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 \[=\left[ \pi {{\left( \frac{1}{2} \right)}^{2}}+\pi {{\left( \frac{1}{2} \right)}^{2}} \right]-1\] \[=\left( \frac{\pi }{4}+\frac{\pi }{4} \right)-1=\frac{2\pi }{4}-1\] \[=\left( \frac{\pi }{2}-1 \right)\ sq\ cm\] |
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