Problem 2. The area of trapezoid $ABCD$ is $164 \mathrm{~cm}^{2}$. Its height is $8 \mathrm{~cm}$, and the lengths of the legs $BC$ and $AD$ are $17 \mathrm{~cm}$ and $10 \mathrm{~cm}$, respectively. What is the length of the base $CD$?
Solution. From the formula for the area of a trapezoid, we get

$$
\overline{A B}+\overline{C D}=\frac{2 P}{h}=\frac{328}{8}=41 \mathrm{~cm}
$$

Let $D E$ and $C F$ be the heights in

![](https://cdn.mathpix.com/cropped/2024_06_05_a1dcf8241dfa57377282g-09.jpg?height=243&width=702&top_left_y=1340&top_left_x=781)
the trapezoid. Then, from the Pythagorean theorem, we get that $\overline{A E}=6 \mathrm{~cm}$ and $\overline{F B}=15 \mathrm{~cm}$, so

$$
\overline{A B}=\overline{A E}+\overline{E F}+\overline{F B}=\overline{A E}+\overline{C D}+\overline{F B}=21+\overline{C D}
$$

From this, $21+\overline{C D}+\overline{C D}=41$, hence $\overline{C D}=10 \mathrm{~cm}$.