imc

By symmetry, $WXYZ$ is a square. Draw $\overline{BZ}$. $BZ=\frac{1}{2}AH=1$, so $△BWZ$ is a $45-45-90△$. Hence $WZ=\frac{1}{\sqrt{2}}$, and $[WXYZ]={\left(\frac{1}{\sqrt{2}}\right)}^2=\boxed{\frac{1}{2}}$. There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.