Two equal circles intersect so that there centres, and the point at which they intersect from a square of side 1 cm. The area (in sq. cm) of the portion that is common to the circles is

Solution (By Examveda Team)

Now, Area of arc AC₁B $$ = \pi {r^2}.\frac{{90}}{{360}} = \frac{\pi }{4}{\left( 1 \right)^2} = \frac{\pi }{4}$$
And area of arc AC₂B = $$\frac{\pi }{4}$$
Area of square = (side)² = 1
Area of common portion = area of arc (AC₁B + AC₂B) - Area of square
$$\eqalign{ & = \frac{\pi }{4} + \frac{\pi }{4} - 1 \cr & = \frac{\pi }{2} - 1\,{\text{sq}}{\text{. m}} \cr} $$