Surface Area

1. 10-4 Surface Area of Prisms and Cylinders10-4
Course 2
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Surface Areas of Prisms and Cylinders

2. 10-4 Surface Area of Prisms and Cylinders
Warm Up
Find the volume of each figure to the nearest
tenth. Use 3.14 for π.
1. rectangular pyramid 7 ft by 8 ft by 10 ft tall
186.7 ft3
12.6 ft3
162 ft3
2. cone with radius 2 ft and height 3 ft
3. triangular pyramid with base area of 54 ft2
and height 9 ft

3. 10-4 Surface Area of Prisms and Cylinders
Problem of the Day
The volume of a 10-meter-tall square
pyramid is 120 m3
. What is the length of
each side of the base?
6 m

4. 10-4 Surface Area of Prisms and Cylinders
Learn to find the surface area of prisms
and cylinders.

5. 10-4 Surface Area of Prisms and Cylinders
Vocabulary
net
surface area

6. 10-4 Surface Area of Prisms and Cylinders
If you remove the surface from a three-
dimensional figure and lay it out flat, the pattern
you make is called a net.
Nets allow you to see all
the surfaces of a solid at
one time. You can use nets
to help you find the
surface area of a
three-dimensional figure.
Surface area is the sum
of the areas of all of the
surfaces of a figure.

7. 10-4 Surface Area of Prisms and Cylinders
You can use nets to write formulas for the
surface area of prisms. The surface area S is the
sum of the areas of the faces of the prism. For
the rectangular prism shown,
S = lw + lh + wh + lw + lh + wh
= 2lw + 2lh + 2wh

8. 10-4 Surface Area of Prisms and Cylinders
SURFACE AREA OF A PRISM
The surface area of a rectangular prism is the sum
of the areas of each face.
S = 2lw + 2lh + 2wh

9. 10-4 Surface Area of Prisms and Cylinders
Find the surface area
of the prism formed
by the net.
Additional Example 1: Finding the Surface Area of a
Prism
= 2lw + 2lh + 2whS
= (2 · 15 · 9) + (2 · 15 · 7) + (2 · 9 · 7)S Substitute.
S = 270 + 210 + 126
S = 606
The surface area of the prism is 606 in2
.
Multiply.
Add.

10. 10-4 Surface Area of Prisms and Cylinders
Find the surface area
of the prism formed
by the net.
= 2lw + 2lh + 2whS
= (2 · 4 · 6) + (2 · 4 · 3) + (2 · 6 · 3)S Substitute.
S = 48 + 24 + 36
S = 108
The surface area of the prism is 108 in2
.
Multiply.
Add.
Check It Out: Example 1
4 in.
6 in.
3 in. 3 in.
4 in.

11. 10-4 Surface Area of Prisms and Cylinders
If you could remove the lateral surface from a
cylinder, like peeling a label from a can, you would
see that it has the shape of a rectangle when
flattened out.
You can draw a net for a cylinder by drawing the
circular bases (like the ends of a can) and the
rectangular lateral surface as shown below. The
length of the rectangle is the circumference, 2πr, of
the cylinder. So the area of the lateral surface is 2πrh.
The area of each base is πr2
.
r
Circumference
of cylinder (2πr)
h

12. 10-4 Surface Area of Prisms and Cylinders
SURFACE AREA OF A CYLINDER
The surface area S of a cylinder is the sum of the
areas of its bases, 2πr2
, plus the area of its lateral
surface, 2πrh.
S= 2πr2
+ 2πrh

13. 10-4 Surface Area of Prisms and Cylinders
Find the surface area of the cylinder formed by
the net to the nearest tenth. Use 3.14 for π.
Additional Example 2: Finding the Surface Area of a
Cylinder
6 ft
6 ft
8.3 ft
S = 2πr2
+ 2πrh
S ≈ (2 · 3.14 · 62
) + (2 · 3.14 · 6 · 8.3) Substitute.
Use the formula.
S ≈ 226.08 + 312.744
S ≈ 538.824
S ≈ 538.8
The surface area of the cylinder is about 538.8 ft2
.
Multiply.
Add.
Round.

14. 10-4 Surface Area of Prisms and Cylinders
Find the surface area of the cylinder formed by
the net to the nearest tenth. Use 3.14 for π.
9 ft
9 ft
20 ft
S = 2πr2
+ 2πrh
S ≈ (2 · 3.14 · 92
) + (2 · 3.14 · 9 · 20) Substitute.
Use the formula.
S ≈ 508.68 + 1130.4
S ≈ 1,639.08
S ≈ 1,639.1
The surface area of the cylinder is about 1,639.1 ft2
.
Multiply.
Add.
Round.
Check It Out: Example 2

15. 10-4 Surface Area of Prisms and Cylinders
Course 2
Additional Example 3: Problem Solving Application
What percent of the total surface area
of the soup can is covered by the
label?
Soup
2 in.
5 in.
6 in.

16. 10-4 Surface Area of Prisms and Cylinders
Course 2
Additional Example 3 Continued
11 Understand the Problem
Rewrite the question as a statement.
• Find the area of the label, and compare to the
total surface area of the can.
List the important information:
• The can is approximately cylinder shaped.
• The height for the can is 6 inches.
• The radius of the can is 2 inches.
• The height of the label is 5 inches.

17. 10-4 Surface Area of Prisms and Cylinders
Course 2
Additional Example 3 Continued
Find the surface area of the can and the
area of the label. Divide to find the percent
of the surface area covered by the label.
22 Make a Plan
Soup
6
in.
5
in.
2 in.

18. 10-4 Surface Area of Prisms and Cylinders
Course 2
Additional Example 3 Continued
Solve33
S = 2πr2
+ 2πrh
≈ 2(3.14)(2)2
+ 2(3.14)(2)(6) Substitute for r and h.
≈ 100.48 in.2
A = lw
= (2πr)w Substitute 2πr for l.
≈ (2)(3.14)(2)(5) Substitute for r and w.
≈ 62.8 cm2
Percent of the surface covered by the label:
= 62.5 %. About 62.5% of the can’s surface is covered by
the label.
62.8 cm2
100.48 cm2

19. 10-4 Surface Area of Prisms and Cylinders
Course 2
Additional Example 3 Continued
Look Back44
Estimate the areas of the two rectangles in
the net.
Label: 2(3)(2)(5) = 60 in2
Can: 2(3)(2)(6) = 77 in2
60 in.2
77 in.2
= 78%
The answer should be less than 78% because
you did not consider the area of the two
circles. So 62.5% is reasonable.

20. 10-4 Surface Area of Prisms and Cylinders
Course 2
Check it Out: Example 3
What percent of the total surface area
of the oil can is covered by the label?
6 in.
4 in.
10 in.

21. 10-4 Surface Area of Prisms and Cylinders
Course 2
Check It Out: Example 3 Continued
11 Understand the Problem
Rewrite the question as a statement.
• Find the area of the label, and compare to the
total surface area of the can.
List the important information:
• The can is approximately cylinder shaped.
• The height for the can is 10 inches.
• The diameter of the can is 6 inches.
• The height of the label is 4 inches.

22. 10-4 Surface Area of Prisms and Cylinders
Course 2
Check It Out: Example 3 Continued
Find the surface area of the can and the
area of the label. Divide to find the percent
of the surface area covered by the label.
22 Make a Plan
10 in. 4 in.
6 in.

23. 10-4 Surface Area of Prisms and Cylinders
Course 2
Check It Out: Example 3 Continued
Solve33
S = 2πr2
+ 2πrh
≈ 2(3.14)(3)2
+ 2(3.14)(3)(10)
Substitute for r
and h.
≈ 244.92 in2
A = lw
= (2πr)w Substitute 2πr for l.
≈ (2)(3.14)(3)(4) Substitute for r and w.
≈ 75.36 cm2
Percent of the surface covered by the label:
= 30.7 %. About 30.7% of the can’s surface is covered by
the label.
75.36 in.2
244.92 in.2

24. 10-4 Surface Area of Prisms and Cylinders
Course 2
Check It Out: Example 3 Continued
Look Back44
Estimate the areas of the two rectangles in
the net.
Label: 2(3)(3)(4) = 72 in2
Can: 2(3)(3)(10) = 180 in2
72 in.2
180 in.2
= 40%
The answer should be less than 40% because
you did not consider the area of the two
circles. So 30.7% is reasonable.

25. 10-4 Surface Area of Prisms and Cylinders
Lesson Quiz
Find the surface area of each figure to the
nearest tenth.
100.5 ft2
352 ft2
25%
3. A drum is cylindrical, and its 14 in. width fits
into a drum stand. What percent of the total
surface area of the drum is covered by the 3 in.
red stripe? Use 3.14 for π.
1. 2.