Question:
Two identical circles intersect so that their centers and intersection points form a square of side 1 cm. The area in sq. cm of the portion common to both circles is:
Two identical circles intersect so that their centers and intersection points form a square of side 1 cm. The area in sq. cm of the portion common to both circles is:
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For intersecting circles, break overlap into two identical circular segments.
Updated On: Jul 31, 2025
- $\frac{\pi}{4}$
- $\frac{\pi}{2} - 1$
- $\frac{\pi}{5}$
- $\sqrt{2} - 1$
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The Correct Option is B
Solution and Explanation
Radius = $\frac{\sqrt{2}}{2}$ cm from geometry of square. Overlap area = sum of two identical circular segments:
\[
\text{Area} = 2\left[\frac{r^2}{2}(\theta - \sin\theta)\right], \quad \theta = \frac{\pi}{2}, \ r^2=\frac12
\]
This simplifies to $\frac{\pi}{2} - 1$.
\[
\boxed{\frac{\pi}{2} - 1}
\]
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