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.

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 7 + 10 = E + 2 15 = E Euler’s Theorem Substitute known values. Solve for E. The frame of the house has 15 edges. ANSWER

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 8 + 6 = 12 + 2 14 = 14 Euler’s Theorem Substitute. This is a true statement. So, the solution checks.

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.

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

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 12 + 20 = 30 + 2 32 = 32 Check Euler’s theorem Substitute This is a true statement so, the solution check

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

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

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

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.

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.

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

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. ≈ 488.57 Simplify. The surface area of the right pentagonal prism is about 488.57 square feet. ANSWER

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

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: 7 4 = 28 in.2 Top and bottom faces: 3 4 = 12 in.2 Front and back faces: 3 7 = 21 in.2 S = 2(28) + 2(12) + 2(21) = 122 in.2

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) + 14 7 = 122 in.2

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?

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 = 2πr2 + 2πrh Surface area of a cylinder. = 2π(60)2 + 2π(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

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 157.08 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.

Find the height of a cylinder EXAMPLE 4 Find the height of a cylinder 157.08 = 2π(2.5)2 + 2π(2.5)h Substitute known values. 157.08 = 12.5π + 5πh Simplify. 157.08 – 12.5π = 5πh Subtract 12.5π from each side. 117.81 ≈ 5πh Simplify. Use a calculator. 7.5 ≈ h Divide each side by 5π. The height of the cylinder is about 7.5 meters. ANSWER

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. 1759.29 cm2 ANSWER

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