In the diagram, $P Q R S$ is a rectangle with $P Q=5$ and $Q R=3 . \quad P R$ is divided into three segments of equal length by points $T$ and $U$. The area of quadrilateral $S T Q U$ is
(A) $\frac{17}{3}$
(B) 5
(C) $\frac{5}{2}$
(D) $\frac{\sqrt{34}}{3}$
(E) $\sqrt{34}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-081.jpg?height=273&width=420&top_left_y=2303&top_left_x=1297)
Since $P Q R S$ is a rectangle, then $P Q$ is perpendicular to $Q R$.

Therefore, the area of $\triangle P Q R$ is $\frac{1}{2}(P Q)(Q R)=\frac{1}{2}(5)(3)=\frac{15}{2}$.

Since $P T=T U=U R$, then the areas of $\triangle P T Q, \triangle T U Q$ and $\triangle U R Q$ are equal. (These triangles have bases $P T, T U$ and $U R$ of equal length and each has height equal to the distance between $Q$ and line segment $P R$.)

Therefore, the area of $\triangle T U Q$ is $\frac{1}{3}\left(\frac{15}{2}\right)=\frac{5}{2}$.

Similarly, the area of $\triangle T U S$ is $\frac{5}{2}$.

The area of quadrilateral $S T Q U$ is the sum of the areas of $\triangle T U Q$ and $\triangle T U S$, or $\frac{5}{2}+\frac{5}{2}=5$.