$\left[\begin{array}{l}\text { [Midline of the trapezoid]} \\ {[\text { Area of the trapezoid }}\end{array}\right]$

The midline of the trapezoid is 10 and divides the area of the trapezoid in the ratio 3:5. Find the bases of the trapezoid.
Let $x$ and $y$ be the bases of the trapezoid and set up a system of equations with respect to $x$ and $y$.

## Solution

Let $x$ and $y$ be the bases of the trapezoid $(x < y)$. Then

$$
x + y = 2 \cdot 10 = 20
$$

If $h$ is the height of the trapezoid, then the height of each of the two trapezoids into which the median line divides the given trapezoid is $\frac{h}{2}$. Therefore,

$$
\frac{x + 10}{y + 10} = \frac{3}{5}
$$

From the system

$$
\left\{\begin{array}{l}
x + y = 20 \\
\frac{x + 10}{y + 10} = \frac{3}{5}
\end{array}\right.
$$

we find that $x = 5$ and $y = 15$.

## Answer

15 and 5.