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S URFACE A REA OF P YRAMIDS AND C ONES Section 9.3
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G OAL Find the surface areas of pyramids and cones.
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K EY V OCABULARY Pyramid Height of a pyramid Slant height of a pyramid Cone Height of a cone Slant height of a cone
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A pyramid is a polyhedron in which base, that is a polygon, and lateral faces which are triangles. base vertex P YRAMID
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height base vertex The height of a pyramid is the perpendicular distance from the vertex to its base. H EIGHT OF A P YRAMID
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height base vertex The slant height of a pyramid, represented by l, is the height of any of its lateral faces. S LANT H EIGHT OF A PYRAMID
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Example 1 Find the Slant Height Find the slant height of the Rainforest Pyramid. Round your answer to the nearest whole number. = 10,000 + 10,000 Simplify. Use the Pythagorean Theorem. (slant height) 2 = (height) 2 + 2 1 side 2 Substitute 100 for height and 200 for base side length. = 100 2 + 2 1 · 200 2 To find the slant height, use the right triangle formed by the height and half of the base. SOLUTION
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Example 1 Find the Slant Height ≈ 141.42 Use a calculator. ANSWER The slant height is about 141 feet. Take the positive square root. slant height = 20,000
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S URFACE A REA OF A P YRAMID
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Words: Surface area = (base area) + ½ (base perimeter) ( slant height). Symbols: S. A. = B + ½ P l + l S URFACE A REA OF A P YRAMID
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1.Find the area of the base. 2.Find the area of one of the lateral sides. 3.Multiply the area of the lateral sides by the number of sides. 4.Add the area of the base and the area found in step 3. S URFACE AREA OF A PYRAMID, A SECOND METHOD Area of Base (B) + Lateral Area (LA) S.A. = B + #LA(LA)
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Example 2 Find Surface Area of a Pyramid B = 6 6 = 36 Find the surface area of the pyramid. Find the perimeter of the base.2. P = 6 + 6 + 6 + 6 = 24 Find the slant height. 3. = 4 2 + 3 2 Substitute. Half of 6 is 3. = 16 + 9 Simplify powers. Use the Pythagorean Theorem. (slant height) 2 = (height) 2 + 2 1 side 2 Find the area of the base. 1. SOLUTION
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Example 2 Find Surface Area of a Pyramid = 36 + 2 1 (24) (5) Substitute. = 25 Simplify. Substitute values into the formula for surface area of a pyramid. 4. 2 1 S = B + Pl Write the formula for surface area. Simplify. = 96 ANSWER The surface area of the pyramid is 96 square feet. slant height = = 5 25 Take positive square root.
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Your Turn: ANSWER 175 in. 2 ANSWER about 197.1 cm 2 ANSWER 360 ft 2 Find the surface area of the pyramid. 1. 2. 3.
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C ONE A cone has a circular base and a vertex that is not in the same plane as the base. r vertex base
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vertex r height base The height of a cone is the perpendicular distance between the vertex and the base H EIGHT OF A C ONE
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vertex r base slant height The slant height of a cone is the distance between the vertex and a point on the base edge. S LANT H EIGHT OF A C ONE
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S URFACE A REA OF A C ONE
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Words: Surface Area = (base area) + (base radius) (slant height) Symbols: S.A. = B + r or S.A. = r 2 + r r r + S URFACE A REA OF A C ONE Base Area (B) Lateral Area ( π r ℓ )
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S URFACE A REA OF A C ONE
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Example 3 Find Surface Area of a Cone Find the surface area of the cone to the nearest whole number. b. a. Write the formula for surface area of a cone. S = πr 2 + πrl = 40π Simplify 16 π + 24 π. = π(4) 2 + π(4)(6) Substitute 4 for r and 6 for l. ≈ 126 Multiply. The radius of the base is 4 inches and the slant height is 6 inches. a. SOLUTION
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Example 3 Find Surface Area of a Cone First find the slant height.b. The surface area is about 126 square inches. ANSWER (slant height) 2 = r 2 + h 2 Use the Pythagorean Theorem. = (12) 2 + (5) 2 Substitute 12 for r and 5 for h. = 169 Simplify 144 + 25. = 13 Simplify. Find the positive square root. slant height = 169
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Example 3 Find Surface Area of a Cone Write the formula for surface area. S = πr 2 + πrl = 300π Simplify 144 π + 156 π. = π(12) 2 + π(12)(13) Substitute. The surface area is about 942 square feet. ANSWER ≈ 942 Multiply. Next substitute 12 for r and 13 for l in the formula for surface area.
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Your Turn: ANSWER 691 in. 2 ANSWER 151 ft 2 ANSWER 75 cm 2 Find the surface area of the cone to the nearest whole number. 1. 2. 3.
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A SSIGNMENT Pg. 495 – 499: #1 – 5 all, 7 – 45 odd
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