Let $ABCD$ be a [rhombus](https://artofproblemsolving.com/wiki/index.php/Rhombus) with $AC = 16$ and $BD = 30$. Let $N$ be a point on $\overline{AB}$, and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$, respectively. Which of the following is closest to the minimum possible value of $PQ$?


$\mathrm{(A)}\ 6.5 \qquad\mathrm{(B)}\ 6.75  \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 7.25 \qquad\mathrm{(E)}\ 7.5$
Let $\overline{AC}$ and $\overline{BD}$ intersect at $O$. Since $ABCD$ is a rhombus, then $\overline{AC}$ and $\overline{BD}$ are [perpendicular bisectors](https://artofproblemsolving.com/wiki/index.php/Perpendicular_bisector). Thus $\angle POQ = 90^{\circ}$, so $OPNQ$ is a [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle). Since the diagonals of a rectangle are of equal length, $PQ = ON$, so we want to minimize $ON$. It follows that we want $ON \perp AB$.
Finding the area in two different ways,
\[\frac{1}{2}   AO \cdot BO = 60 = \frac{1}{2} ON \cdot AB = \frac{\sqrt{8^2 + 15^2}}{2} \cdot ON \Longrightarrow ON = \frac{120}{17} \approx 7.06 \Rightarrow \mathrm{(C)}\]