In trapezoid $A B C D$, the base $A D$ is 16, the sum of the diagonals $A C$ and $B D$ is 36, and the angle $C A D$ is $60^{\circ}$. The ratio of the areas of triangles $A O D$ and $B O C$, where $O$ is the intersection point of the diagonals, is 4. Find the area of the trapezoid.
Let $K$ be the intersection point of the line passing through vertex $B$ parallel to $AC$, with the extension of the base $AD$. Apply the Law of Cosines to triangle $KBD$.

## Solution

From the similarity of triangles $AOD$ and $COB$, it follows that $BC=8$. Draw a line through vertex $B$ parallel to diagonal $AC$, intersecting the extension of base $AD$ at point $K$. Let $BK=AC=x$. Then

$$
BD=36-x, AK=BC=8, \angle BKD=60^{\circ}
$$

By the Law of Cosines,

$$
BD^{2}=KB^{2}+KD^{2}-2KB \cdot KD \cos 60^{\circ},
$$

or

$$
(36-x)^{2}=x^{2}+24^{2}-24x
$$

From this equation, we find that $x=15$. Therefore,

$$
\begin{aligned}
S_{\mathrm{ABCD}} & =S_{\triangle \mathrm{KBD}}=\frac{1}{2} KB \cdot KD \sin 60^{\circ}= \\
\quad & \frac{1}{2} \cdot 15 \cdot 24 \cdot \frac{\sqrt{3}}{2}=90 \sqrt{3}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-41.jpg?height=325&width=828&top_left_y=1461&top_left_x=615)

## Answer

$90 \sqrt{3}$.