In the diagram, $\triangle P Q R$ is right-angled at $Q$ and point $S$ is on $P R$ so that $Q S$ is perpendicular to $P R$. If the area of $\triangle P Q R$ is 30 and $P Q=5$, the length of $Q S$ is
(A) $\frac{60}{13}$
(B) 5
(C) $\frac{30}{13}$
(D) 4
(E) 3

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-022.jpg?height=257&width=439&top_left_y=457&top_left_x=1298)
Since $\triangle P Q R$ is right-angled at $Q$, its area equals $\frac{1}{2} \cdot P Q \cdot Q R$.

Since its area is 30 and $P Q=5$, then $\frac{1}{2} \cdot 5 \cdot Q R=30$ and so $Q R=30 \cdot \frac{2}{5}=12$.

By the Pythagorean Theorem, we know that

$$
P R^{2}=P Q^{2}+Q R^{2}=5^{2}+12^{2}=25+144=169
$$

Since $P R>0$, then $P R=\sqrt{169}=13$.

If we now consider $\triangle P Q R$ as having base $P R$ and perpendicular height $Q S$, we see that its area equals $\frac{1}{2} \cdot P R \cdot Q S$.

Since its area is 30 and $P R=13$, then $\frac{1}{2} \cdot 13 \cdot Q S=30$ which gives $Q S=30 \cdot \frac{2}{13}=\frac{60}{13}$.