3D Geometry
Equilateral triangle between two parallel lines
nta_pyq_2025_apr
Grade 12
Question:
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 ________.
Step-by-Step Solution
Key Concept: Set up the geometry with $P$ at distance 1 from one line and 4 from the other, use the equilateral condition $PR=PQ=QR=d$ and the sine rule in the resulting triangles to express $d$ in terms of a single angle $\theta$.
Let $PR=\operatorname{cosec}\theta$ (with $R$ at distance 5 from the nearer line), $PQ=4\sec(30°+\theta)$.
For equilateral: $PR=PQ \Rightarrow \cos(\theta+30°)=4\sin\theta$.
$\tfrac{\sqrt{3}}{2}\cos\theta-\tfrac{1}{2}\sin\theta=4\sin\theta \Rightarrow \tan\theta=\dfrac{1}{3\sqrt{3}}$.
$\sec^2\theta=1+\dfrac{1}{27}=\dfrac{28}{27}$, $\operatorname{cosec}^2\theta=1+9\cdot3=28$.
$(QR)^2=d^2=\operatorname{cosec}^2\theta=28$.
Correct Answer: 28