5. The lengths of the diagonals of a trapezoid are 3 and 5, and the segment connecting the midpoints of the bases is 2 units long. Calculate the area of the trapezoid.
Solution. From point $C$, we draw a line parallel to the diagonal BD. The intersection of this line with the line on which the base $AB$ lies is marked as $M$. Triangles $ACD$ and $BMC$ have equal bases ($\overline{BM}=\overline{CD}=b$) and heights, so $P_{\triangle ACD}=P_{\triangle BMC}$. Therefore, $P_{ABCD}=P_{\triangle AMC}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_ee54e896908301490c8ag-8.jpg?height=457&width=603&top_left_y=1639&top_left_x=857)
Thus, $P_{ABCD}=P_{\triangle AMC}$.

From point $C$, we draw a line parallel to the segment connecting the midpoints of the bases of the trapezoid. Let the intersection of this line with the base $AB$ be point $N$. On the same line, we choose point $P$ such that $\overline{CN}=\overline{NP}$. Then, triangles $ANC$ and $PMN$ are congruent ($\overline{CN}=\overline{NP}$, $\measuredangle PNM=\measuredangle ANC$, $\overline{AN}=\overline{NM}=\frac{a+b}{2}$), so their areas are equal. For triangles $AMC$ and $PMC$, $P_{\triangle AMC}=P_{\triangle PMC}$, because $P_{\triangle AMC}=P_{\triangle ANC}+P_{\triangle NMC}$, and $P_{\triangle PMC}=P_{\triangle PMN}+P_{\triangle NMC}$. The sides of triangle $PMC$ are $\overline{PM}=\overline{AC}=5$, $\overline{MC}=\overline{BD}=3$, and $\overline{PC}=2\overline{CN}=4$, forming a Pythagorean triple. Therefore, triangle $PMC$ is a right triangle, so $P_{\triangle PMC}=\frac{4 \cdot 3}{2}=6$. According to this, $P_{ABCD}=6$.