Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY=90^\circ$. Let $M,N$ be the midpoints of legs $OX,OY$, respectively. Given $XN=31$ and $YM=22$, find $XY$.

1. $OM^2+OY^2=YM^2=22^2$ and $ON^2+OX^2=XN^2=31^2$ (theorem 18), where $OM=\frac12OX$ and $ON=\frac12OY$.
2. Adding: $\frac14OX^2+OY^2+\frac14OY^2+OX^2=22^2+31^2=1445$, so $OX^2+OY^2=\frac45(1445)=1156$.
3. $XY=\sqrt{1156}=\boxed{34}$.