Trapezoid $ABCD$, where side $AB$ is parallel to side $CD$, is divided by its diagonals, which intersect at point $M$, into four parts. Determine its area, given that triangle $AMD$ has an area of $8 \mathrm{~cm}^{2}$ and triangle $DCM$ has an area of $4 \mathrm{~cm}^{2}$.

(M. Volfová)
Angles $B M A$ and $D M C$ have the same size, as they are vertical angles. Angles $A B M$ and $C D M$ have the same size, because they are alternate interior angles. Triangles $A B M$ and $C D M$ are therefore similar (by the AA similarity criterion). We will determine the ratio of the areas of these two triangles step by step.

![](https://cdn.mathpix.com/cropped/2024_04_17_1e8c778e46dcaf241e4bg-1.jpg?height=400&width=740&top_left_y=2227&top_left_x=661)

Let the height of the trapezoid be $v$ and the height of triangle $C D M$ perpendicular to side $C D$ be $x$, as shown in the figure. For the areas of triangles $C D M$ and $C D A$, we have

$$
\begin{aligned}
S_{C D M} & =\frac{1}{2}|C D| \cdot x=4\left(\mathrm{~cm}^{2}\right) \\
S_{C D A} & =\frac{1}{2}|C D| \cdot v=12\left(\mathrm{~cm}^{2}\right) .
\end{aligned}
$$

By comparing both expressions, we find that $v=3 x$. The height of triangle $A B M$ perpendicular to side $A B$ is, according to the figure, equal to $v-x=3 x-x=2 x$. Similar triangles $C D M$ and $A B M$ thus have corresponding heights in the ratio $1: 2$, and the areas of these triangles are therefore in the ratio $1^{2}: 2^{2}$, i.e., $1: 4$. The area of triangle $A B M$ is

$$
S_{A B M}=4 \cdot S_{C D M}=4 \cdot 4=16\left(\mathrm{~cm}^{2}\right) .
$$

To solve the problem, it remains to determine the area of triangle $M B C$. From the figure, we can easily derive the following relationships:

$$
\begin{aligned}
& S_{A M D}=S_{A B D}-S_{A B M}, \\
& S_{M B C}=S_{A B C}-S_{A B M},
\end{aligned}
$$

and since $S_{A B C}=S_{A B D}$, we have

$$
S_{M B C}=S_{A M D}=8\left(\mathrm{~cm}^{2}\right) .
$$

The area of trapezoid $A B C D$ is obtained by summing the areas of the individual triangles:

$$
4+8+16+8=36\left(\mathrm{~cm}^{2}\right)
$$

Another solution. Triangles $C D A$ and $C D M$ share the side $C D$. Since their areas are in the ratio $3: 1$, their heights perpendicular to side $C D$ must also be in the ratio $3: 1$. If we denote the first of these heights by $v$, the second will be $\frac{1}{3} v$. The height of triangle $A B M$ perpendicular to side $A B$ is equal to the difference of the mentioned heights, i.e., $v-\frac{1}{3} v=\frac{2}{3} v$.

Triangles $A B D$ and $A B M$ share the side $A B$ and we have just shown that their heights perpendicular to this side are in the ratio $3: 2$. The areas of these triangles must therefore also be in the same ratio. From the given information, we know that the difference in areas is $8 \mathrm{~cm}^{2}$, so the area of triangle $A B D$ is $3 \cdot 8=24\left(\mathrm{~cm}^{2}\right)$ and the area of triangle $A B M$ is $2 \cdot 8=16\left(\mathrm{~cm}^{2}\right)$.

To determine the area of trapezoid $A B C D$, we need to know the area of triangle $M B C$. Triangles $C D A$ and $C D B$ share the side $C D$ and also have the same height perpendicular to this side, so their areas must also be the same. Triangle $C D M$ forms the common part of these triangles, and the remaining part of triangle $C D A$ must have the same area as the remaining part of triangle $C D B$. In other words, the area of triangle $D A M$, which is given as $8 \mathrm{~cm}^{2}$, is equal to the area of triangle $M B C$.

We know the areas of all four parts of trapezoid $A B C D$; the area of this trapezoid is

$$
4+8+16+8=36\left(\mathrm{~cm}^{2}\right)
$$

Evaluation. 2 points for the area of triangle $M B C$; 1 point for determining the ratio of the heights of triangles $C D M$ and $C D A$; 1 point for correctly determining the height of triangle $A B M$; 1 point for the area of triangle $A B M$; 1 point for the correct conclusion.