Q28 of 37 Page 13

Find the area of the shaded region in the figure, if PQ=24 cm, PR = 7cm and O is the centre of the circle.


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



So,






Now, Area of ΔPQR





= 12 × 7


= 84cm2


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





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