Question

AB, CD and EF are three parallel lines, in that order. Let d1 and d2 be the distances from CD to AB and EF respectively. d1 and d2 are integers, where d1 : d2 = 2 : 1. P is a point on AB, Q and S are points on CD and R is a point on EF. If the area of the quadrilateral PQRS is 30 square units, what is the value of QR when value of SR is the least?

Accepted Answer

E. SR is minimum when SR is perpendicular to the parallel lines, so $SR = d2 = x$. Quadrilateral PQRS consists of $	riangle PQS$ and $	riangle SRQ$ on common base $QS = b$. Height of $	riangle PQS = d1 = 2x$, height of $	riangle SRQ = d2 = x$. Total Area $= \frac{1}{2} b (2x) + \frac{1}{2} b x = \frac{3}{2} b x = 30 \implies b x = 20 \implies b = \frac{20}{x}$. Since $d1, d2$ are integers, $x \ge 1$. To minimize $x$, set $x = 1 \implies b = 20$. Since $SR = 1$ is perpendicular to $CD$, in right $	riangle SRQ$, $QR = \sqrt{QS^2 + SR^2} = \sqrt{20^2 + 1^2} = \sqrt{401} \approx 20.025$. Thus, $QR$ is slightly greater than 20 units.

Quadrilateral Area between Parallel Lines

Explanation

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