Quadrilateral Area between Parallel Lines

SR is minimum when SR is perpendicular to the parallel lines, so $SR = d2 = x$. Quadrilateral PQRS consists of $\triangle PQS$ and $\triangle SRQ$ on common base $QS = b$. Height of $\triangle PQS = d1 = 2x$, height of $\triangle SRQ = d2 = x$. 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 $\triangle SRQ$, $QR = \sqrt{QS^2 + SR^2} = \sqrt{20^2 + 1^2} = \sqrt{401} \approx 20.025$. Thus, $QR$ is slightly greater than 20 units.