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Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side $1$ cm. The area in sq. cm of the portion that is common to the two circles is

  1. $\pi / 4$
  2. $\frac{\pi}{2} – 1$
  3. $\frac{\pi}{5}$
  4. $\sqrt{2} – 1$
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We have to find area of red portion.

Since side of square = 1, So radius of both the circles = 1.

Since ABCD is an square, so angle at centers of circle = 90.

Area of sector ACD = ($\pi$r$^{2}$)/4.

So area of red portion = 2 * area of sector ACD( equal to sector BCD) - area of square.

                                         = 2 * $\pi$/4 - 1

                                         = $\pi$/2 - 1.

Ans- B. 

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