9-4

1. 9-4 Surface Areas of Prisms and Cylinders Warm Up Problem of the Day Lesson Presentation Course 2

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

3. 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. Learn to find the surface area of prisms and cylinders.

5. Vocabulary net surface area

6. 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. S = lw +lh + wh + lw + lh + wh = 2lw + 2lh + 2wh 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,

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

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

11. Circumference of cylinder (2r) r h 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, 2r, of the cylinder. So the area of the lateral surface is 2rh. The area of each base is r2.

13. Additional Example 2: Finding the Surface Area of a Cylinder Find the surface area of the cylinder formed by the net to the nearest tenth. Use 3.14 for . 6 ft 8.3 ft 6 ft S = 2r2 + 2rh Use the formula. S  (2 · 3.14 · 62) + (2 · 3.14 · 6 · 8.3) Substitute. S  226.08 + 312.744 Multiply. Add. S 538.824 Round. S 538.8 The surface area of the cylinder is about 538.8 ft2.

14. Check It Out: Example 2 Find the surface area of the cylinder formed by the net to the nearest tenth. Use 3.14 for . 9 ft 20 ft 9 ft S = 2r2 + 2rh Use the formula. S  (2 · 3.14 · 92) + (2 · 3.14 · 9 · 20) Substitute. S  508.68 + 1130.4 Multiply. Add. S 1,639.08 Round. S 1,639.1 The surface area of the cylinder is about 1,639.1 ft2.

15. Soup Additional Example 3: Problem Solving Application What percent of the total surface area of the soup can is covered by the label? 2 in. 6 in. 5 in. Course 2

16. 1 Understand the Problem Additional Example 3 Continued 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. Course 2

17. Make a Plan 2 in. 5 in. 6 in. Soup 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. Course 2

18. 3 Solve 62.8 cm2100.48 cm2 Percent of the surface covered by the label: = 62.5 %. About 62.5% of the can’s surface is covered by the label. Additional Example 3 Continued S = 2r2 + 2rh Substitute for r and h.  2(3.14)(2)2 + 2(3.14)(2)(6)  100.48 in.2 A = lw Substitute 2r for l. = (2r)w  (2)(3.14)(2)(5) Substitute for r and w.  62.8 cm2 Course 2

19. 4 Additional Example 3 Continued Look Back 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. Course 2

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

21. 1 Understand the Problem Check It Out: Example 3 Continued 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. Course 2

22. Make a Plan 6 in. 4 in. 10 in. 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. Course 2

23. 3 Solve 75.36 in.2244.92 in.2 Percent of the surface covered by the label: = 30.7 %. About 30.7% of the can’s surface is covered by the label. Check It Out: Example 3 Continued S = 2r2 + 2rh Substitute for r and h.  2(3.14)(3)2 + 2(3.14)(3)(10)  244.92 in2 A = lw Substitute 2r for l. = (2r)w  (2)(3.14)(3)(4) Substitute for r and w.  75.36 cm2 Course 2

24. 4 Check It Out: Example 3 Continued Look Back 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. Course 2

25. Lesson Quiz Find the surface area of each figure to the nearest tenth. 1. 2. 100.5 ft2 352 ft2 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 p. 25%