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

A quadrilateral is divided into four triangles by its diagonals. The areas of three of them are 10, 20, and 30, and each is less than the area of the fourth triangle. Find the area of the given quadrilateral.

#
Let $M$ be the point of intersection of the diagonals $AC$ and $BD$ of quadrilateral $ABCD$. If

$$
S_{DAMD}=30, S_{DAMB}=10, S_{DCMD}=20,
$$

then

$$
\begin{gathered}
\frac{BM}{MD}=\frac{S_{\triangle AMB}}{S_{\triangle AMD}}=\frac{1}{3} \\
S_{DBMC}=\frac{1}{3} S_{DCMD}=\frac{20}{3}<10 .
\end{gathered}
$$

By examining other possible cases, we can confirm that only two cases are possible:

$$
S_{DAMB}=20, S_{DAMD}=10, S_{DCMD}=30
$$

or

$$
S_{DAMB}=30, S_{DAMD}=10, S_{DCMD}=20
$$

In each of these possible cases, $S_{DBMC}=60$. Therefore, $S_{ABCD}=10+20+30+60=120$.

## Answer

120.