Find the area of the shaded region in the figure if $PQ = 24 \, cm$,$PR = 7 \, cm$ and $O$ is the centre of the circle. [Use $\pi = \frac{22}{7}$]

(N/A) $1$. Since $RQ$ is a diameter of the circle,the angle in the semicircle is a right angle. Therefore,$\angle QPR = 90^{\circ}$.
$2$. In the right-angled triangle $\triangle QPR$,by Pythagoras theorem:
$RQ^2 = PQ^2 + PR^2$
$RQ^2 = 24^2 + 7^2 = 576 + 49 = 625$
$RQ = \sqrt{625} = 25 \, cm$.
$3$. The radius of the circle $r = \frac{RQ}{2} = \frac{25}{2} = 12.5 \, cm$.
$4$. The area of the semicircle is $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (12.5)^2 = \frac{11}{7} \times 156.25 = \frac{1718.75}{7} \approx 245.536 \, cm^2$.
$5$. The area of $\triangle QPR = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 24 = 84 \, cm^2$.
$6$. The area of the shaded region = Area of semicircle - Area of $\triangle QPR = \frac{1718.75}{7} - 84 = \frac{1718.75 - 588}{7} = \frac{1130.75}{7} \approx 161.54 \, cm^2$.