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M is the centre of the circle. QS = 10√2. PR = RS and PR || QS. Find the approximate area of the shaded region. (Take π = 3.14)

(a) 100 sq. units 

(b) 114 sq. units 

(c) 60 sq. units 

(d) 200 sq. units

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Answer : (c) 60 sq. units 

∠PQS = ∠PRS = 90º (Angle in a semicircle) 

∵ QS || PR, ∴ ∠QPR = ∠QSR = 90º 

∴ PQSR is a rectangle as a pair of opposite sides is parallel and all angles are 90º

⇒ PR = QS and PQ = RS

⇒ PQ = QS = RS = PR = 10√2 cm. Hence PQSR is a square

∴ Diagonal PS of the square = Diameter of the circle

= 10 √2 × √2 = 20 cm.

∴ The radius of the circle = 10 cm. 

Hence, Area of shaded portion = Area of semi-circle – Area of △PQS

= (157 - 100) cm2 = 57 cm2

≈ 60 cm2

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