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

$$\because$$ ROQ is a diameter

$$\therefore ∠ \ RPQ \ = \ 90°$$

In $$∆ \ PRQ$$ ,
$$RQ^2 \ = \ RP^2 \ + \ PQ^2$$

$$\Rightarrow \ RQ^2 \ = \ 7^2 \ + \ 24^2 \$$
$$= \ 49 \ + \ 576 \$$
$$\Rightarrow RQ^2 = \ 625$$

$$\Rightarrow \ RQ \ = \ 25$$ cm

$$\therefore$$ Radius $$r \ = \ \frac{1}{2} RQ \ = \ \frac{25}{2}$$cm

Area of the semi circle $$= \ \frac{1}{2} \pi r^2 \$$
$$= \ \frac{1}{2} \ × \ \frac{22}{7} \ × \ ( \frac{25}{2})^2 \$$
$$= \ \frac{6875}{28} cm^2$$

and area of $$∆ \ RPQ \ = \ \frac{1}{2} \ × \ RP \ × \ PQ$$

$$= \ 12 \ × \ 7 \ × \ 24 \ = \ 84$$ cm2

= Area of the semi circle - Area ( $$∆ \ RPQ$$ )
$$= \ \frac{6875}{28} \ - \ 84 \ = \ \frac{4523}{28}$$ cm2