10.348. A convex quadrilateral is divided by its diagonals into four triangles; the areas of three of them are 10, 20, and $30 \mathrm{~cm}^{2}$, and each is less than the area of the fourth triangle. Find the area of the given quadrilateral.
## Solution.

Let $O$ be the point of intersection of the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ (Fig. 10.131), $S_{\triangle AOB}=10 \text{ cm}^2, S_{\triangle AOD}=20 \text{ cm}^2, S_{\triangle BOC}=30 \text{ cm}^2$.

Drop perpendiculars $AM$ and $CN$ to $BD$. We have

\[
\frac{S_{\triangle AOB}}{S_{\triangle AOD}}=\frac{\frac{1}{2} BO \cdot AM}{\frac{1}{2} DO \cdot AM}=\frac{BO}{DO}=\frac{1}{2}, \frac{S_{\triangle BOC}}{S_{\triangle COD}}=\frac{\frac{1}{2} BO \cdot CN}{\frac{1}{2} DO \cdot CN}=\frac{BO}{DO}. \text{ Then }
\]

$S_{\triangle COD}=2 S_{\triangle BOC}=60 \text{ cm}^2, S_{ABCD}=S_{\triangle AOB}+S_{\triangle AOD}+S_{\triangle BOC}+S_{\triangle COD}=$ $=120 \text{ cm}^2$.

Answer: $120 \text{ cm}^2$.