Find 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.
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Given :

In the given figure, $PQ=24 cm, PR=7 cm$ and $O$ is the center of the circle.

To do :

We have to find the area of the shaded region.

Solution :

Area of the shaded region $=$ Area of the semi-circle $-$ Area of $\triangle PQR$

In $\triangle PQR$,

$\angle QPR = 90°$                [Diameter subtends $90^o$ on any point on the circle]

Therefore,

$QR^2 = PQ^2+PR^2$

$QR^2 = 24^2 + 7^2$

$QR^2 = 576+49$

$QR^2 = 625$

$QR = 25 cm$

Diameter $= 25 cm$

Radius $r= \frac{25}{2} cm$

Base of triangle (b)$= 7 cm$, height (h)$= 24cm$.

Area of the shaded region $= \frac{1}{2} \pi r^2 - \frac{1}{2} \times b \times h$

$= \frac{1}{2}(\pi r^2- b \times h)$

$= \frac{1}{2}( \frac{22}{7}\times \frac{25}{2}\times \frac{25}{2}- 7 \times 24)$

$ = \frac{1}{2}(\frac{13750}{28} - 168)$

$=\frac{1}{2}(491-168)$ 

$= \frac{1}{2}(323)$

$ = \frac{323}{2}$

$= 161.5\ cm^2$.

Therefore, the area of the shaded region is $161.5\ cm^2$.

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Updated on: 10-Oct-2022

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