1. Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

2. Ch 12.2 Surface Areas of Prisms & Cylinders Standard 8.0 Students know, derive, and solve problems involving the area of common geometric figures. Learning Target: I will be able to solve problems involving areas of prisms and cylinders. Ch 10.5 Ch 12.2

3. Theorem 12-1 Ch 12.2 Lateral Area: The sum of the areas of the lateral faces of a solid. Lateral Faces: The faces that are not bases. In a prism, the lateral faces are parallelograms.

4. Example 1 Lateral Area of a Prism Find the lateral area of the regular hexagonal prism. The bases are regular hexagons. So the perimeter of one base is 6(5) or 30 centimeters. Answer:The lateral area is 360 square centimeters. Lateral area of a prism P = 30, h = 12 Multiply. Ch 12.2

5. Example 1 A.162 cm 2 B.216 cm 2 C.324 cm 2 D.432 cm 2 Find the lateral area of the regular octagonal prism. Ch 12.2 Lateral area of a prism P = 24, h = 9 Multiply. = (3 * 8) (9) = 216

6. Concept Ch 12.2 Theorem 12-2 Surface Area: The sum of the areas of all surfaces of a solid figure. Solid Figure: A figure that encloses a part of space. In a prism, the lateral faces are parallelograms.

7. Surface Area of a Prism Find the surface area of the rectangular prism. Ch 12.2 Surface area of a prism P = 24, h = 12, B = 36 Simplify. = [(6 * 4) (12)] + 2 (6 * 6) = 360 S = L + 2B

8. Example 2 A.320 units 2 B.512 units 2 C.368 units 2 D.416 units 2 Find the surface area of the triangular prism. Ch 12.2 Surface area of a prism P = 32, h=10, B = 48 Simplify. = [(10+10+12)(10)] + 2 (½)(12*8) = 416 S = L + 2B

9. Concept Ch 12.2 Theorem 12-3 & 12-4

10. Example 3 Lateral Area and Surface Area of a Cylinder Find the lateral area and the surface area of the cylinder in terms of π L = PhLateral area of a cylinder =2  rhP = 2πr (circumference of a circle) =2  (14)(18) r = 14, h = 18. ≈504πSimplify. Ch 12.2 S = L + 2BSurface area of a cylinder =504π + 2  r 2 L = 504π, B = π r 2 ≈504π + 2  (14) 2 r = 14 ≈896πSimplify.

11. Example 3 A.lateral area ≈ 480π ft 2 and surface area ≈ 768π ft 2 B.lateral area ≈ 480π ft 2 and surface area ≈ 384π ft 2 C.lateral area ≈ 240π ft 2 and surface area ≈ 768π ft 2 D.lateral area ≈ 240π ft 2 and surface area ≈ 384π ft 2 Find the lateral area and the surface area of the cylinder in terms of π. Ch 12.2

12. Example 4 Find Missing Dimensions MANUFACTURING A soup can is covered with the label shown. What is the radius of the soup can? L = PhLateral area of a cylinder =2  rhP = 2π r (circumference of a circle) 125.6=2  r(8) L = 15.7 × 8, h = 8. 125.6=16  rSimplify. 7.85/π=rDivide each side by 16 . Ch 12.2

13. Example 4 A.12 inches B.16 inches C.18 inches D.24 inches Find the diameter of a base of a cylinder if the surface area is 480  square inches and the height is 8 inches. Ch 12.2