Answer:2*sqrt(130)
Step-by-step explanation:[tex]From right triangle $AXC$, we have $\angle XAC = 90^\circ - \angle XCA$. We also must have $\angle YCB + \angle ACB + \angle XCA = 180^\circ$, so $\angle YCB = 180^\circ - 90^\circ - \angle XCA = 90^\circ - \angle XCA$, which means $\angle YCB = \angle XAC$. Combining this with $\angle X = \angle Y$, we have $\triangle XCA \sim \triangle YBC$ by AA Similarity, and\[\frac{CX}{AX} = \frac{BY}{CY},\]so\[\frac{CX}{4} = \frac{18}{2CX}.\]\\\\This gives us $2CX^2 = 72$, so $CX^2 =36$ and $CX=6$. \\\\\\\\[/tex][tex]Therefore, $CY = 12$.Applying the Pythagorean Theorem to right triangles $XCA$ and $CYB$ gives us\begin{align*}CA^2 &= XA^2 + XC^2 = 16 + 36 = 52,\\BC^2 &= CY^2 + BY^2 = 144 + 324 = 468.\end{align*}Applying the Pythagorean Theorem to right triangle $ABC$ gives\[AB = \sqrt{AC^2 + BC^2} = \sqrt{52 + 468} = \sqrt{520} = \boxed{2\sqrt{130}}.\](We can also compute $AB$ by dropping a perpendicular from $A$ to $\overline{YZ}$, which creates a right triangle.)[/tex]
