In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints
of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown 
in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is  

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 1+2\sqrt{3}\qquad \textbf{(C)}\ 2\sqrt{5}\qquad \textbf{(D)}\ \frac{8+4\sqrt{3}}{3}\qquad \textbf{(E)}\ 5$
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 $\triangle AFG$, 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 $\boxed{\text{E}}$.