(8) = 384 cm2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
The lateral surface of a cylinder is the curved surface
that connects the two bases. The axis of a cylinder is
the segment with endpoints at the centers of the
bases. The axis of a right cylinder is perpendicular to
its bases. The axis of an oblique cylinder is not
perpendicular to its bases. The altitude of a right
cylinder is the same length as the axis.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 2A: Finding Lateral Areas and Surface Areas
of Right Cylinders
Find the lateral area and surface area of the
right cylinder. Give your answers in terms of .
The radius is half the diameter,
or 8 ft.
L = 2rh = 2(8)(10) = 160 in2
S = L + 2r2 = 160 + 2(8)2
= 288 in2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 2B: Finding Lateral Areas and Surface Areas
of Right Cylinders
Find the lateral area and surface area of a right
cylinder with circumference 24 cm and a height
equal to half the radius. Give your answers in
terms of .
Step 1 Use the circumference to find the radius.
C = 2r
24 = 2r
r = 12
Holt Geometry
Circumference of a circle
Substitute 24 for C.
Divide both sides by 2.
10-4 Surface Area of Prisms and Cylinders
Example 2B Continued
Find the lateral area and surface area of a right
cylinder with circumference 24 cm and a height
equal to half the radius. Give your answers in
terms of .
Step 2 Use the radius to find the lateral area and
surface area. The height is half the radius, or 6 cm.
L = 2rh = 2(12)(6) = 144 cm2
Lateral area
S = L + 2r2 = 144 + 2(12)2
Surface area
= 432 in2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 2
Find the lateral area and surface area of a
cylinder with a base area of 49 and a height
that is 2 times the radius.
Step 1 Use the circumference to find the radius.
A = r2
49 = r2
r=7
Holt Geometry
Area of a circle
Substitute 49 for A.
Divide both sides by  and take the
square root.
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 2 Continued
Find the lateral area and surface area of a
cylinder with a base area of 49 and a height
that is 2 times the radius.
Step 2 Use the radius to find the lateral area and
surface area. The height is twice the radius, or 14 cm.
L = 2rh = 2(7)(14)=196 in2
Lateral area
S = L + 2r2 = 196 + 2(7)2 =294 in2 Surface area
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 3: Finding Surface Areas of Composite
Three-Dimensional Figures
Find the surface area of the composite figure.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 3 Continued
The surface area of the rectangular prism is
.
A right triangular prism is added to the
rectangular prism. The surface area of the
triangular prism is
.
Two copies of the rectangular prism base are
removed. The area of the base is B = 2(4) = 8 cm2.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 3 Continued
The surface area of the composite figure is the sum
of the areas of all surfaces on the exterior of the
figure.
S = (rectangular prism surface area) + (triangular
prism surface area) – 2(rectangular prism base area)
S = 52 + 36 – 2(8) = 72 cm2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 3
Find the surface area of the composite figure.
Round to the nearest tenth.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 3 Continued
Find the surface area of the composite figure.
Round to the nearest tenth.
The surface area of the rectangular prism is
S =Ph + 2B = 26(5) + 2(36) = 202 cm2.
The surface area of the cylinder is
S =Ph + 2B = 2(2)(3) + 2(2)2 = 20 ≈ 62.8 cm2.
The surface area of the composite figure is the sum
of the areas of all surfaces on the exterior of the
figure.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 3 Continued
Find the surface area of the composite figure.
Round to the nearest tenth.
S = (rectangular surface area) +
(cylinder surface area) – 2(cylinder base area)
S = 202 + 62.8 — 2()(22) = 239.7 cm2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Remember!
Always round at the last step of the problem. Use
the value of  given by the  key on your
calculator.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 4: Exploring Effects of Changing Dimensions
The edge length of the cube is tripled. Describe
the effect on the surface area.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 4 Continued
24 cm
original dimensions:
S = 6ℓ2
= 6(8)2 = 384 cm2
edge length tripled:
S = 6ℓ2
= 6(24)2 = 3456 cm2
Notice than 3456 = 9(384). If the length, width, and
height are tripled, the surface area is multiplied by 32,
or 9.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 4
The height and diameter of the cylinder are
multiplied by
surface area.
Holt Geometry
. Describe the effect on the
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 4 Continued
11 cm
7 cm
original dimensions:
S = 2(112) + 2(11)(14)
= 550 cm2
height and diameter halved:
S = 2(5.52) + 2(5.5)(7)
= 137.5 cm2
Notice than 550 = 4(137.5). If the dimensions are
halved, the surface area is multiplied by
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 5: Recreation Application
A sporting goods company sells tents in two
styles, shown below. The sides and floor of each
tent are made of nylon.
Which tent requires less nylon to manufacture?
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 5 Continued
Pup tent:
Tunnel tent:
The tunnel tent requires less nylon.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Check It Out! Example 5
A piece of ice shaped like a 5 cm by 5 cm by 1 cm
rectangular prism has approximately the same
volume as the pieces below. Compare the surface
areas. Which will melt faster?
The 5 cm by 5 cm by 1 cm prism has a surface area of
70 cm2, which is greater than the 2 cm by 3 cm by
4 cm prism and about the same as the half cylinder. It
will melt at about the same rate as the half cylinder.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Lesson Quiz: Part I
Find the lateral area and the surface area of
each figure. Round to the nearest tenth, if
necessary.
1. a cube with edge length 10 cm
L = 400 cm2 ; S = 600 cm2
2. a regular hexagonal prism with height 15 in.
and base edge length 8 in.
L = 720 in2; S  1052.6 in2
3. a right cylinder with base area 144 cm2 and a
height that is
the radius
L  301.6 cm2; S = 1206.4 cm2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Lesson Quiz: Part II
4. A cube has edge length 12 cm. If the edge
length of the cube is doubled, what happens to
the surface area?
The surface area is multiplied by 4.
5. Find the surface area of the composite figure.
S = 3752 m2
Holt Geometry