8,9

In trapezoid $ABCD$, points $K$ and $M$ are the midpoints of the bases $AB=5$ and $CD=3$. Find the area of the trapezoid if triangle $AMB$ is a right triangle and $DK$ is the height of the trapezoid.
Draw the height of the trapezoid from point $M$.

## Solution

Notice that in triangle $A M B \angle M=90^{\circ}$ (otherwise, $M A$ or $M B$ would be perpendicular to $A B$, which is impossible).

Let $M P$ be another height of the trapezoid. Then

\[
\begin{gathered}
K P=D M=\frac{1}{2} D C=\frac{3}{2}, B P=K B-K P=\frac{5}{2}-\frac{3}{2}=1 \\
A P=A K+K P=\frac{5}{2}+\frac{3}{2}=4, M P^{2}=A P \cdot P B=4
\end{gathered}
\]

Therefore,

\[
S_{\mathrm{ABCD}}=\frac{A B+D C}{2} \cdot M P=8
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-37.jpg?height=343&width=780&top_left_y=1858&top_left_x=639)

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