Question : Two identical circles intersect so that their centres and the points at which they intersect form a square of side $2\;\mathrm{cm}$. What is the area (in$\;\mathrm{cm^2}$) of the portion common to the two circles?
Option 1: $2\pi -4$
Option 2: $4\pi -8$
Option 3: $3\pi -4$
Option 4: $\pi -2$
Correct Answer: $2\pi -4$
Solution :
The length of each side of the formed square is $2\;\mathrm{cm}$.
The length of the radii of the circles = $2\;\mathrm{cm}$
The area of square $=(2\;\mathrm{cm})^2 = 4\;\mathrm{cm^2}$
The area of each circle $=\pi \times 2^2 = 4\pi\;\mathrm{cm^2}$
The area of each sector of each circle $=\frac{90^\circ}{360^\circ} \times 4\pi = \pi\;\mathrm{cm^2}$
The area of the common portion $=2 \times$ The area of each sector of each circle $–$ The area of square
The area of the common portion $=\pi \times 2 - 4 = (2\pi - 4)\;\mathrm{cm^2}$
Hence, the correct answer is $2\pi - 4$.
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