4. In trapezoid $A B C D$ with bases $A D / / B C$

diagonals intersect at point $E$. It is known that

the areas $S(\triangle A D E)=12$ and $S(\triangle B C E)=3$. Find

the area of the trapezoid.
Answer: 27

![](https://cdn.mathpix.com/cropped/2024_05_06_35052e7d512455511f98g-06.jpg?height=303&width=366&top_left_y=254&top_left_x=1462)

Solution: Triangles ADE and CBE are similar, their areas are in the ratio of the square of the similarity coefficient. Therefore, this coefficient is 2. This means that point E divides the diagonals in the ratio 1:2. Therefore, the areas of triangles ABE and CDE are twice the area of BCE and are equal to 6.

We get $S(ABCD)=12+3+6+6=27$.