If ABCD is a square of unit side, 4 – circles of unit radius are described with centres at A, B, C, D then area common to 4 – circles is

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$1 + \frac{π}{3} - \sqrt{3}$

$1 - \frac{π}{3} + \sqrt{3}$

$1 - \frac{π}{4} + \sqrt{3}$

$1 + \frac{π}{4} - \frac{\sqrt{3}}{2}$

answer is C.

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Detailed Solution

Required are = $4 \times \int_{\frac{1}{2}}^{\sqrt{3} / 2} (\sqrt{1 - x^{2}} - \frac{1}{2}) d x$

$= 1 + \frac{π}{3} - \sqrt{3} Sq.units$ Question Image

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