1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Solution: Here, P is in the semi-circle and so, P = 90° So, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle. ∴ QR = D Using Pythagorean theorem, QR 2 = PR 2 +PQ 2 Or, QR 2 = 7 2 +24 2