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Lesson 12-5, 6, 13-2

Lesson 12-5, 6, 13-2. Cones & Pyramids. Objectives. Find lateral areas of regular pyramids Find surface areas of regular pyramids Find the volume of pyramids Find lateral areas of cones Find surface areas of cones Find the volume of cones. Vocabulary.

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Lesson 12-5, 6, 13-2

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  1. Lesson 12-5, 6, 13-2 Cones & Pyramids

  2. Objectives • Find lateral areas of regular pyramids • Find surface areas of regular pyramids • Find the volume of pyramids • Find lateral areas of cones • Find surface areas of cones • Find the volume of cones

  3. Vocabulary • Regular Pyramid – a pyramid with a regular polygon for a base and a vertex perpendicular to the base • Slant Height – the height of each lateral face of the pyramid • Circular Cone – tepee shaped cone • Right Cone – a cone with an axis that is also an altitude • Oblique Cone – any other cone (non-right cone)

  4. r l Cones – Surface Area & Volume Cone Net l – slant height h – height l h r Surface Area = Lateral Area + Base(s) Area Cone – A solid with circular base and a vertex. Volume (V) = 1/3 * B * h Base Area (B) = π * r2 V = 1/3 * π * r2 * h LA = π * r * l Base Area = π * r2 SA = LA + BA SA = π * r * l + π * r2

  5. Example 1 Find the surface area and the volume of the cone to the right 5 4 SA = LA + BA LA = π r l and Base Area = π r2 SA = π r l + π r²  need to find r and l 3 SA = π r l + π r² = π (3) (4) + π (3)² = 12π + 9π = 21π ≈ 65.97 V = 1/3 B h = 1/3 πr² h V = 1/3 π(3)² 4 = 1/3(9)(4)π = 12π ≈ 37.70

  6. Example 2 Find the surface area and the volume of the cone to the right l 12 SA = LA + BA LA = π r l and Base Area = π r2 SA = π r l + π r²  need to find r and l 10 r = ½ d = ½ 10 = 5 and use Pythagorean theorem to find ll ² = 12² + 5² l = 13 SA = π r l + π r² = π (5) (13) + π (5)² = 65π + 25π = 90π ≈ 282.74 V = 1/3 B h = 1/3 πr² h V = 1/3 π(5)² 12 = 1/3(25)(12)π = 100π ≈ 314.16

  7. l l h s B s Pyramids – Surface Area & Volume Pyramid (Square) Net Surface Area = Lateral Area + Base(s) Area l – slant height ½ perimeter Volume (V) = 1/3 * B * h Base area (B) = area of the base example above V = 1/3 * s2 * h LA = 4 * ½ s * l Bases Area = s * s = s2 SA = LA + BA SA = 2 * s * l + s2 In general: Pyramid LA = ½ P * l

  8. Example 1 Find the surface area and the volume of the square pyramid to the right 10 8 SA = LA + BA LA = 4·½ s·l and Base Area = s·s = s² SA = 2·s·l + s2 6 s = 2(6) = 12 l = 10 SA = 2·s·l + s2 = 2 (12) (10) + (10)² = 240 + 100 = 340 square units V = 1/3 B h = 1/3 s² h V = 1/3 (6)² 8 = 1/3(36)(8) = 96 cubic units

  9. Example 2 Find the surface area and the volume of the square pyramid to the right l 12 SA = LA + BA LA = 4·½ s·l and Base Area = s·s = s² SA = 2·s·l + s2 5 s = 2(5) = 10 to find l we need to solve l ² = 5² + 12² so l = 13 SA = 2·s·l + s2 = 2 (10) (13) + (10)² = 260 + 100 = 360 square units V = 1/3 B h = 1/3 s² h V = 1/3 (10)² 12 = 1/3(100)(12) = 400 cubic units

  10. Example 3 Find the surface area and the volume of the square pyramid to the right l 12 SA = LA + BA LA = 4·½ s·l and Base Area = s·s = s² SA = 2·s·l + s2 a 18 a = ½ (18) = 9 to find l we need to solve l ² = 9² + 12² so l = 15 SA = 2·s·l + s2 = 2 (18) (15) + (18)² = 540 + 324 = 874 square units V = 1/3 B h = 1/3 s² h V = 1/3 (18)² 12 = 1/3(324)(12) = 1296 cubic units

  11. Example 6-2a Find the surface area of the cone. Round to the nearest tenth. Surface area of a cone Use a calculator. Answer: The surface area is approximately 20.2 sq cm.

  12. Use a calculator. Example 2-2a Find the volume of the cone to the nearest tenth. Volume of a cone r = 5, h = 12 Answer: The volume of the cone is approximately 314.2 in³.

  13. Example 5-2a Find the surface area of the regular pyramid to the nearest tenth. To find the surface area, first find the slant height of the pyramid. The slant height is the hypotenuse of a right triangle with legs that are the altitude and a segment with a length that is one-half the side measure of the base. Pythagorean Theorem Use a calculator.

  14. Now find the surface area of a regular pyramid. The perimeter of the base is and the area of the base is Example 5-2c Surface area of a regular pyramid Use a calculator. Answer: The surface area is 179.4 square meters to the nearest tenth.

  15. s 3, h 7 21 Example 2-1a Teofilo has a solid clock that is in the shape of a square pyramid. The clock has a base of 3 inches and a height of 7 inches. Find the volume of the clock. Volume of a pyramid Multiply. Answer: The volume of the clock is 21 cubic inches.

  16. Summary & Homework • Summary: • The slant height l is the length of an altitude of a lateral face in a pyramid (and the side of a cone) • P is the perimeter of the base of the pyramid • Surface area = Lateral Area + Base Area • Pyramid • Volume: V = ⅓ Bh Surface Area: SA = LA + B=½ Pl + B • Square pyramids are the most common • Cone • Volume: V= ⅓πr² h Surface Area: SA = πrl + πr² = πr(r+l) • Homework: • pg 699-701; 9, 11-13, 15-16, 33

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