By the Pythagorean Theorem in $\triangle PRS$, $PS=\sqrt{RS^{2}-PR^{2}}=\sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}=5$ since $PS>0$. Thus, $PQ=PS+SQ=5+11=16$. By the Pythagorean Theorem in $\triangle PRQ$, $RQ=\sqrt{PR^{2}+PQ^{2}}=\sqrt{12^{2}+16^{2}}=\sqrt{144+256}=\sqrt{400}=20$ since $RQ>0$. Therefore, the perimeter of $\triangle QRS$ is $RS+SQ+RQ=13+11+20=44$.
If $\triangle PQR$ is right-angled at $P$ with $PR=12$, $SQ=11$, and $SR=13$, what is the perimeter of $\triangle QRS$?