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  • Find the area of the shaded region in Fig, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

1. Find the area of the shaded region in Fig, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

                    

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We know that \angleRPQ  is 90o as ROQ is the diameter.

RQ can be found using Pythagoras theorem.

                                             RP^2\ +\ PQ^2\ =\ QR^2

or                                          7^2\ +\ 24^2\ =\ QR^2

or                                          QR\ =\ \sqrt{625}\ =\ 25\ cm

Now, the area of the shaded region is given by : = Area of semicircle   -   Area of \DeltaPQR

Area of semicircle is :- 

                                            =\ \frac{1}{2}\pi r^2

or                                        =\ \frac{1}{2}\pi \times \left ( \frac{25}{2} \right )^2

or                                        =\ 245.53\ cm^2 

And, the area of triangle PQR is :     

                                                     =\ \frac{1}{2}\times 24\times 7\ =\ 84\ cm^2

Hence the area of the shaded region is   :      =\ 245.53\ -\ 84\ =\ 161.53\ cm^2

Posted by

Devendra Khairwa

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