Question
The distance between the two parallel lines is 1 unit. A point ' $A$ ' is chosen to lie between the lines at a distance ' $d$ ' from one of them. Triangle $ABC$ is equilateral with $B$ on one line and $C$ on the other parallel line. The length of the side of the equilateral triangle is:
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Solution
Sol. [Key Concept: Equilateral triangle across two parallels $1$ apart: $lsin\ \theta = d,\ lsin(60{^\circ} - \theta) = 1 - d$ ⇒ $l = \frac{2}{\sqrt{3}}\sqrt{d^{2 -}d + 1}$.]
$$\begin{array}{r} lsin\theta = d\#(1) \end{array}$$
$$\begin{matrix} & & l\ sin(60 - \theta) = l\frac{\sqrt{3}}{2}cos\theta - \frac{l}{2}sin\theta \\ & & \ = 1 - d \\ \Rightarrow \ lcos\theta & \ = \frac{2 - d}{\sqrt{3}} & \text{(2)} \end{matrix}$$
Squaring and adding Eqns. (1) and (2)
$${\Rightarrow l^{2} = d^{2} + \frac{4 + d^{2} - 4d}{33} = \frac{4}{3}\left( d^{2} - d + 1 \right) }{\Rightarrow \ l = 2\sqrt{\frac{d^{2} - d + 1}{3}}}$$
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