The area of a convex quadrilateral is equal to half the product of the diagonals by the sine of the angle between them.

## Solution

Let $R-$ be the radius of the circle circumscribed around triangle $K L M \left(R=\frac{5}{2}\right), \angle K M L=\alpha, L M K L=\beta$. Then

$$
\sin \alpha=\frac{K L}{2 R}=\frac{4}{5}, \sin \beta=\frac{L M}{2 R}=\frac{3}{5}
$$

Since $M LK L$. Therefore

$$
\cos \alpha=-\frac{3}{5} K M=\frac{16-9}{5}=\frac{7}{5}
$$

Let $O$ be the point of intersection of the diagonals $A C$ and $B D$ of quadrilateral $A B C D$. Since $\angle B O C=\angle K L M$, then

$$
\begin{aligned}
S_{\mathrm{ABCD}} & =\frac{1}{2} A C \cdot B D \sin \left(180^{\circ}-\alpha-\beta\right)= \\
& =\frac{1}{2} A C \cdot B D \sin (\alpha+\beta)= \\
= & \frac{1}{2} \cdot \frac{3}{2} K L \cdot \frac{3}{2} L M \cdot \frac{7}{25}=\frac{189}{25} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-13.jpg?height=440&width=464&top_left_y=1458&top_left_x=797)

Answer

$\frac{189}{25}$.

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Points $K, L, M$ divide the sides of the convex quadrilateral $A B C D$ in the ratio $A K: B K=C L: B L=C M: D M=1$ : 2. The radius of the circle circumscribed around triangle $K L M$ is $\frac{5}{2}, K L=4, L M=3$. What is the area of quadrilateral $A B C D$, given that $K M<K L$?