Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm.

Question

Two identical circles intersect so that their centers, and the points at which they intersect, form a square of side 1 cm. The area in sq cm of the portion that is common to the two circles is:

(a) \(\frac{π }{4}\)

(b)  \(\frac{π }{2}\)−1

(c)  \(\frac{π }{5}\)

(d) \(\sqrt{2}\) - 1

Answer 1

Answer : (b) \(\frac{​​\pi}{2} -1\) 

Since each side of the square AOBO′ = 1 cm, so radius OA = O′A = 1 cm

∴ Area of each circle = πr²  = π sq. cm

Area of sector AOB 

= \(\frac{90^o}{360^o}\) × π = \(\frac{π}{4}\) 

⇒ Area of sector AO'B = \(\frac{π}{4}\) 

∴ Common area = Area of sector AOB + Area of sector 

AO'B – Area of square = \(\frac{π}{4}\) + \(\frac{π}{4}\) -1 

= \(\frac{​​\pi}{2} -1\)