Question

In the given figure, PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Find the area of the shaded region. [Take $\pi =3.14$]

Answer 1

For the given diagram..

$\angle QPR={90}^{\circ }$ because it is half the reflex of $\angle QOR$ which is ${180}^{\circ }$

By pythagoras theroem....
$Q{R}^2=Q{P}^2+R{P}^2$
= ${24}^2+7^2$
= 576 + 49 = 625
QR = 25cm
$\to radiusofthecircle=\frac{25}{2}$
area of the shaded region = area of semicircle - area of triangle PQR

= $\pi \frac{r^2}{2}-(PR\times \frac{PQ}{2})$

=$(22\times 25\times \frac{25}{2}\times 2\times 2\times 7)-(7\times \frac{24}{2})$
= $(11\times 25\times \frac{25}{2}\times 2\times 7)-(7\times 12)$
= $\frac{6875}{84}-84$
= $\frac{4523}{28}c{m}^2$
=$161.54c{m}^2$