In the diagram, $P R T Y$ and $W R S U$ are squares. Point $Q$ is on $P R$ and point $X$ is on $T Y$ so that $P Q X Y$ is a rectangle. Also, point $T$ is on $S U$, point $W$ is on $Q X$, and point $V$ is the point of intersection of $U W$ and $T Y$, as shown. If the area of rectangle $P Q X Y$ is 30 , the length of $S T$ is closest to
(A) 5
(B) 5.25
(C) 5.5
(D) 5.75
(E) 6

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-016.jpg?height=431&width=423&top_left_y=1129&top_left_x=1312)

Part C: Each correct answer is worth 8.
Let $S T=a$ and let $\angle S T R=\theta$.

Since $\triangle R S T$ is right-angled at $S$, then $\angle T R S=90^{\circ}-\theta$.

Since $P R T Y$ and $W R S U$ are squares, then $\angle P R T=\angle W R S=90^{\circ}$.

Thus, $\angle Q R W+\angle W R T=\angle W R T+\angle T R S$ and so $\angle Q R W=\angle T R S=90^{\circ}-\theta$.

Since $P Q X Y$ is a rectangle, then $\angle P Q X=90^{\circ}$, which means that $\triangle W Q R$ is right-angled at $Q$.

This means that $\angle Q W R=90^{\circ}-\angle Q R W=90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-032.jpg?height=431&width=417&top_left_y=571&top_left_x=911)

Consider $\triangle R S T$.

Since $S T=a$ and $\angle S T R=\theta$, then $\cos \theta=\frac{a}{R T}$ and so $R T=\frac{a}{\cos \theta}$.

Also, $\tan \theta=\frac{R S}{a}$ and so $R S=a \tan \theta$.

Since $P R T Y$ is a square, then $P Y=P R=R T=\frac{a}{\cos \theta}$.

Since $W R S U$ is a square, then $R W=R S=a \tan \theta$.

Next, consider $\triangle Q R W$.

Since $R W=a \tan \theta$ and $\angle Q W R=\theta$, then $\sin \theta=\frac{Q R}{a \tan \theta}$ and so $Q R=a \tan \theta \sin \theta$.

This means that

$$
\begin{aligned}
P Q & =P R-Q R \\
& =\frac{a}{\cos \theta}-a \tan \theta \sin \theta \\
& =\frac{a}{\cos \theta}-a \cdot \frac{\sin \theta}{\cos \theta} \cdot \sin \theta \\
& =\frac{a}{\cos \theta}-\frac{a \sin ^{2} \theta}{\cos \theta} \\
& =\frac{a\left(1-\sin ^{2} \theta\right)}{\cos \theta} \\
& =\frac{a \cos ^{2} \theta}{\cos \theta} \quad\left(\text { since } \sin ^{2} \theta+\cos ^{2} \theta=1\right) \\
& =a \cos \theta \quad(\text { since } \cos \theta \neq 0)
\end{aligned}
$$

Since the area of rectangle $P Q X Y$ is 30 , then $P Q \cdot P Y=30$ and so $a \cos \theta \cdot \frac{a}{\cos \theta}=30$ which gives $a^{2}=30$.

Since $a>0$, we get $a=\sqrt{30} \approx 5.48$. Of the given choices, $S T=a$ is closest to 5.5.

ANSWER: (C)