Find the area of the shaded region, if $PQ=24 cm, PR=7 cm$, and O is the center 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° 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 $= \frac{1}{2} πr^2 - \frac{1}{2} \times b \times h$

                                              $= \frac{1}{2}( πr^2- b \times h)$

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

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

                                           $=\frac{1}{2}(491-168)$  

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

                                          $ = \frac{323}{2} = 161.53 cm^2$.

Therefore, the area of the shaded region is $161.53 cm^2$.

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Updated on: 10-Oct-2022

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