$3+$

Find the area of trapezoid $ABCD (AD \| BC)$, if its bases are in the ratio $5:3$, and the area of triangle $ADM$ is 50, where $M$ is the point of intersection of lines $AB$ and $CD$.
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.

## Solution

Let ${ }^{B C} /{ }_{A D}=3 / 5$. The similarity coefficient of triangles $M B C$ and $M A D$ is $3 / 5$. Therefore, $S_{M B C}=9 / 25 S_{M A D}=18$. Hence, $S_{A B C D}=S_{M A D}-S_{M B C}=50-18=32$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-50.jpg?height=412&width=598&top_left_y=1529&top_left_x=730)

## Answer

32.