Question

Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius $10cm.$ each.
(use $\pi =3.14$)

Answer 1

Area of $DBC=\frac{\pi }{2}{r}^2=25\pi$

area of $ABD=\frac{\pi }{2}{r}^2=25\pi$

$\therefore$ area of designed region $=50\pi -100$

$=50\times 3.14-100$

$=\frac{314}{2}-100$

$=157-100$

$=57$ ${cm}^2$