In $\triangle P Q R, \angle R P Q=90^{\circ}$ and $S$ is on $P Q$. If $S Q=14, S P=18$, and $S R=30$, then the area of $\triangle Q R S$ is
(A) 84
(B) 168
(C) 210
(D) 336
(E) 384

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-058.jpg?height=431&width=528&top_left_y=224&top_left_x=1167)
Solution 1

Since $\triangle R P S$ is right-angled at $P$, then by the Pythagorean Theorem, $P R^{2}+P S^{2}=R S^{2}$ or $P R^{2}+18^{2}=30^{2}$.

This gives $P R^{2}=30^{2}-18^{2}=900-324=576$, from which $P R=24$, since $P R>0$.

Since $P, S$ and $Q$ lie on a straight line and $R P$ is perpendicular to this line, then $R P$ is actually a height for $\triangle Q R S$ corresponding to base $S Q$.

Thus, the area of $\triangle Q R S$ is $\frac{1}{2}(24)(14)=168$.

## Solution 2

Since $\triangle R P S$ is right-angled at $P$, then by the Pythagorean Theorem, $P R^{2}+P S^{2}=R S^{2}$ or $P R^{2}+18^{2}=30^{2}$.

This gives $P R^{2}=30^{2}-18^{2}=900-324=576$, from which $P R=24$, since $P R>0$.

The area of $\triangle Q R S$ equals the area of $\triangle R P Q$ minus the area of $\triangle R P S$.

Since $\triangle R P Q$ is right-angled at $P$, its area is $\frac{1}{2}(P R)(P Q)=\frac{1}{2}(24)(18+14)=12(32)=384$.

Since $\triangle R P S$ is right-angled at $P$, its area is $\frac{1}{2}(P R)(P S)=\frac{1}{2}(24)(18)=12(18)=216$.

Therefore, the area of $\triangle Q R S$ is $384-216=168$.