The base of a triangle is of length $b$, and the latitude 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) $
1. **Label the triangle and inscribed rectangle:**
   - Let the triangle be \( \triangle ABC \) with base \( BC = b \) and height \( h \).
   - Let the rectangle \( DEFG \) be inscribed in the triangle such that \( D \) and \( E \) lie on \( AB \) and \( AC \) respectively, and \( G \) and \( F \) lie on \( BC \).

2. **Drop a perpendicular from \( A \) to \( DE \):**
   - Let the intersection of this perpendicular with \( DE \) be \( Y \).
   - The height of the rectangle is \( x \), so \( AY = h - x \).

3. **Use similar triangles to find the lengths \( BG \) and \( FC \):**
   - Consider \( \triangle AYD \sim \triangle DGB \):
     \[
     \frac{BG}{s} = \frac{x}{h - x} \implies BG = \frac{sx}{h - x}
     \]
   - Similarly, for \( \triangle AYE \sim \triangle EFC \):
     \[
     \frac{FC}{t} = \frac{x}{h - x} \implies FC = \frac{tx}{h - x}
     \]

4. **Express the total base \( BC \) in terms of \( s \) and \( t \):**
   - Since \( BC = b \), we have:
     \[
     b = BG + GF + FC = \frac{sx}{h - x} + s + t + \frac{tx}{h - x}
     \]
   - Combine terms:
     \[
     b = s + t + \frac{sx + tx}{h - x} = s + t + \frac{(s + t)x}{h - x}
     \]
   - Factor out \( s + t \):
     \[
     b = (s + t) \left( 1 + \frac{x}{h - x} \right) = (s + t) \left( \frac{h}{h - x} \right)
     \]

5. **Solve for \( s + t \):**
   - Rearrange the equation:
     \[
     s + t = \frac{b(h - x)}{h}
     \]

6. **Calculate the area of the rectangle \( DEFG \):**
   - The area of the rectangle is given by:
     \[
     \text{Area} = (s + t) \cdot x = \left( \frac{b(h - x)}{h} \right) x = \frac{bx(h - x)}{h}
     \]

Conclusion:
\[
\boxed{\frac{bx(h - x)}{h}}
\]