12.2
Warm UpWarm Up
Lesson QuizLesson Quiz
Lesson PresentationLesson Presentation
Surface Area of Prisms and Cylinders
12.2 Warm-Up
1. Evaluate 2xy + 2yz + 2xz for x = 9, y = 6, and z = 4.
2. Evaluate 2πx2 + 2πxy for x = 4 and y = 7. Use 3.14 for π .
ANSWER 228
ANSWER 276.32
12.2 Warm-Up
3. Find the area of a circle with radius 8 cm. Use 3.14 for π.
4. Find the area of a right triangle with hypotenuse 13 in.and a side length of 5 in.
ANSWER 30 in.2
ANSWER 200.96 cm2
12.2 Warm-Up
5. Find the area of a regular pentagon with apothem4.13 ft and side length 6 ft.
ANSWER 61.95 ft2
12.2 Example 1
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.
12.2 Example 1
STEP 2 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 cm2.
12.2 Example 2
SOLUTION
STEP 1 Find the area of a pentagonal base. Use the formula of a regular polygon from page 793.
Find the surface area and lateral area of the right pentagonal prism.
Apothem a = √ 62 –3.5252 ≈ 4.86
Perimeter P = 5(7.05) = 35.25
Area of base B = aP
= (0.5)(4.86)(35.25) 85.66
12.2 Example 2
STEP 2 Use the formula for the surface area of a right prism.
S = 2B + Ph
≈ 2(85.66) + (35.25)(9)
≈ 488.57
The surface area of the right pentagonal prism is about 488.57 square feet. The lateral area is Ph ≈ (35.25)(9) ≈ 317.25 square feet.
12.2 Guided Practice
2. Find the surface area and lateral 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.
ANSWER
a. 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
Lateral area = 2(28) + 2(21) = 98 in.2
b. S = 2B + Ph = 2(3 · 4 ) + 14 · 7 = 122 in.2
Lateral area = Ph = 14 · 7 = 98 in.2
12.2 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?
12.2 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 = 2πr2 + 2πrh Surface area of a cylinder.
= 2π(60)2 + 2π(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.
12.2 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 = 2πr2 + 2πrh Surface area of a cylinder.
157.08 = 2π(2.5)2 + 2π(2.5)h Substitute known values.
157.08 = 12.5π + 5πh Simplify.
157.08 – 12.5π = 5πh Subtract 12.5π from each side.
117.81 ≈ 5πh Simplify. Use a calculator.
7.5 ≈ h Divide each side by 5π.
12.2 Guided Practice
3. Find the surface area and lateral area of a right cylinder with height 18 centimeters and radius 10 centimeters. Round your answer to two decimal places.
1759.29 cm2; 1130.97 cm2ANSWER
4. Find the radius of a right cylinder with height 5 feet and surface area 208π square feet.
8 ft ANSWER
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