question_answer

Two equal circles intersect so that their centres and the points at which they intersect form a square of side 1 cm. The area of the portion that is common to the circles is [SSC CGL Tier II, 2017]

A) \[\frac{\pi }{4}\,sq\,cm\]

B) \[\frac{\pi }{2}-1\,sq\,cm\]

C) \[\frac{\pi }{5}sq\,cm\]

D) \[(\sqrt{2}-1)\ sq\ cm\]

Correct Answer: B

Solution :

[b] In the adjoining figure\[\angle ABO=\angle BMO=90{}^\circ \] and AM = BM and area of square \[AOBO'=1\,sq\,cm\]             \[\therefore \]      \[AM=BM=\frac{1}{2}\] \[\therefore \]Area of common place of circles \[=\left[ \pi {{\left( \frac{1}{2} \right)}^{2}}+\pi {{\left( \frac{1}{2} \right)}^{2}} \right]-1\] \[=\left( \frac{\pi }{4}+\frac{\pi }{4} \right)-1=\frac{2\pi }{4}-1\] \[=\left( \frac{\pi }{2}-1 \right)\ sq\ cm\]