Task B-2.5. The difference in the lengths of the bases of an isosceles trapezoid is $10 \mathrm{~cm}$. The length of its leg is the arithmetic mean of the lengths of the bases. Calculate the lengths of the sides of the trapezoid if the ratio of the height to the larger base is $2: 3$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_06_03_ade005c4283eff439485g-10.jpg?height=371&width=640&top_left_y=243&top_left_x=713)

Given

$$
\begin{aligned}
a-c & =10 \\
\frac{a+c}{2} & =b \\
\frac{v}{a} & =\frac{2}{3}
\end{aligned}
$$

From this system of equations, it follows that

$$
b=a-5 \text { and } v=\frac{2}{3} a
$$

The Pythagorean theorem gives

$$
b^{2}=\left(\frac{a-c}{2}\right)^{2}+v^{2}
$$

so we have

$$
(a-5)^{2}=5^{2}+\left(\frac{2}{3} a\right)^{2}
$$

This leads to the quadratic equation

$$
a^{2}-18 a=0
$$

whose only possible solution is $a=18 \text{ cm}$.

Then, $b=a-5=13 \text{ cm}, c=a-10=8 \text{ cm}$.