Question : Three circles of equal radius '$a$' cm touch each other. The area of the shaded region is:
Option 1: $\left ( \frac{\sqrt{3}+\pi}{2} \right )a^{2}$ sq. cm
Option 2: $\left ( \frac{6\sqrt{3}-\pi }{2} \right )a^{2}$ sq. cm
Option 3: $\left ( \sqrt{3}-\pi \right )a^{2}$ sq. cm
Option 4: $\left ( \frac{2\sqrt{3}- \pi}{2} \right )a^{2}$ sq. cm
Correct Answer: $\left ( \frac{2\sqrt{3}- \pi}{2} \right )a^{2}$ sq. cm
Solution : The radius of all circles = $a$ cm Side length of triangle ABC = $2a$ cm Area of equilateral $\triangle ABC$ = $\frac{\sqrt3}{4}$ × side 2 = $\frac{\sqrt3}{4}\times (2a)^2$ = $\sqrt3a^2$ Area of 3 sectors = $3\times \frac{60^\circ}{360^\circ}\times \pi \times a^2$ =$\frac{\pi a^2}{2}$ Area of shaded region = $\sqrt3 a^2-\frac{\pi a^2}{2}$ = $(\frac{2\sqrt3-\pi}{2})a^2$ sq. cm Hence, the correct answer is $(\frac{2\sqrt3-\pi}{2})a^2$ sq. cm.
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