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Find the surface area of each.

Find the surface area of each. S = (P ℓ )/2 + B = 2π(8)(√(15 2 +8 2 )/2 + π(8 2 ) = 200π ft 2. S = (P ℓ )/2 + B = (20×3)(15)/2 + 20√(10 2 +20 2 )/2 = 623.2 in 2. 8.7. S = (P ℓ )/2 + B = (10×6)(14)/2 + (8.7)(10×6)/2 = 679.8 ft 2. S = (P ℓ )/2 + B

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Find the surface area of each.

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  1. Find the surface area of each. S = (Pℓ)/2 + B = 2π(8)(√(152+82)/2 + π(82) = 200π ft2 S = (Pℓ)/2 + B = (20×3)(15)/2 + 20√(102+202)/2 = 623.2 in2 8.7 S = (Pℓ)/2 + B = (10×6)(14)/2 + (8.7)(10×6)/2 = 679.8 ft2 S = (Pℓ)/2 + B = 2π(√(82-62)(8)/2 + π(6.92) = 70.3π m2

  2. Learning Target: I will be able to solve problems involving the volume of pyramids and cones. Standard 9.0 Students compute the volumes of pyramids and cones and commit to memory the formulas for pyramids. Ch 12.5Volumes of Pyramids & Cones

  3. Theorem 12-11 Concept

  4. s 3, h 7 21 Volume of a Pyramid Find the volume of the square pyramid. Volume of a pyramid Multiply. Answer: The volume of the pyramid is 21 cubic inches. Example 1

  5. A.416 ft3 B. C. D. Brad is building a model pyramid for a social studies project. The model is a square pyramid with a base edge of 8 feet and a height of 6.5 feet. Find the volume of the pyramid. 1 3 V = Bh Volume of a pyramid 1 3 B = s2 , s = 8 , h =6.5 = (64)(6.5) = 138.7 Multiply. Example 1

  6. Theorem 12-12 Concept

  7. Volume of a Cone A. Find the volume of the oblique cone in terms of π. 1 3 V = Bh Volume of a cone B = πr2 r = 9.1, h = 25 = 690π Simplify Example 2A

  8. Volume of a Cone B. Find the volume of the cone in terms of π. 1 3 V = Bh Volume of a cone B = πr2 r = 5, h = 12 = 100π Simplify. Example 2B

  9. A. Find the volume of the oblique cone in terms of π. A. 141π m3 B. 8746π m3 C. 112π m3 D. 2915π m3 1 3 V = Bh Volume of a cone 1 3 B = π r2 , r = 20.6 , h =20.6 = π(424.36)(20.6) = 2915π Multiply. Example 2A

  10. B. Find the volume of the cone in terms of π. A. 960π m3 B. 40π m3 C. 320π m3 D. 880π m3 1 3 V = Bh Volume of a cone 1 3 B = π r2 , r = 8 , h =15 = π(64)(15) = 320π Multiply. Example 2B

  11. 1 3 = s2 h Find Real-World Volumes SCULPTUREAt the top of a stone tower is a pyramidion in the shape of a square pyramid. This pyramid has a height of 52.5 centimeters and the base edges are 36 centimeters. What is the volume of the pyramidion? Round to the nearest tenth. Volume of a pyramid B = s2 B = 36 ● 36, h = 52.5 Simplify. Example 3

  12. SCULPTUREIn a botanical garden is a silver pyramidion in the shape of a square pyramid. This pyramid has a height of 65 centimeters and the base edges are 30 centimeters. What is the volume of the pyramidion? Round to the nearest tenth. A. 18,775 cm3 B. 19,500 cm3 C. 20,050 cm3 D. 21,000 cm3 1 3 V = Bh Volume of a pyramid 1 3 B = s2 , s = 30 , h =65 = π(900)(65) = 19500 Multiply. Example 3

  13. Concept

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