jmc

Each of $△PSQ$ and $△RSQ$ is right-angled at $S$, so we can use the Pythagorean Theorem in both triangles.

In $△RSQ$, we have $Q{S}^2=Q{R}^2-S{R}^2=25^2-20^2=625-400=225$, so $QS=\sqrt{225}=15$ since $QS>0$.

In $△PSQ$, we have $P{Q}^2=P{S}^2+Q{S}^2=8^2+225=64+225=289$, so $PQ=\sqrt{289}=17$ since $PQ>0$.

Therefore, the perimeter of $△PQR$ is $PQ+QR+RP=17+25+(20+8)=\boxed{70}$.