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in Areas Related To Circles by (44.5k points)
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Find the area of the shaded region in the figure, if PQ=24 cm, PR = 7cm and O is the centre of the circle.

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Given: PQ = 24cm and PR = 7cm

Since QR is a diameter, it forms a semicircle

We know that angle in a semicircle is a right angle.

Hence, ∠RPQ = 90°

Hence, ΔRPQ is a right triangle

In ΔRPQ, by Pythagoras theorem

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

(QR)2 = (PQ)2 + (PR)2

⇒ (QR)2 = (24)2 + (7)2

⇒ (QR)2 = 576 + 49

⇒ (QR)2 = 625

⇒ (QR)2 = (25)2

⇒ QR = 25cm

∴ Diameter, QR = 25cm

Radius = 25/2

So,

= 12 × 7

= 84cm2

Area of shaded region = Area of semicircle – Area of ΔPQR

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