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Q.

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR)² is equal to ________.

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answer is 28.

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Detailed Solution

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PR = cosec $θ$ , PQ = 4sec(30 + $θ$ )
For equilateral
d = PR = PQ
$\begin{array}{l} \Rightarrow \cos (θ + 30^{\circ}) = 4 \sin θ \\ \Rightarrow \frac{\sqrt{3}}{2} \cos θ - \frac{1}{2} \sin θ = 4 \sin θ \\ \Rightarrow \tan θ = \frac{1}{3 \sqrt{3}} \\ {QR}^{2} = d^{2} = {cosec}^{2} θ = 28 \end{array}$

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Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR)2 is equal to ________.