smc

Within $WXYZ$, the parallelogram piece has vertical side $BF=\sqrt{1^2+2^2}=\sqrt{5}$, and diagonal side $FD=1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $△AFG$, so we deduce $YZ=AG=\frac{\frac{1}{2}\times 1\times 2}{\frac{1}{2}\times \sqrt{5}}=\frac{2}{\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put together without gaps or overlaps, so its area is $2^2=4$, and thus the vertical side $WZ$ is $\frac{4}{\frac{2}{\sqrt{5}}}=2\sqrt{5}$, so the required ratio is $\frac{2\sqrt{5}}{\frac{2}{\sqrt{5}}}=\sqrt{5}\times \sqrt{5}=5$, which is $\boxed{5}$.