A trapezium has vertices marked as P, Q, R and S (in that order anticlockwise). The side PQ is parallel to side SR. Further, it is given that PQ = 11 cm, QR = 4 cm, RS = 6 cm and SP = 3 cm. What is the shortest distance between PQ and SR (in cm)? A. 1.80 B. 2.40 C. 4.20 D. 5.76

GATE 2022 · General Aptitude · Geometry · medium

Answer: The shortest distance between PQ and SR is 2.40 cm.

1. Set up coordinates: Place P at the origin and Q at (11,0). Since PQ || SR, points S and R lie at height h. With SR=6, we have R at x-coordinate a+6 if S is at x-coordinate a.
2. Write distance equations for the legs: SP = 3 gives a^2 + h^2 = 9. QR = 4 gives (a-5)^2 + h^2 = 16. Expanding: a^2 - 10a + 25 + h^2 = 16.
3. Solve for a by subtracting the two equations: Subtracting: a^2 - 10a + 25 - a^2 = 7 => -10a = -18 => a = 1.8 cm.
4. Compute the height h: Substituting a = 1.8 into SP^2 = 9: h^2 = 9 - (1.8)^2 = 9 - 3.24 = 5.76, so h = 2.4 cm.