The base of a triangle is of length $b$, and the altitude is of length $h$. 
A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:
$\textbf{(A)}\ \frac{bx}{h}(h-x)\qquad \textbf{(B)}\ \frac{hx}{b}(b-x)\qquad \textbf{(C)}\ \frac{bx}{h}(h-2x)\qquad \textbf{(D)}\ x(b-x)\qquad \textbf{(E)}\ x(h-x)$
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, $\triangle CXY \sim \triangle CAB$.  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 $\boxed{\textbf{(A)}}$.