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A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14, √3 = 1.73)

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Given: Radius (r) =15 cm, central angle (θ) = 60° 

To find: Areas of major and minor segments. 

Let chord PQ subtend ∠POQ = 60° at centre. 

∴ θ = 60°

= 225 [0.0908] 

= 20.43 cm2 

∴ area of minor segment = 20.43 cm2 

Area of circle = πr2 

= 3.14 × 15 × 15 

= 3.14 × 225

= 706.5 cm2 

Area of major segment = Area of circle – area of minor segment

= 706.5 – 20.43 

= 686.07 cm2 

Area of major segment 686.07 cm2 

∴ The area of minor segment Is 20.43 cm2 and the area of major segment is 686.07 cm2 .

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