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2. r Area = π x radius A =A = π x rx r π = 3.14 [2 d.p.] special number it has its own name x radius x rx r A =A = π x r 2x r 2 How do we find the area of a circle? C =C = 2 x πx π x rx r Compare with

3. © T Madas Why is the area of a circle given by: A = π r 2 ?

4. r r Start with a circle of radius r Divide it into many equal sectors Place the sectors as shown The area of the circle is almost equal to the area of a parallelogram If we had taken more sectors the parallelogram would be almost a rectangle of base: and height: Some quick area calculations:

5. Examples

6. © T Madas A circle has a radius of 1.2 m. Find its area. 3 cm 1.2 m A =A = π x r 2x r 2 c A =A = π x 3 2x 3 2 c A = 28.3 cm 2 [1 d.p.] A =A = π x r 2x r 2 c A =A = π x 1.2 2 c A = 4.5 m2m2 [1 d.p.] A circle has a radius of 3 cm. Find its area.

7. © T Madas 3 m 1 m A =A = π x r 2x r 2 c A =A = π x 1.5 2 c A = 7.1 m2m2 [1 d.p.] A =A = π x r 2x r 2 c A =A = π x 0.5 2 c A = 0.79 m2m2 [2 d.p.] d = 3 m r = 1.5 m d = 1 m r = 0.5 m A circle has a diameter of 1 m. Find its area A circle has a diameter of 3 m. Find its area.

8. © T Madas 50.92 r A circle has an area of 160 cm 2. Find its radius. A =A = π x r 2x r 2 c 160 = π x r 2x r 2 c r = 7.1 cm [1 d.p.] (÷π)(÷π) (÷π)(÷π) 50.92 = r 2r 2 c ( ) 94 ÷ π ≈ 50.92 [2 d.p.] = 7.1 [1 d.p.]

9. © T Madas 0.4775 r A circle has an area of 1.5 m 2. Find its radius. A =A = π x r 2x r 2 c 1.5 = π x r 2x r 2 c r = 0.69 m [2 d.p.] (÷π)(÷π) (÷π)(÷π) 0.4775 = r 2r 2 c ( ) 1.5 ÷ π ≈ 0.4775 [4 d.p.] = 0.69 [2 d.p.]

10. © T Madas Worded Problems

11. © T Madas In the local park a circular region of radius 4.5 metres, is to have a new lawn laid. Turf costs £ 2.20 a square metre. How much would it cost to turf this region? 4.5 m A =A = π x r 2x r 2 c A =A = π x 4.5 2 c A = 63.62 m2m2 [2 d.p.] The area of this circular region: The total cost for the turf: 63.62 x 2.2 ≈ £139.95

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13. A circle and a semicircle, shown below have the same area. If the semicircle has a radius of 8 cm, calculate the radius of the circle, correct to 3 significant figures. 8 cm x A = π r 2r 2 area of a circle A = π x x 2 A = π r 2r 2 area of a circle A = π x 8 2 area of the semicircle A = 1212 π x 8 2 x = 32 π = πx 2= πx 2 x 2x 2 =32x≈5.66cm[3 s.f.]if, then

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15. The figure below shows two circles sharing the same centre O and having radii R and r with R > r. The shaded area is ¾ of the area of the larger circle. Express R in terms of r. R r O π R 2R 2 Area of big circle: π r 2r 2 Area of small circle: π R 2R 2 Shaded Area: π r 2r 2 – π R 2R 2 π r 2r 2 –= 3434 π R 2R 2 = ⇔ π R 2R 2 – 3434 π R 2R 2 π r 2r 2 ⇔ = 1414 π R 2R 2 π r 2r 2 ⇔ = 4 π R 2R 2 π r 2r 2 ⇔ = 4r 24r 2 R 2R 2 ⇔ = 4r 24r 2 R ⇔ = 2r2r R

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17. The figure below shows two circles sharing the same centre O and having radii R and r with R > r. AB is a tangent to the smaller circle and has a length of 8 cm. 1. Write down an expression for the area of the shaded region. 2. Calculate the area of the shaded region, to 3 s.f. R r O π R 2R 2 Area of big circle = π r 2r 2 Area of small circle = π R 2R 2 Shaded Area = π r 2r 2 – A B 4 r 2r 2 +4242 =R 2R 2 ⇔ r 2r 2 –16=R 2R 2 π R 2R 2 =r 2r 2 – π x = ≈ 50.3 cm 2

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