Let a point A lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}

Question

Let a point A lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines $L_{1}$ and $L_{2}$ respectively, is :

Accepted Answer

Sol.$${sin	heta = \frac{3}{a} }{sin\left( 60^{\circ} + 	heta ight) = \frac{9}{a} }{\frac{\sqrt{3}}{2}cos	heta + \frac{1}{2}sin	heta = \frac{9}{a} }{\sqrt{3}\sqrt{1 - \frac{9}{a^{2}}} + \frac{3}{a} = \frac{18}{a} }{a = \sqrt{84} }$$Area of $\bigtriangleup ABC = \frac{\sqrt{3}}{4}a^{2} = \frac{\sqrt{3}}{4} 	imes 84 = 21\sqrt{3}$

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