Notice the angle created by the diagonal to the horizontal and vertical lines is $45^{\circ}$. Now it's just simple length chase. And as far as I remember $XY = \frac{9\sqrt{2}}{2}$.

I think the best solution for this is using Cartesian Coordinates, especially because everything is just rectangles, and also, the $45^\circ$ everywhere. If ever there's a synthetic solution out there, its most likely just a Cartesian solution in disguise.

Draw the entire square for clarity. You will find a lot of isosceles right triangles. Easy length calculations which on adding give $\frac{9}{\sqrt2}$