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

Answer: Option B

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
$$\begin{array}{rl}& =\frac{\pi }{4}+\frac{\pi }{4}-1\\ & =\frac{\pi }{2}-1\text{sq}\text{. m}\end{array}$$