Question

Question 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.

Answer 1

PQ = 24 cm and PR = 7 cm
$\angle P={90}^{\circ }$ (Angle in the semi-circle)
$\therefore QR$ is hypotenuse of the circle = Diameter of the circle.
By Pythagoras theorem,
$Q{R}^2=P{R}^2+P{Q}^2$
$\Rightarrow Q{R}^2=7^2+{24}^2\Rightarrow Q{R}^2=49+576\Rightarrow Q{R}^2=625\Rightarrow QR=25cm$
$\therefore Radiusofthecircle=\frac{25}{2}cm$
Area of the semicircle $=\frac{\left(\pi R^2\right)}{2}$
$=\left(\frac{\frac{22}{7}\times \frac{25}{2}\times \frac{25}{2}}{2}\right)c{m}^2=\frac{13750}{56}c{m}^2=245.54c{m}^2$
Area of the $\Delta PQR=\frac{1}{2}\times PR\times PQ$
$=\frac{1}{2}\times 7\times 24c{m}^2=84c{m}^2$
Area of the shaded region
= Area of the semicircle - Area of the $\Delta PQR$
$=245.54c{m}^2-84c{m}^2=161.54c{m}^2$