The diagonals of the trapezoid are 3 and 5, and the segment connecting the midpoints of the bases is 2. Find the area of the trapezoid.

#
Through the vertex of the smaller base of the trapezoid, draw a line parallel to the diagonal.

## Solution

Let $M$ and $K$ be the midpoints of the bases $BC$ and $AD$ of trapezoid $ABCD$. Through vertex $C$ of the smaller base $BC (AC=3, BD=5)$, draw a line parallel to diagonal $BD$, intersecting line $AD$ at point $P$, and draw a line parallel to $MK$, intersecting line $AD$ at point $Q$. Then

$$
\begin{gathered}
A Q=A K+K Q=A K+M C=\frac{1}{2} A D+\frac{1}{2} B C= \\
=\frac{1}{2}(A D+B C)=\frac{1}{2}(A D+D P)
\end{gathered}
$$

Therefore, $CQ$ is the median of triangle $ACP$,

$$
C Q=M K=2, A C=3, C P=B D=5, S_{\mathrm{ABCD}}=S_{\triangle \mathrm{ACP}}
$$

Extend median $CQ$ beyond point $Q$ and lay off segment $QF$ equal to $CQ$. The sides of triangle $CFP$ are:

$$
C F=2 C Q=4, C P=B D=5, F P=A C=3 .
$$

This triangle is right-angled ( $C P^{2}=C F^{2}+P F^{2}$ ). Therefore,

$$
S_{\triangle \mathrm{CFP}}=\frac{1}{2} C F \cdot P F=6
$$

Thus,

$$
S_{\mathrm{ABCD}}=S_{\triangle \mathrm{ACP}}=S_{\triangle \mathrm{CFP}}=6
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-33.jpg?height=463&width=918&top_left_y=1698&top_left_x=570)

## Answer

6.