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Ch 12.2

Ch 12.2. Find the area of the circle. A = (π)(12 2 ) = 144π. Find the area of the sector. A = (π)(4 2 ) = 8π 2. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360. Ch 10.5. Ch 12.2. Learning Target:

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Ch 12.2

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  1. Ch 12.2 Find the area of the circle. A = (π)(122) = 144π Find the area of the sector. A = (π)(42) = 8π 2 Find the area of the sector. A = (π)(222)135 = 1089π 360

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

  3. 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. Theorem 12-1

  4. Ch 12.2 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. Lateral area of a prism P=30, h=12 Multiply. Answer: The lateral area is 360 square centimeters. Example 1

  5. Ch 12.2 Find the lateral area of the regular octagonal prism. A. 162 cm2 B. 216 cm2 C. 324 cm2 D. 432 cm2 Lateral area of a prism = (3 * 8) (9) P=24, h=9 = 216 Multiply. Example 1

  6. Ch 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. Theorem 12-2 Concept

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

  8. Ch 12.2 Find the surface area of the triangular prism. A. 320 units2 B. 512 units2 C. 368 units2 D. 416 units2 S = L + 2B Surface area of a prism = [(10+10+12)(10)] + 2 (½)(12*8) P=32, h=10, B = 48 = 416 Simplify. Example 2

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

  10. Ch 12.2 Lateral Area and Surface Area of a Cylinder Find the lateral area and the surface area of the cylinder in terms of π L = Ph Lateral area of a cylinder = 2rh P = 2πr (circumference of a circle) = 2(14)(18) r = 14 , h = 18. ≈ 504π Simplify. S = L + 2B Surface area of a cylinder = 504π + 2r2 L = 504π , B = π r2 ≈ 504π + 2(14)2r = 14 ≈ 896π Simplify. Example 3

  11. Ch 12.2 Find the lateral area and the surface area of the cylinder in terms of π. A. lateral area ≈ 480π ft2 andsurface area ≈ 768π ft2 B. lateral area ≈ 480π ft2 andsurface area ≈ 384π ft2 C. lateral area ≈ 240π ft2 andsurface area ≈ 768π ft2 D. lateral area ≈ 240π ft2 andsurface area ≈ 384π ft2 Example 3

  12. Ch 12.2 Find Missing Dimensions MANUFACTURINGA soup can is covered with the label shown. What is the radius of the soup can? L = Ph Lateral area of a cylinder = 2rh P = 2π r (circumference of a circle) 125.6 = 2r(8) L = 15.7 × 8 , h = 8. 125.6 = 16r Simplify. 7.85/π = r Divide each side by 16. Example 4

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

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