12. A right trapezoid $P Q R S$ has been assigned, with right angles at $P$ and $S$, where $\overline{P Q}>\overline{R S}$ and $\overline{P S}=2 \overline{R S}=62 / 5$. Let $K$ be the point on side $P S$ such that $\overline{P K}=8$. Given that $\widehat{S K R}=\widehat{P Q R}$, what will be the measure of $K Q$?
(A) 17
(B) 13
(C) 15
(D) 16
(E) 20
(12) The correct answer is $(A)$.

Let $T$ be the foot of the perpendicular from $R$ to $P Q$. From the data given in the problem, it follows that the right triangles $K S R$ and $Q T R$ are similar, and the sides of the latter are twice the sides of the former. Therefore, $\overline{T Q}=2 \overline{K S}=2 \cdot(\overline{P S}-\overline{P K})=2 \cdot\left(\frac{62}{5}-8\right)=\frac{44}{5}$.

From this, it follows that $\overline{P Q}=\overline{P T}+\overline{T Q}=\frac{31}{5}+\frac{44}{5}=15$.

Using the Pythagorean theorem, we conclude that $\overline{K Q}=\sqrt{\overline{P K}^{2}+\overline{P Q}^{2}}=\sqrt{8^{2}+15^{2}}=17$. Problem proposed by Camilla Casamento Tumeo.