2003 AMC 12B Problem 22

Let $O$ be the intersection of the diagonals. Then $\triangle AOB$ is right-angled at $O$ with legs $OA = 8$ and $OB = 15.$ Quadrilateral $OPNQ$ has right angles at $O,$ $P,$ and $Q,$ so it is a rectangle and $PQ = ON.$ The minimum of $ON$ is the altitude from $O$ to $\overline{AB}$ in $\triangle AOB.$ Since $AB = \sqrt{8^2 + 15^2} = 17,$ equating the two area expressions gives $$\begin{aligned} ON &= \frac{OA \cdot OB}{AB} \\ &= \frac{8 \cdot 15}{17} \\ &= \frac{120}{17} \approx 7.06. \end{aligned}$$ This is closest to $7.$ Thus, the correct answer is C.