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16. Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

                                     

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It is clear from the figure that the required area (designed area) is the area of the intersection of two sectors.

Area of the sector is:-  

                                         =\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 8^2

                                      =\ \frac{352}{7}\ cm^2

And, area of the triangle:- 

                                       =\ \frac{1}{2}\times 8\times 8\ =\ 32\ cm^2

Hence the area of the designed region is : 

                                        =\ 2\left ( \frac{352}{7}\ -\ 32 \right )

                                        =\ \frac{256}{7}\ cm^2 

Posted by

Devendra Khairwa

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