1. EXAMPLE 2
Use Euler’s Theorem in a real-world situation
House Construction
Find the number of edges on the frame of the house.
SOLUTION
The frame has one face as its
foundation, four that make up its walls, and two that make up its roof, for a total of 7 faces.

2. Use Euler’s Theorem in a real-world situation
EXAMPLE 2
Use Euler’s Theorem in a real-world situation
To find the number of vertices, notice that there are 5 vertices around each pentagonal wall, and there are no other vertices. So, the frame of the house has 10 vertices.
Use Euler’s Theorem to find the number of edges.
F + V = E + 2
= E + 2
15 = E
Euler’s Theorem
Substitute known values.
Solve for E.
The frame of the house has 15 edges.
ANSWER

3. Use Euler’s Theorem with Platonic solids
EXAMPLE 3
Use Euler’s Theorem with Platonic solids
Find the number of faces, vertices, and edges of the regular octahedron.
Check your answer using Euler’s Theorem.
SOLUTION
By counting on the diagram, the octahedron has 8 faces, 6 vertices, and 12 edges. Use Euler’s Theorem to check.
F + V = E + 2 =
14 = 14
Euler’s Theorem
Substitute.
This is a true statement. So, the solution checks.

4. EXAMPLE 4
Describe cross sections
Describe the shape formed by the intersection of the
plane and the cube. a. b.
SOLUTION
a. The cross section is a square.
b. The cross section is a rectangle.

5. EXAMPLE 4
Describe cross sections c.
SOLUTION
c. The cross section is a trapezoid.

6. GUIDED PRACTICE for Examples 2, 3, and 4
4. Find the number of faces, vertices, and edges of the regular dodecahedron on page 796. Check your answer using Euler’s Theorem.
SOLUTION
Counting on the diagram, the dodecahedron has 12 faces, 20 vertices, and 30 edges. Use Euler’s theorem to
F + V = E + 2 =
32 = 32
Check
Euler’s theorem
Substitute
This is a true statement so, the solution check

7. GUIDED PRACTICE
for Examples 2, 3, and 4
Describe the shape formed by the intersection of the
plane and the solid. 5.
ANSWER
The cross section is a triangle

8. GUIDED PRACTICE
for Examples 2, 3, and 4 6.
ANSWER
The cross section is a circle

9. GUIDED PRACTICE
for Examples 2, 3, and 4 7.
ANSWER
The cross section is a hexagon

10. EXAMPLE 1
Use the net of a prism
Find the surface area of a rectangular prism with height 2 centimeters, length 5 centimeters, and width 6 centimeters.
STEP 1
SOLUTION
Sketch the prism. Imagine unfolding it to
make a net.

11. EXAMPLE 1
Use the net of a prism
Find the areas of the rectangles that form the faces of the prism.
STEP 2
STEP 3
Add the areas of all the faces to find the
surface area.
The surface area of the prism is
S = 2(12) + 2(10) + 2(30)
= 104 cm2.

12. Find the surface area of a right prism
EXAMPLE 2
EXAMPLE 2
Find the surface area of a right prism
Find the surface area of the right pentagonal prism.
SOLUTION
STEP 1
Find the perimeter and area of a base of the prism.
Each base is a regular pentagon.
Perimeter P = 5(7.05)
= 35.25
Apothem a = √ 62 –3.5252
≈ 4.86

13. Find the surface area of a right prism
EXAMPLE 2
Find the surface area of a right prism
STEP 2
Use the formula for the surface area that
uses the apothem.
S = aP + Ph
Surface area of a
right prism
≈ (4.86)(35.25) + (35.25)(9)
Substitute known
values. ≈
Simplify.
The surface area of the right pentagonal prism is
about square feet.
ANSWER

14. GUIDED PRACTICE
for Examples 1 and 2
1. Draw a net of a triangular prism.
SOLUTION

15. GUIDED PRACTICE
for Examples 1 and 2
2. Find the surface area of a right rectangular prism with height 7 inches, length 3 inches, and width 4 inches using (a) a net and (b) the formula for the surface area of a right prism.
ANSWER
(a) NET: Left and right faces:
= 28 in.2
Top and bottom faces:
= 12 in.2
Front and back faces:
= 21 in.2
S = 2(28) + 2(12) + 2(21) = 122 in.2

16. GUIDED PRACTICE
for Examples 1 and 2
2. Find the surface area of a right rectangular prism with height 7 inches, length 3 inches, and width 4 inches using (a) a net and (b) the formula for the surface area of a right prism.
ANSWER
(b) S = 2B + Ph =
2(3 4)
= 122 in.2

17. EXAMPLE 3
Find the height of a cylinder
COMPACT DISCS
You are wrapping a stack of 20 compact discs using a shrink wrap. Each disc is cylindrical with height 1.2 millimeters and radius 60 millimeters. What is the minimum amount of shrink wrap needed to cover the stack of 20 discs?

18. Find the height of a cylinder
EXAMPLE 3
Find the height of a cylinder
SOLUTION
The 20 discs are stacked, so the height of the stack will be 20(1.2) = 24 mm. The radius is 60 millimeters. The minimum amount of shrink wrap needed will be equal to the surface area of the stack of discs.
S = 2πr πrh
Surface area of a cylinder.
= 2π(60) π(60)(24)
Substitute known values.
≈ 31,667
Use a calculator.
You will need at least 31,667 square millimeters, or about 317 square centimeters of shrink wrap.
ANSWER

19. Find the height of a cylinder
EXAMPLE 4
Find the height of a cylinder
Find the height of the right cylinder shown, which has a surface area of square meters.
SOLUTION
Substitute known values in the formula for the surface area of a right cylinder and solve for the height h.
S = 2πr2 + 2πrh
Surface area of a
cylinder.

20. Find the height of a cylinder
EXAMPLE 4
Find the height of a cylinder
= 2π(2.5)2 + 2π(2.5)h
Substitute known
values.
= 12.5π + 5πh
Simplify.
– 12.5π = 5πh
Subtract 12.5π from
each side.
≈ 5πh
Simplify. Use a
calculator.
7.5 ≈ h
Divide each side by
5π.
The height of the cylinder is about 7.5 meters.
ANSWER

21. GUIDED PRACTICE
for Examples 3 and 4
3. Find the surface area of a right cylinder with height 18 centimeters and radius 10 centimeters. Round your answer to two decimal places.
cm2
ANSWER

22. GUIDED PRACTICE
for Examples 3 and 4
4. Find the radius of a right cylinder with height 5 feet and surface area 208π square feet.
The radius of cylinder is 8 feet.
ANSWER