CHAPTER Surface Area - School District #35

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Name: _____________________________________________________ Date: ______________ 5.4 Surface Area of a Cylinder cylinder • a 3-D object with 2 parallel and congruent circular bases Congruent means the exact same size. Working Example 1: Determine the Surface Area of a Right Cylinder cylinder a) Estimate the surface area of the can. 11 cm Solution 7.6 cm Surface area of can = area of 2 circles + area of 1 rectangle To estimate, use approximate values: d ≈ 8, so r = d ÷ 2 7.6 cm 7.6 cm w = 11 cm ≈ l = circumference π ≈ 3 w ≈ 10 length of rectangle = circumference l = π × d Area of circle = π × r 2 Area of rectangle = l × w C = π × d ≈ 3 × ≈ 3 × × ≈ 2 = circumference × w ≈ (π × d) × w ≈ 3 × 8 × 10 ≈ cm 2 There are 2 circles: 2 × 48 = Estimated surface area ≈ area of 2 circles + area of 1 rectangle ≈ + ≈ The estimated surface area is cm 2 . 256 MHR ● Chapter 5: Surface Area
Name: _____________________________________________________ Date: ______________ b) What is the actual surface area of the can Round your answer to the nearest hundredth of a square centimetre (2 decimal places). Solution Method 1: Use a Net Step 1: Draw the net and label the measurements. Step 2: Find the radius. diameter = 7.6 cm radius = 7.6 ÷ 2 top side bottom 7.6 cm = Step 3: Find the area of 1 circle. A = π × r 2 = 3.14 × 3.8 2 3.14 3.8 3.8 11 cm = Step 4: Find the area of 2 circles. 2 × 45.3416 = Step 5: Find the area of the rectangle using the circumference. A = l × w A = (π × d) × w length of rectangle = circumference A ≈ 3.14 × 7.6 × 11 A ≈ Step 6: Total surface area = area of 2 circles + area of 1 rectangle = + = The total surface area is approximately cm 2 . Round your answer to the nearest hundredth (2 decimal places). 5.4 Surface Area of a Cylinder ● MHR 257
Page 1 and 2: CHAPTER 5 Surface Area GET READY 22
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CHAPTER Surface Area - School District #35

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