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

## Presentation on theme: "Find the surface area of each. S = (Pℓ)/2 + B = (20×3)(15)/2 + 20√(10 2 +20 2 )/2 = 623.2 in 2 S = (Pℓ)/2 + B = (10×6)(14)/2 + (8.7)(10×6)/2 = 679.8 ft."— Presentation transcript:

Find the surface area of each. S = (Pℓ)/2 + B = (20×3)(15)/2 + 20√(10 2 +20 2 )/2 = 623.2 in 2 S = (Pℓ)/2 + B = (10×6)(14)/2 + (8.7)(10×6)/2 = 679.8 ft 2 8.7 S = (Pℓ)/2 + B = 2π(8)(√(15 2 +8 2 )/2 + π(8 2 ) = 200π ft 2 S = (Pℓ)/2 + B = 2π(√(8 2 -6 2 )(8)/2 + π(6.9 2 ) = 70.3π m 2

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

Concept Theorem 12-11

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

Example 1 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. A.416 ft 3 B. C. D. Volume of a pyramid B = s 2, s = 8, h = 6.5 Multiply. = (64)(6.5) = 138.7 V = Bh 1313 1313

Concept Theorem 12-12

Example 2A Volume of a Cone A. Find the volume of the oblique cone in terms of π. Simplify B = π r 2 r = 9.1, h = 25 = 690π Volume of a cone V = Bh 1313

Example 2B Volume of a Cone B. Find the volume of the cone in terms of π. Simplify. Volume of a cone r = 5, h = 12 = 100π B = π r 2 V = Bh 1313

Example 2A A.141π m 3 B.8746π m 3 C.112π m 3 D.2915π m 3 A. Find the volume of the oblique cone in terms of π. Volume of a cone B = π r 2, r = 20.6, h = 20.6 Multiply. = π(424.36)(20.6) = 2915π V = Bh 1313 1313

Example 2B A.960π m 3 B.40π m 3 C.320π m 3 D.880π m 3 B. Find the volume of the cone in terms of π. Volume of a cone B = π r 2, r = 8, h = 15 Multiply. = π(64)(15) = 320π V = Bh 1313 1313

Example 3 Find Real-World Volumes SCULPTURE At 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 = 36 ● 36, h = 52.5 Simplify. = s 2 h 1313 B = s 2

Example 3 A.18,775 cm 3 B.19,500 cm 3 C.20,050 cm 3 D.21,000 cm 3 SCULPTURE In 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. Volume of a pyramid B = s 2, s = 30, h = 65 Multiply. = π(900)(65) = 19500 V = Bh 1313 1313

Concept

Download ppt "Find the surface area of each. S = (Pℓ)/2 + B = (20×3)(15)/2 + 20√(10 2 +20 2 )/2 = 623.2 in 2 S = (Pℓ)/2 + B = (10×6)(14)/2 + (8.7)(10×6)/2 = 679.8 ft."

Similar presentations