[The ratio of the areas of triangles with a common base or common height] Class

The diagonals $AC$ and $BD$ of trapezoid $ABCD$ with bases $AD$ and $BC$ intersect at point $O$. It is known that $AD=2BC$ and the area of triangle $AOB$ is 4. Find the area of the trapezoid.
Triangle BOC is similar to triangle DOA with a coefficient of $\frac{1}{2}$, so $O C=\frac{1}{2} A O$, therefore

$$
\begin{aligned}
& S_{\triangle B O C}=\frac{1}{2} S_{\triangle A O B}=\frac{1}{2} \cdot 4=2 \\
& S_{\triangle D O A}=2 S_{\triangle A O B}=2 \cdot 4=8
\end{aligned}
$$

since triangles $ABD$ and $ACD$ are equal in area (they have the same base and equal heights), then

$$
S_{\triangle C O D}=S_{\triangle A C D}-S_{\triangle D O A}=S_{\triangle A B D}-S_{\triangle D O A}=S_{\triangle A O B}=4
$$

Therefore,

$$
S_{A B C D}=2 S_{\triangle A O B}+S_{\triangle B O C}+S_{\triangle D O A}=2 \cdot 4+2+8=18
$$

## Answer

18.00