In the trapezoid $A B C D$ with bases $A B$ and $C D$ of lengths $7 \mathrm{~cm}$ and $4 \mathrm{~cm}$, points $S$ and $T$ are the midpoints of sides $A D$ and $B C$, see the figure. Point $X$ is the intersection of segments $A C$ and $S T$, point $Y$ is the intersection of segment $A B$ and line $D X$. The area of quadrilateral $A Y C D$ is $12 \mathrm{~cm}^{2}$. Calculate the area of trapezoid $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2a0500418bf8c778174dg-6.jpg?height=460&width=865&top_left_y=1392&top_left_x=630)

(M. Dillingerová)
Segment $S T$ connects the midpoints of the non-parallel sides of trapezoid $A B C D$, so it must be parallel to its base $C D$. Segment $S X$, which lies on segment $S T$, is thus parallel to side $C D$ of triangle $C D A$. Furthermore, we know that its endpoint $S$ is the midpoint of side $D A$. Segment $S X$ is therefore the midline of triangle $C D A$. Similarly, it can be proven that segment $S X$ is the midline of triangle $A Y D$. For the midline of a triangle, it generally holds that it is half the length of the side of the triangle parallel to it. The length of segment $S X$ is half the length of side $C D$ and at the same time half the length of side $A Y$. Therefore, the lengths of segments $C D$ and $A Y$ must be the same, both measuring $4 \mathrm{~cm}$. Since the parallel sides $A Y$ and $C D$ of quadrilateral $A Y C D$ have the same length, it must be a rhombus. The area of this rhombus is given by $S_{1}=|C D| \cdot v$, where $v$ is the length of its height to side $C D$. From the problem statement, we know $S_{1}$ and $|C D|$, and we can calculate the length $v$:

$$
v=S_{1}:|C D|=12: 4=3(\mathrm{~cm}) .
$$

This length $v$ is also equal to the distance between the bases of trapezoid $A B C D$. The area of this trapezoid is therefore

$$
S_{2}=\frac{v}{2} \cdot(|A B|+|C D|)=\frac{3}{2}(7+4)=16.5\left(\mathrm{~cm}^{2}\right) .
$$

Another solution. We will prove that triangles $A X Y$ and $C X D$ are congruent. Their angles $A X Y$ and $C X D$ are vertical angles and thus have the same measure. The parallel sides $A B$ and $D C$ are cut by the diagonal $A C$, so the angles $Y A X$ and $D C X$ of the mentioned triangles are alternate interior angles and also have the same measure. Therefore, these two triangles are similar. Moreover, the heights of these triangles to the corresponding sides $A Y$ and $C D$ have the same length, as both represent the distance between the bases of trapezoid $A B C D$ and its midline. Since triangles $A X Y$ and $C X D$ are congruent in the three elements mentioned, they must be congruent. The corresponding sides $A Y$ and $C D$ therefore have the same length, namely $4 \mathrm{~cm}$. The rest of the solution is the same as the one mentioned above.