[

## Law of Cosines

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

In trapezoid $A B C D$, the base $A D$ is 16, the sum of the lateral side $A B$ and the diagonal $B D$ is 40, and the angle $C B D$ is $60^{\circ}$. The ratio of the areas of triangles $A B O$ and $B O C$, where $O$ is the point of intersection of the diagonals, is 2. Find the area of the trapezoid.

#
Apply the Law of Cosines.

## Solution

Let $A B=x$. Then $B D=40-x$. By the Law of Cosines,

$$
A B^{2}=A D^{2}+B D^{2}-2 A D \cdot B D \cos \angle A D B
$$

or

$$
x^{2}=16^{2}+(40-x)^{2}-16(40-x)
$$

From this equation, we find that $x=19$. Then $B D=40-x=21$. Therefore,

$$
S_{\triangle \mathrm{ADB}}=\frac{1}{2} A D \cdot D B \sin 60^{\circ}=\frac{1}{2} \cdot 16 \cdot 21 \cdot \frac{\sqrt{3}}{2}=84 \sqrt{3}
$$

Since triangles $A O D$ and $C O B$ are similar, then

$$
\frac{B C}{A D}=\frac{C O}{A O}=\frac{S_{\triangle B O C}}{S_{\triangle B O A}}=\frac{1}{2}
$$

Therefore, $B C=8$, and

$$
S_{\triangle \mathrm{BDC}}=\frac{1}{2} S_{\triangle \mathrm{ABD}}=42 \sqrt{3}
$$

$$
S_{\mathrm{ABCD}}=S_{\Delta \mathrm{ABD}}+S_{\Delta \mathrm{BDC}}=126 \sqrt{3} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-11.jpg?height=288&width=686&top_left_y=176&top_left_x=686)

Answer

$126 \sqrt{3}$.