smc

Let $AB=b$, $DE=h$, and $WX=YZ=x$. Since $CD$ is perpendicular to $AB$, $ND=WX$. That means $CN=h-x$. The sides of the rectangle are parallel, so $XY\parallel WZ$. That means by AA Similarity, $△CXY\sim △CAB$. Letting $n$ be the length of the base of the rectangle, that means $$\frac{h-x}{n}=\frac{h}{b}$$ $$n=\frac{b(h-x)}{h}$$ Thus, the area of the rectangle is $\frac{bx}{h}(h-x)$, which is answer choice $\boxed{\frac{bx}{h}(h-x)}$.