### Instruction: 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:
$\textbf{(A)}\ \frac{bx}{h}(h-x)\qquad \textbf{(B)}\ \frac{hx}{b}(b-x)\qquad \textbf{(C)}\ \frac{bx}{h}(h-2x)\qquad \textbf{(D)}\ x(b-x)\qquad \textbf{(E)}\ x(h-x)$.### Response: I start by drawing a diagram of the situation.
[Insert a sketch of a triangle with base $b$ and height $h$, and a rectangle of height $x$ and base $y$ inscribed in the triangle, as shown here: https://imgur.com/a/CLyufDw]
I label the base of the rectangle as $y$, since it is not given in the problem.
I notice that the area of the rectangle is simply $xy$, so I need to find an expression for $y$ in terms of $b$, $h$, and $x$.
I see that there are similar triangles involved in the diagram: the big triangle with base $b$ and height $h$, and the small triangle with base $y$ and height $h-x$.
I can use the fact that the ratio of corresponding sides of similar triangles is equal to apply it to the bases and heights of these two triangles.
I get the equation $\frac{y}{b} = \frac{h-x}{h}$, which I can cross-multiply and solve for $y$.
I get $y = \frac{b(h-x)}{h}$.
I substitute this expression for $y$ into the formula for the area of the rectangle, $xy$.
I get $xy = x \cdot \frac{b(h-x)}{h} = \frac{bx}{h}(h-x)$.
I compare this expression with the answer choices, and I see that it matches choice A.