The area of the shaded region in the given figure, if every circle is of unit radius is

(\sqrt{3} - \frac{π}{2})

sq. units

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(\sqrt{3} + π)

sq. units

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(2 \sqrt{3} - \frac{π}{2})

sq. units

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(\sqrt{3} - \frac{π}{4})

sq. units

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Solution

The correct option is A $(\sqrt{3} - \frac{π}{2})$ sq. units
Join the centres of the circles, such that it forms an equilateral triangle of side 2 cm and cuts a sector of $60^{\circ}$
Required area = Area of equilateral triangle - 3 (Area of sector)
= $\frac{\sqrt{3}}{4} (a)^{2} - 3 (\frac{θ}{360} π (r)^{2})$
= $\frac{\sqrt{3}}{4} (2)^{2} - 3 (\frac{60}{360} π (1)^{2})$
= $\sqrt{3} - \frac{π}{2}$

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Similar questions

x^{2} + y^{2} = 1

| y | = x + 1

y = cos x, x = 0,

x = π

The area (in sq. units) of the region ${(x, y) : y^{2} \geq 2 x a n d x^{2} + y^{2} \leq 4 x, x \geq 0, y \geq 0}$ is :

f (x) = sin x, \forall x \in [0, \frac{π}{2}],

f (x) + f (π - x) = 2, \forall x \in (\frac{π}{2}, π]

f (x) = f (2 π - x), \forall x \in (π, 2 π],

y = f (x)

x

\frac{1}{2}

\sqrt{2}

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