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P.O.D. #3

P.O.D. #3. basic. advanced. A rotating sprinkler that sprays water at a radius of 11 ft is used to water a lawn. Find the area of the lawn that is watered. (Round to the nearest tenth.). Find the area of a circle with a radius of 12 cm. (Round to the nearest tenth.). A = πr 2

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P.O.D. #3

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  1. P.O.D. #3 basic advanced A rotating sprinkler that sprays water at a radius of 11 ft is used to water a lawn. Find the area of the lawn that is watered. (Round to the nearest tenth.) Find the area of a circle with a radius of 12 cm. (Round to the nearest tenth.) A = πr2 A ≈ 3.14 (12 cm)2 A ≈ 3.14 144 cm2 A ≈ 452.16 cm2 A ≈ 452.2 cm2 A = πr2 A ≈ 3.14 (11 ft)2 A ≈ 3.14 121 ft2 A ≈ 379.94 ft2 A ≈ 379.9 ft2

  2. Area of Composite Figures

  3. A composite figure is made up of two or more shapes. To find the area of a composite figure, break the figure into shapes with areas you know. Then find the sum of these areas.

  4. 2 in 2 in 3 in A = ½πr2 A ≈ ½  3.14 (1 in)2 A ≈ ½  3.14  1 in2 A ≈ 1.57 in2 A = lw A = 2 in  3 in A = 6 in2 A = ½πr2 A ≈ ½  3.14 (1 in)2 A ≈ ½  3.14  1 in2 A ≈ 1.57 in2 Total Area = 1.57 in2+ 6 in2+ 1.57 in2 = 9.14 in2

  5. To find the area of some composite figures, you must subtract the area of one shape from another. Area = minus

  6. 1 in 1 in 3 in 3 in Area of Square A = lw A = 3 in  3 in A = 9 in2 Area of Triangle A = ½ lw A = ½  1 in 1 in A = ½ in2 Area of Blue Figure Area = 9 in2 – ½in2 = 8½ in2

  7. Whiteboard:

  8. Whiteboard: Several Possible Answers

  9. Whiteboard: Mr. DiNardo decided to install a new floor in his living room and hallway. Below is a diagram of the space. Determine how much flooring he will need to purchase. Area of Hallway: A = lw A = 1.5 m  3.2 m A = 4.8 m2 Area of Living Room: A = lw A = 4.2 m  6 m A = 25.2 m2 Total Area = 4.8 m2 + 25.2 m2 = 30 m2

  10. Group Ponder / Whiteboards: Mrs. Sprinkle wants to build a round patio similar to the one shown in the picture. How many stone tiles will she need to purchase if each tile is 1 ft2? 20 ft Area of Outer Circle: A = πr2 A ≈ 3.14 (10 ft)2 A ≈ 3.14 100 ft2 A ≈ 314 ft2 Area of Inner Circle: A = πr2 A ≈ 3.14 (4 ft)2 A ≈ 3.14 16 ft2 A ≈ 50.24 ft2 Total Area of Stone ≈ 314 ft2 – 50.24 ft2 ≈ 263.76 ft2 ≈ 264 ft2 8ft

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