Triangles $PRS$ and $QPR$ are similar because $\angle PSR = \angle QRP$ (since $PR =PS$) and $\angle PRS = \angle QPR$ (since $QP =QR$).

Hence $\frac{SR}{RP} = \frac{RP}{PQ}$, that is $\frac{SR}{6} = \frac{6}{9}$, that is $SR = 4$.

*This problem is taken from the UKMT Mathematical Challenges.*