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Question

In the figure m POQ = 30° and radius OP = 12 cm, Find the following (Given π = 3.14)
(i) Area of sector O-PRQ
(ii) Area of OPQ
(iii) Area of segment PRQ

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Solution

(i) Given:
Radius of the circle (OP) = 12 cm
Angle subtended at the centre θ = 30°
We know:
Area of the sector O-PRQ = θ360×πr2

∴ Area of the sector O-PRQ = 30360×3.14×122

= 3.14×12
= 37.68 cm2


(ii) Area of the triangle = 12absinθ

ArOPM = 12×OP×OM×sin30° = 12×12×12×12 =36 cm2

(iii) Area of the segment = Area of sector O-PRQ - Area of OPQ

= 37.68 -36
= 1.68 cm2

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