Question : Two equal circles intersect so that their centres and the points at which they intersect form a square of side 1 cm. The area (in cm2) of the portion that is common to the circle is:
Option 1: $\frac{\pi }{4}$
Option 2: $\frac{\pi }{2}-1$
Option 3: $\frac{\pi }{5}$
Option 4: $(\sqrt{2}-1)$
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Correct Answer: $\frac{\pi }{2}-1$
Solution :
Given: The side of the square = 1 cm
The area of the square = $\operatorname{(side)^2}$
$A_{\text{square}}= 1$ cm
2
The radius of each circle = 1 cm
The angle of the sector = $\frac{\pi}{2}$
The area of the sector = $\frac{\theta}{360^{\circ}}×\pi r^2$
$ A_{\text{sector}} = \frac{90^{\circ}}{360^{\circ}}×\pi (1)^2 = \frac{\pi}{4} $ cm
2
The area of the portion that is common to the two circles,
$2A_{\text{sector}} - A_{\text{square}} = \frac{2\pi}{4} - 1 =\frac{\pi}{2} - 1$.
Hence, the correct answer is $\frac{\pi}{2} - 1$.
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