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If $P$ is a point which moves inside an equilateral triangle of side length ‘a’ such that it is nearer to any angular bisector of the triangle than to any of its sides, then the area of the region in which $P$ lies is $K (2 \sqrt{3} - 3)$ sq units for $a = 2$ Then $‘ k ’ =$

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answer is 2.

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Detailed Solution

Shaded area is the region traced by $P$ , its area =

$△ A B C - 3 △ A B D$

$= \frac{\sqrt{3}}{4} a^{2} - \frac{3}{2} a \times \frac{a}{2} \tan 15^{\circ}$

$= \frac{\sqrt{3}}{4} a^{2} (\frac{\sqrt{3} - 1}{\sqrt{3} + 1})$

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