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 $P Q R$ is formed such that $Q$ lies on one of the parallel lines, while $R$ lies on the other. Then $(Q R)^2$ is equal to _______ -.

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Aicharya · Accepted Answer

$P R=\operatorname{cosec} 	heta, P Q=4 \sec (30+	heta)$ For equilateral $\begin{aligned} & \mathrm{d}=\mathrm{PR}=\mathrm{PQ} \ & \Rightarrow \cos \left(	heta+30^{\circ}ight)=4 \sin 	heta \ & \Rightarrow \frac{\sqrt{3}}{2} \cos 	heta-\frac{1}{2} \sin 	heta=4 \sin 	heta \ & \Rightarrow 	an 	heta=\frac{1}{3 \sqrt{3}} \ & \mathrm{QR}^2=\mathrm{d}^2=\operatorname{cosec}^2 	heta=28 \end{aligned}$

Let the distance between two parallel lines be 5 units and a point $P$ lie between the lines at a unit…

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