10-3 Surface Area of Cylinders and Prisms #1: Find the lateral area of the cylinder to the nearest tenth with d = 10 cm and h = 2 cm. L = 2π rh d 10 = =5 2 2 L = 2π ( 5 ) ( 2 ) ≈ 62.8 cm 2 r= Find (a) the lateral area and (b) the surface area of the prism. Round to the nearest whole number. #7: L = Ph Since the base is a right triangle with legs 8 and 15, we can find the third side using Pythagorean theorem: 8 2 + 15 2 = x 2 #11: L = Ph Since the base is a right triangle with legs 15 and 20, we can find the third side using Pythagorean theorem: 15 2 + 20 2 = x 2 64 + 225 = x 2 225 + 400 = x 2 289 = x 2 625 = x 2 x 2 = 289 = 17 P = 8 + 15 + 17 = 40 m x 2 = 625 = 25 P = 15 + 20 + 25 = 60 m L = 40 ( 8 ) = 320 m 2 L = 60 ( 25 ) = 1500 m 2 SA = 2B + L 1 1 B = bh = ( 8 ) (15 ) = 60 m 2 2 2 SA = 2 ( 60 ) + 320 = 120 + 320 = 440 m 2 SA = 2B + L 1 1 B = bh = (15 ) ( 20 ) = 150 m 2 2 2 SA = 2 (150 ) + 1500 = 300 + 1500 = 1800 m 2 Find the surface area of the cylinder in terms of π. #13: r = 1 m; h = 3 m SA = 2π r 2 + 2π rh SA = 2π (1) + 2π (1) ( 3) 2 2π + 6π 8π m 2 Homework: Finish the Practice 10-3 Worksheet. 10-4 Surface Area of Cones and Pyramids #1: Find the lateral area of the cone with h = 10 m and r = 2 m to the nearest whole number. L = π rl l is the slant height, which we can find using Pythagorean theorem: 2 2 + 10 2 = l 2 #4: Find the surface area of the cone with h = 3 cm and r = 4 cm in terms of π. 4 + 100 = l l = 25 = 5 104 = l 2 2 SA = π rl + π r 2 32 + 4 2 = l 2 9 + 16 = l 2 25 = l 2 SA = π ( 4 ) ( 5 ) + π ( 4 ) l = 104 L = π ( 2 ) 104 ≈ 64 m 2 = 20π + 16π 2 = 36π cm 2 #7: Find the lateral area of the regular pyramid to the nearest tenth. 1 L = Pl 2 The base is a square with sides of 7m, #10: Find the surface area of the regular pyramid to the nearest tenth. 1 SA = B + Pl 2 The base is a square with sides of 5 m, so P = 4 ( 7 ) = 28 m. so P = 4 ( 5 ) = 20 m and B = 5 ( 5 ) = 25 m 2 . h = 7 and half the base = 3.5 h = 12 and half the base = 2.5 l = 7 + 3.5 = 49 + 12.25 = 61.25 l 2 = 12 2 + 2.5 2 = 144 + 6.25 = 150.25 l = 61.25 1 L = ( 28 ) 61.25 ≈ 109.6 m 2 2 l = 150.25 1 SA = 25 + ( 20 ) 150.25 ≈ 147.6 m 2 2 2 2 2 Homework: Finish the Practice 10-4 Worksheet. 10-5 Volume of Cylinders and Prisms #1: Find the volume of a cylinder with diameter 12 m and height 10 m to the nearest tenth. V = Bh = π r 2 h 12 r= =6m 2 2 V = π ( 6 ) (10 ) ≈ 1131.0 m 2 Find the volume of the prism to the nearest whole number. #7: length = 8, width = 5, height = 3 V = lwh = 8 ( 5 ) ( 3) = 120 in 3 #8: V = Bh The base is an equilateral triangle with side length 3 ft. s 2 3 32 3 9 3 = = 4 4 4 ⎛ 9 3⎞ 3 V =⎜ ⎟ ( 9 ) ≈ 35 ft 4 ⎝ ⎠ B= Find the volume of the composite figure to the nearest whole number. #13: This solid is made of two rectangular prisms. The bottom one’s dimensions are 6x2x3, and the top’s dimensions are 4x2x5. (You get 5 by doing 8 – 3.) Vbottom = 6 ( 2 ) ( 3) = 36 ft 3 Vtop = 4 ( 2 ) ( 5 ) = 40 ft 3 Vtotal = 36 + 40 = 76 ft 3 Homework: Finish the Practice 10-5 Worksheet. 10-6 Volume of Pyramids and Cones Find the volume of the pyramid. #1: Base sides = 54 cm, l = 45 cm. To find height, we need to use Pythagorean Thm: 54 ÷ 2 = 27 27 2 + h 2 = 45 2 729 + h 2 = 2025 #4: B = 400 yd2 and h = 36 yd 1 V = Bh 3 1 V = ( 400 ) ( 36 ) = 4800 yd 3 3 h 2 = 2025 − 729 = 1296 h = 1296 = 36 cm 1 V = Bh 3 B = 54 2 = 2916 cm 2 1 V = ( 2916 ) ( 36 ) = 34992 cm 3 3 Find the volume of the cone. Round to the nearest tenth. #7: d = 10 cm, h = 24 cm 10 r= = 5 cm 2 1 V = π r 2h 3 1 2 V = π ( 5 ) ( 24 ) ≈ 628.3 cm 3 3 Find the variable. 1 #13: V = Bh 3 1 1500 = (15 ) (15 ) x 3 1500 = 75x x= 1500 = 20 units 75 Homework: Finish the Practice 10-6 Worksheet. #9: d = 26 m, h = 28 m 26 r= = 13 cm 2 1 V = π r 2h 3 1 2 V = π (13) ( 28 ) ≈ 4955.3 m 3 3 10-7 Surface Area and Volume of Spheres #1: Find the surface area of a sphere with radius 14 in. to the nearest tenth. SA = 4π r 2 SA = 4π (14 ) ≈ 2463.0 in 2 2 #7: Find the volume of a sphere with a diameter of 14 mi to the nearest tenth. 4 3 πr 3 14 r= = 7 mi 2 4 3 V = π ( 7 ) ≈ 1436.8 mi3 3 V= #11: Find the volume of a sphere with a surface area of 90,790 cm2 to the nearest tenth. SA = 4π r 2 90790 = 4π r 2 90790 r2 = 4π 90790 r= ≈ 84.99905 4π 3 4 ⎛ 90790 ⎞ V = π⎜ ≈ 2, 572, 354.6 cm 3 3 ⎝ 4π ⎟⎠ #13: Find the surface area of a sphere with V = 1200 ft3. Round to the nearest whole number. 4 1200 = π r 3 3 1200 r3 = 4 π 3 1200 r= ≈ 6.592208 3 4 π 3 SA = 4π ( 6.592208 ) ≈ 546 ft 2 2 Homework: Finish the Practice 10-7 Worksheet. #16: Find the volume of a baseball with C = 24 cm. C = 2π r 24 = 2π r 24 12 r= = 2π π 3 4 ⎛ 12 ⎞ V = π ⎜ ⎟ ≈ 233 cm 3 3 ⎝π⎠
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