# 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:**

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

∴ ∠RPQ = 90°

Thus, ΔRQP is a right-angled triangle.

Given, PQ = 24 cm, PR = 7 cm

∴ Using Pythagoras theorem,

RQ^{2} = PR^{2 }+ PQ^{2}

RQ = √7² + 24²

= √49 + 576

= √625

Thus, PQ = 25 cm which is the diameter

∴ Radius (r) = 25/2 cm

Area of shaded region = Area of semicircle RPQ - Area of ΔRQP

= 1/2 × πr^{2 }- 1/2 × PQ × RP

= 1/2 [(22/7 × 25/2 × 25/2) - (24 × 7)]

= 1/2 [6875/14 - 168]

= 1/2 [(6875 - 2352)/14]

= 1/2 × 4523/14

= 4523/28 cm^{2}

= 161.54 cm^{2 }(approximately)

**Video Solution:**

## 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.

### NCERT Solutions Class 10 Maths - Chapter 12 Exercise 12.3 Question 1:

**Summary:**

The area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle is 161.54 cm^{2}.