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P.O.D. #2. basic. advanced. If a frisbee is 11.25 inches across, what is the circumference of the frisbee ?. If a circle has a radius of 5 ft , what is the circumference of the circle?. C = 2πr C = 2 π 5 ft C ≈ 2 3.14 5 ft C ≈ 31.4 ft. C = πd C = π 11.25 in
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P.O.D. #2 basic advanced If a frisbee is 11.25 inches across, what is the circumference of the frisbee? If a circle has a radius of 5 ft, what is the circumference of the circle? C = 2πr C = 2 π 5 ft C ≈ 2 3.14 5 ft C ≈ 31.4 ft C = πd C = π 11.25 in C ≈ 3.14 11 in C ≈ 35.4 in
The formula for the area of a circle is A = πr2. Example: Find the area of a circle with a radius of 2 inches. A = πr2 A ≈ 3.14 (2 in)2 A ≈ 3.14 4 in2 A ≈ 12.56 in2
A semicircle is half of a circle. The formula for the area of a semicircle is A = ½πr2. Example: 16 Find the area of the semicircle. A = ½πr2 A ≈ ½ 3.14 (8 in)2 A ≈ ½ 3.14 64 in2 A ≈ 100.48 in2
Group Ponder: whiteboards A restaurant is selling one 14” pizza for $10 or two 9” pizzas for for the same price. Which is the better deal? 14” Pizza A = πr2 A = π (7 in)2 A ≈ 3.14 7 in 7 in A ≈ 153.86 in2 9” Pizzas A = πr2 2 A = 2 π (4.5 in)2 2 A = 2 π (4.5 in)2 2 A ≈ 2 3.14 4.5 in 4.5 in 2 A ≈ 127.17 in2 The 14” pizza is the better deal, because the area of the 14” pizza is larger than the area of the two 9” pizzas. This means you would get more pizza if you bought the 14” pizza.
Whiteboard: Brad is putting an 8-foot diameter circular flower garden in his yard. He will put plastic edging along the flowerbed. How many feet of edging will Brad need to enclose his flower garden? How many square feet of land are in Brads garden? C = πd C = 3.14 × 8 C = 25.12 ft A = πr2 A = 3.14 × (4ft)2 A = 3.14 × 16ft2 A = 50.24ft2
Whiteboard: Find the area of the face of the Wisconsin state quarter. A = πr2 24 mm A = 3.14 × 122 A = 3.14 × 144 A = 452mm2
Review: Area of Other Shapes h b h h b b Rectangle Triangle A = ½ b h A = b h
Review: Area of Other Shapes h b h b Rectangle Parallelogram A = b h A = b h
Review: Area of Other Shapes h b Rectangle Trapezoid b1 A = b h A = ½ (b1 + b2) h h b2