A quadrilateral $A B C D$ is given, see the figure. Point $T_{1}$ is the centroid of triangle $B C D$, point $T_{2}$ is the centroid of triangle $A B D$, and points $T_{1}$ and $T_{2}$ lie on segment $A C$. The length of segment $T_{1} T_{2}$ is $3 \mathrm{~cm}$ and point $D$ is $3 \mathrm{~cm}$ away from segment $A C$.

Determine the area of quadrilateral $A B C D$.

(E. Patáková)

![](https://cdn.mathpix.com/cropped/2024_04_17_7dc1166ff4d250faa543g-2.jpg?height=666&width=546&top_left_y=1089&top_left_x=755)
Let $E$ be the intersection of the diagonals of quadrilateral $A B C D$. Points $T_{1}$ and $T_{2}$ are the centroids of triangles $B C D$ and $A B D$, so segments $C E$ and $A E$ are medians in these triangles, and thus point $E$ is the midpoint of segment $B D$. Since $|D E|=|E B|$, the areas of triangles $D E C$ and $E B C$ are equal, as are the areas of triangles $D E A$ and $E B A$. Therefore, triangles $A C D$ and $A C B$ have the same area. From the problem statement, we know the height of triangle $A C D$ from vertex $D$, and to express its area, we need to determine the length of segment $A C$.

From the properties of the centroid, we know that

$$
|C E|=3 \cdot\left|T_{1} E\right|, \quad|A E|=3 \cdot\left|T_{2} E\right| .
$$

The length of segment $A C$ is

$$
|A C|=|A E|+|E C|=3 \cdot\left(\left|T_{2} E\right|+\left|E T_{1}\right|\right)=3 \cdot\left|T_{2} T_{1}\right|=3 \cdot 3=9(\mathrm{~cm})
$$

The height of triangle $A C D$ from vertex $D$ is $3 \mathrm{~cm}$, so the area of the triangle is

$$
S_{A C D}=\frac{9 \cdot 3}{2}=\frac{27}{2}\left(\mathrm{~cm}^{2}\right) .
$$

The area of quadrilateral $A B C D$ is

$$
S_{A B C D}=2 \cdot S_{A C D}=27\left(\mathrm{~cm}^{2}\right)
$$

Evaluation. 3 points for determining the length of segment $A C$; 2 points for justifying the equality of the areas of triangles $A C D$ and $A C B$; 1 point for determining the desired area.