## Task B-1.4.

In an isosceles trapezoid, a square is inscribed such that the shorter base of the trapezoid is also a side of the square, and the two remaining vertices of the square lie on the longer base of the trapezoid. The square covers $25 \%$ of the area of the trapezoid. What is the ratio of the lengths of the longer and shorter bases of the trapezoid?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7713364b1a15726862ddg-03.jpg?height=203&width=471&top_left_y=618&top_left_x=724)

Let the longer base of the trapezoid be denoted by $a$, and the shorter base by $c$.

According to the conditions of the problem, $c$ is the side length of the square, and also the height of the trapezoid.

The area of the trapezoid is $P=\frac{a+c}{2} \cdot c$, and the area of the square is $c^{2}$.

$$
\begin{aligned}
& \frac{1}{4} \frac{a+c}{2} \cdot c=c^{2} \\
& a c+c^{2}=8 c^{2} \\
& a c=7 c^{2} \\
& \frac{a}{c}=\frac{7}{1}
\end{aligned}
$$

The bases of the trapezoid are in the ratio $7: 1$.