Question ID: #643
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:
- (1) $15\sqrt{6}$
- (2) 27
- (3) $21\sqrt{3}$
- (4) $12\sqrt{2}$
Solution:
Let the side length of the equilateral triangle $ABC$ be $a$.
Let the angle made by side $AC$ with the line $L_2$ be $\theta$.
Since the triangle is equilateral, $\angle A = 60^\circ$.
Given the distance of $A$ from $L_2$ is 3, we have the projection of $AC$ perpendicular to $L_2$:
$$a \sin \theta = 3 \Rightarrow \sin \theta = \frac{3}{a}$$
And
$$a \sin(60^\circ + \theta) = 9$$
Expanding using the sine addition formula:
$$a (\sin 60^\circ \cos \theta + \cos 60^\circ \sin \theta) = 9$$
$$a \left( \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \right) = 9$$
From $\sin \theta = \frac{3}{a}$, we find $\cos \theta = \sqrt{1 – \frac{9}{a^2}}$.
Substituting these values:
$$a \left( \frac{\sqrt{3}}{2} \sqrt{1 – \frac{9}{a^2}} + \frac{1}{2} \cdot \frac{3}{a} \right) = 9$$
$$\frac{\sqrt{3}}{2} \sqrt{a^2 – 9} + \frac{3}{2} = 9$$
$$\sqrt{3} \sqrt{a^2 – 9} = 18 – 3 = 15$$
$$\sqrt{a^2 – 9} = \frac{15}{\sqrt{3}} = 5\sqrt{3}$$
Squaring both sides:
$$a^2 – 9 = 75 \Rightarrow a^2 = 84$$
Area of equilateral triangle $ABC$:
$$\text{Area} = \frac{\sqrt{3}}{4} a^2$$
$$\text{Area} = \frac{\sqrt{3}}{4} (84) = 21\sqrt{3}$$
Ans. (3)
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