1. Surface Area of Prisms and Cylinders 8-7 A. S = 2r 2 + 2rh = 2(4 2 ) + 2(4)(6) = 80 in 2  251.2 in 2 B. S = 2B + Ph = 204 ft 2 = 2( 8 3) + (18)(10) 1212 Additional Example 1: Finding Surface Area Find the surface area of each figure to the nearest tenth. Use 3.14 for .

2. Surface Area of Prisms and Cylinders 8-7 A. S = 2r 2 + 2rh = 2(15 2 ) + 2(15)(3) = 540 in 2  1695.6 cm 2 B. S = 2B + Ph = 252 cm 2 = 2( 7 6) + (21)(10) 1212 Check It Out: Example 1 15 cm 3 cm 7 cm 10 cm 6 cm Find the surface area of each figure to the nearest tenth. Use 3.14 for .

3. Surface Area of Prisms and Cylinders 8-7 Additional Example 3: Application A cylindrical soup can is 7.6 cm in diameter and 11.2 cm tall. What is the area of the label that covers the side of the can? Only the lateral surface needs to be covered. Diameter = 7.6 cm, so r = 3.8 cm. L = 2rh = 2(3.8)(11.2) ≈ 267.3 cm 2

4. Surface Area of Prisms and Cylinders 8-7 Check It Out: Example 3 A cylindrical storage tank that is 6 ft in diameter and 12 ft tall needs to be painted. The paint will cover 100 square feet per gallon. How many gallons will it take to paint the tank? The diameter is 6 ft, so r = 3 ft. S = 2r 2 + 2rh = 2(3 2 ) + 2(3)(12) ≈ 282.6 ft 2 Move the decimal point 2 places to the left to divide by 100. ≈ 2.826 gal

5. Surface Area of Prisms and Cylinders 8-7 Lesson Quiz Find the surface area of each figure to the nearest tenth. Use 3.14 for . 1. the triangular prism 2. the cylinder 320.3 in 2 360 cm 2

6. Surface Area of Prisms and Cylinders 8-7 2. Identify the surface area of the cylinder rounded to the nearest tenth. Use 3.14 for . Lesson Quiz for Student Response Systems