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