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 _________.

Solution

Maths sir · 4 steps

Put P at the origin, say the side is a at angle θ, and turn PQ by 60° to reach R.

1. Set up axes. Take P at the origin and the lines as y = 1 and y = -4. They are then 5 units apart, with P 1 unit from the first line. Let Q lie on y = 1 and R on y = -4, and let each side of the triangle be a.
2. Height of Q. If PQ makes angle θ with the x-axis, then Q = (a cos θ, a sin θ). Q is on y = 1, so a sin θ = 1.
3. Height of R. PR is PQ turned clockwise by 60°, so R is at height a sin(θ - 60°). R is on y = -4, so a sin(60° - θ) = 4.

   √3/2 a cos θ - 1/2 a sin θ = 4 ⇒ √3/2 a cos θ = 9/2 ⇒ a cos θ = 3√3

4. So. a² = (a sin θ)² + (a cos θ)² = 1 + 27 = 28. QR is a side of the equilateral triangle, so (QR)² = 28.

The idea to keep. Write one side as a(cos θ, sin θ) and turn it by 60° for the other. The two distance conditions give a sin θ and a cos θ; square and add for a².

Answer from the NTA final answer key. Solution by PrepWiser.