Let a point A lie between the parallel lines L 1 a n d L 2 such that its distances from L 1 a n d 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

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Solution

Q.
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: [2026]

1 27

2 $15 \sqrt{6}$

3 $21 \sqrt{3}$

4 $12 \sqrt{2}$

Ans.
(3)

$\sin θ = \frac{3}{a}$

$\sin (60 ° + θ) = \frac{9}{a}$

$\frac{\sqrt{3}}{2} \cos θ + \frac{1}{2} \sin θ = \frac{9}{a}$

$\sqrt{3} \sqrt{1 - \frac{9}{a^{2}}} + \frac{3}{a} = \frac{18}{a}$

$a = \sqrt{84}$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times 84 = 21 \sqrt{3}$

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