In the diagram, a semi-circle has diameter $X Y$. Rectangle $P Q R S$ is inscribed in the semi-circle with $P Q=12$ and $Q R=28$. Square $S T U V$ has $T$ on $R S, U$ on the semi-circle and $V$ on $X Y$. The area of $S T U V$ is closest to
(A) 12
(B) 13
(C) 16
(D) 14
(E) 15

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-112.jpg?height=255&width=461&top_left_y=2176&top_left_x=1293)
Let $O$ be the centre of the circle, let $r$ be the radius of the circle, and let $s$ be the side length of the square. We want to calculate $s^{2}$.

By symmetry, $O$ is the midpoint of $P S$, so $O P=O S=\frac{1}{2} Q R=14$.

Join $O$ to $R$ and $O$ to $U$. We see that $O R=O U=r$, the radius of the circle.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-223.jpg?height=442&width=848&top_left_y=397&top_left_x=693)

Since $\triangle O S R$ is right-angled at $S$, then $O R^{2}=O S^{2}+S R^{2}$ by the Pythagorean Theorem, or $r^{2}=14^{2}+12^{2}=196+144=340$.

Since $\triangle O V U$ is right-angled at $S$, then $O U^{2}=O V^{2}+V U^{2}$ or $r^{2}=(14+s)^{2}+s^{2}$.

But $r^{2}=340$, so

$$
\begin{aligned}
340 & =s^{2}+28 s+196+s^{2} \\
0 & =2 s^{2}+28 s-144 \\
0 & =s^{2}+14 s-72 \\
0 & =(s+18)(s-4)
\end{aligned}
$$

Since $s$ must be positive, then $s=4$, so the area of $S T U V$ is $s^{2}=16$.

ANSWER: (C)