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  • In Fig, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

15. In Fig, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

 

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Consider \Delta ABC,

BC^2\ =\ AC^2\ +\ AB^2

=\ 14^2\ +\ 14^2

BC\ =\ 14\sqrt{2}\ cm                                          

The area of the triangle is :

=\ \frac{1}{2}\times 14\times 14

=\ 98\ cm^2         

Now, the area of the sector is :  

=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 14^2

=\ 154\ cm^2

The area of the semicircle is: - 

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

=\ \frac{1}{2} \pi \times (7\sqrt{2})^2

=\ 154\ cm^2

Hence the area of the shaded region =\ 154\ -\ (154\ -\ 98)\ =\ 98\ cm^2 

Posted by

Devendra Khairwa

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