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. Cones – Surface Area & Volume h r l Cone Cone – A solid with circular base and a vertex. l – slant height h – height Volume (V) = 1/3 * B * h Base Area (B) = π * r 2 V = 1/3 * π * r 2 * h Surface Area = Lateral Area + Base(s) Area LA = π * r * l Base Area = π * r 2 SA = LA + BA SA = π * r * l + π * r 2 l r Net

5. Example 1 4 3 5 Find the surface area and the volume of the cone to the right SA = LA + BA LA = π r l and Base Area = π r 2 SA = π r l + π r²  need to find r and l 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 12 10 l Find the surface area and the volume of the cone to the right SA = LA + BA LA = π r l and Base Area = π r 2 SA = π r l + π r²  need to find r and l r = ½ d = ½ 10 = 5 and use Pythagorean theorem to find l l 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. Pyramids – Surface Area & Volume Volume (V) = 1/3 * B * h Base area (B) = area of the base example above V = 1/3 * s 2 * h Pyramid (Square) h B l l – slant height Surface Area = Lateral Area + Base(s) Area LA = 4 * ½ s * l Bases Area = s * s = s 2 SA = LA + BA SA = 2 * s * l + s 2 In general: Pyramid LA = ½ P * l s s l ½ perimeter Net

8. Example 1 Find the surface area and the volume of the square pyramid to the right s = 2(6) = 12 l = 10 SA = 2·s· l + s 2 = 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 8 6 10 SA = LA + BA LA = 4·½ s· l and Base Area = s·s = s² SA = 2·s· l + s 2

9. Example 2 Find the surface area and the volume of the square pyramid to the right s = 2(5) = 10 to find l we need to solve l ² = 5² + 12² so l = 13 SA = 2·s· l + s 2 = 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 12 5 l SA = LA + BA LA = 4·½ s· l and Base Area = s·s = s² SA = 2·s· l + s 2

10. Example 3 Find the surface area and the volume of the square pyramid to the right a = ½ (18) = 9 to find l we need to solve l ² = 9² + 12² so l = 15 SA = 2·s· l + s 2 = 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 12 18 l SA = LA + BA LA = 4·½ s· l and Base Area = s·s = s² SA = 2·s· l + s 2 a

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

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

15. 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 Answer:The volume of the clock is 21 cubic inches. Multiply. 21 s 3, h 7

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