University:
Felician University
Course:
MATH-160 | College Algebra
Academic year:

2019
Views:

423

Pages:

3
Author:

SpaceCadet101

**Practice B** **Surface Area of Prisms and Cylinders** Find the lateral area and surface area of each right prism. Round to the nearest tenth if necessary. 1. [Image of a rectangular prism with dimensions 12m x 10m x 4m] Lateral Area (L) = ph = 42(4) = 168 m² Surface Area (S) = 168 + 2(12)(10) = 408 m² 2. [Image of a triangular prism with base edge 1.37 m and height 4 m] Lateral Area (L) = ph = 7(4) = 28 m² Surface Area (S) = 28 + 2(1/2)(1.37)(4) = 33.8 m² 3. A cube with edge length 20 inches Lateral Area (L) = 4(20)(20) = 1600 in² Surface Area (S) = 6(20)² = 2400 in² Find the lateral area and surface area of each right cylinder. Give your answers in terms of π. 4. A cylinder with base radius 5 cm and height twice the radius Radius (r) = 5 cm Height (h) = 2(5) = 10 cm Lateral Area (L) = 2πrh = 2π(5)(10) = 100π cm² Surface Area (S) = 2πr² + 2πrh = 2π(5)² + 100π = 150π cm² 5. A cylinder with base circumference 8π m and a height one-fourth the radius Circumference = 2πr = 8π r = 4 m Height (h) = 1/4(4) = 1 m Lateral Area (L) = 2πrh = 2π(4)(1) = 8π m² Surface Area (S) = 2πr² + 2πrh = 2π(4)² + 8π = 40π m² Find the surface area of each composite figure. Round to the nearest tenth. 7. [Image of a composite figure with a rectangular prism and a triangular prism] See Attached 8. [Image of a composite figure with a cylinder and a cone] See Attached Describe the effect of each change on the surface area of the given figure. 9. The dimensions are multiplied by 12. SA × 12² or 144 10. The dimensions are divided by 4. SA ÷ 4² or 16 Toby has eight cubes with edge length 1 inch. He can stack the cubes into three different rectangular prisms: 2-by-2-by-2, 8-by-1-by-1, and 2-by-4-by-1. Each prism has a volume of 8 cubic inches. 11. Tell which prism has the smallest surface-area-to-volume ratio. 2x2x2 12. Tell which prism has the greatest surface-area-to-volume ratio. 8x1x1 6. C = 8πr = 2πr r = 4 h = ¼r + 1 = 2 L = 2πrh = 2π(4)(2) = 16π m² S = 2πr² + 2πrh = 2π(4)² + 16π = 40π m² 7. S = S(side) + ½S(cyl) - ½(area of missing side) S = rectangular prism + ½ cylinder - ½ cylinder of base P = 2(6) + 2(2) = 16 S₁ = 16(6) + 2(2)(6) = 120 km² Area of missing side = ½(2πr)(6) = 6π km² S = 120 - 12 + 6π - π = 123.7 km² 8. S = S(rect) + S(cyl) - 2(area of base) P = 2(2) + 2(4) = 12 S₁ = 12(5) + 2(2)(4) = 76 in² S(cyl) = 2π(3/2)(4) + 2π(3/2)² = 12π + 9π/2 = 33π/2 in² S = 76 + 33π/2 - 2(9π/2) = 76 + 12π = 113.7 in²