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Find the area of the shaded region given in Figure.

Answers (1)

Area of square PQRS =(side)2=(14)2

=196 cm2

Area of ABCD (let side a) =(side)2=(a)2

Area of 4 semi-circle \left ( r=\frac{a}{2} \right )=4 \times \frac{1}{2}\pi\left ( \frac{a}{2} \right )^{2}                            

Area of semi-circle=\frac{1}{2}\times \pi \times r^{2}

\\=\frac{2\pi a^{2}}{4}\\ =\frac{\pi a^{2}}{2}

Total inner area = Area of ABCD + Area of 4 semi-circles

 \\=a^{2}+\frac{\pi a^{2}}{2}\\

\\EF=8cm\\ EF=\frac{a}{2}+a+\frac{a}{2}\\ 8=\frac{a+2a+a}{2}\\ 4a=16\\ a=4cm

Area of inner region =4^{2}+\frac{\pi 4^{2}}{2}\\
 \\=16+\frac{ 16 \pi}{2}\\ =16+8 \pi

Area of shaded area = Area of PQRS – inner region area         

\\=196-16-8 \pi\\ =180-8 \times 3.14\\ =180-25.12\\ =154.88 cm^{2}

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