L12.2 New

Geometry:

Geometry Chapter 12 Lesson 2

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EXAMPLE 1 Use the net of a prism Find the surface area of a rectangular prism with height 2 centimeters, length 5 centimeters , and width 6 centimeters . SOLUTION STEP 1 Sketch the prism. Imagine unfolding it to make a net.

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EXAMPLE 1 STEP 2 Use the net of a prism Find the areas of the rectangles that form the faces of the prism. STEP 3 Add the areas of all the faces to find the surface area. The surface area of the prism is S = 2(12) + 2(10) + 2(30) = 104 cm 2 .

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EXAMPLE 2 Find the surface area of a right prism SOLUTION STEP 1 Find the perimeter and area of a base of the prism. Perimeter P = 5(7.05) Apothem a = √ 6 2 –3.525 2 EXAMPLE 2 Find the surface area of the right pentagonal prism. Find the surface area of a right prism Each base is a regular pentagon. ≈ 4.86 = 35.25

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EXAMPLE 2 STEP 2 Use the formula for the surface area that uses the apothem. S = aP + Ph Surface area of a right prism ≈ (4.86)(35.25) + (35.25)(9) Substitute known values. ≈ 488.57 Simplify. Find the surface area of a right prism The surface area of the right pentagonal prism is about 488.57 square feet . ANSWER

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GUIDED PRACTICE for Examples 1 and 2 1. Draw a net of a triangular prism. SOLUTION

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2. Find the surface area of a right rectangular prism with height 7 inches , length 3 inches, and width 4 inches using (a) a net and (b) the formula for the surface area of a right prism. for Examples 1 and 2 GUIDED PRACTICE ANSWER (a) NET: Left and right faces: 7 4 = 28 in. 2 Top and bottom faces: 3 4 = 12 in. 2 Front and back faces: 3 7 = 21 in. 2 S = 2(28) + 2(12) + 2(21) = 122 in.2

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2. Find the surface area of a right rectangular prism with height 7 inches , length 3 inches, and width 4 inches using (a) a net and (b) the formula for the surface area of a right prism. for Examples 1 and 2 GUIDED PRACTICE ANSWER (b) S = 2B + Ph = 122 in. 2 = 2(3 4) + 14 7

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EXAMPLE 3 COMPACT DISCS You are wrapping a stack of 20 compact discs using a shrink wrap. Each disc is cylindrical with height 1.2 millimeters and radius 60 millimeters . What is the minimum amount of shrink wrap needed to cover the stack of 20 discs? Find the height of a cylinder

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EXAMPLE 3 SOLUTION The 20 discs are stacked, so the height of the stack will be 20(1.2) = 24 mm . The radius is 60 millimeters . The minimum amount of shrink wrap needed will be equal to the surface area of the stack of discs. S = 2 πr 2 + 2 πrh Surface area of a cylinder. = 2 π (60) 2 + 2 π (60)(24) Substitute known values. ≈ 31,667 Use a calculator. You will need at least 31,667 square millimeters, or about 317 square centimeters of shrink wrap. ANSWER Find the height of a cylinder

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EXAMPLE 4 SOLUTION Substitute known values in the formula for the surface area of a right cylinder and solve for the height h . Find the height of the right cylinder shown, which has a surface area of 157.08 square meters . S = 2π r 2 + 2π rh Surface area of a cylinder. Find the height of a cylinder

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EXAMPLE 4 157.08 = 2 π (2.5) 2 + 2 π (2.5) h Substitute known values. 157.08 = 12.5 π + 5 πh Simplify. 157.08 – 12.5 π = 5 πh Subtract 12.5 π from each side. 117.81 ≈ 5π h Simplify. Use a calculator. 7.5 ≈ h Divide each side by 5 π . Find the height of a cylinder The height of the cylinder is about 7.5 meters . ANSWER

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GUIDED PRACTICE for Examples 3 and 4 3. Find the surface area of a right cylinder with height 18 centimeters and radius 10 centimeters . Round your answer to two decimal places. S = 2 πr 2 + 2 πrh Surface area of a cylinder. = 2 π (60) 2 + 2 π (10)18 Substitute known values. = 1759.29 cm 2 Use a calculator. SOLUTION

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GUIDED PRACTICE for Examples 3 and 4 4. Find the radius of a right cylinder with height 5 feet and surface area 208 π square feet . S = 2 πr 2 + 2 πrh Surface area of a cylinder. 208 π =2 π ( r ) 2 + 2 πr (5) Substitute known value. 208 π = 2 πr 2 + 10 πr 104 = r 2 +5 r Divide 2 π from each side . Simplify. SOLUTION

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GUIDED PRACTICE for Examples 3 and 4 Simplify. Use a calculator. r = 8 The radius of cylinder is 8 feet . ANSWER