In the diagram, $\triangle P Q R$ is right-angled at $P$ and $P R=12$. If point $S$ is on $P Q$ so that $S Q=11$ and $S R=13$, the perimeter of $\triangle Q R S$ is
(A) 47
(B) 44
(C) 30
(D) 41
(E) 61

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-045.jpg?height=379&width=523&top_left_y=1247&top_left_x=1191)
By the Pythagorean Theorem in $\triangle P R S$,

$$
P S=\sqrt{R S^{2}-P R^{2}}=\sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}=5
$$

since $P S>0$.

Thus, $P Q=P S+S Q=5+11=16$.

By the Pythagorean Theorem in $\triangle P R Q$,

$$
R Q=\sqrt{P R^{2}+P Q^{2}}=\sqrt{12^{2}+16^{2}}=\sqrt{144+256}=\sqrt{400}=20
$$

since $R Q>0$.

Therefore, the perimeter of $\triangle Q R S$ is $R S+S Q+R Q=13+11+20=44$.

ANSWER: (B)