Petra has an equilateral triangle $ABC$. First, she folded the triangle so that point $A$ coincided with point $C$. Then she folded the resulting shape so that point $B$ coincided with point $C$. She then drew the outline of this shape on paper and found that its area is $12 \, \text{cm}^2$. Determine the area of the original triangle.

(E. Novotná)
After the first fold, point $A$ coincided with point $C$. The line along which the fold was made is the axis of the segment $A C$, which in the equilateral triangle $A B C$ passes through point $B$. After this fold, Petra obtained the triangle $B E C$, which is exactly half of the triangle $A B C$ (point $E$ denotes the midpoint of the segment $A C$, see the figure).

![](https://cdn.mathpix.com/cropped/2024_04_17_52f9a35295cef28fb34dg-1.jpg?height=486&width=579&top_left_y=2170&top_left_x=767)

After the second fold, point $B$ also coincided with point $C$. The line along which the fold was made is the axis of the segment $B C$, which in the original triangle passes through point $A$. After this fold, Petra obtained the quadrilateral $O E C D$ ( $D$ denotes the midpoint of the segment $B C$ and $O$ is the intersection of the axes).

![](https://cdn.mathpix.com/cropped/2024_04_17_52f9a35295cef28fb34dg-2.jpg?height=517&width=591&top_left_y=621&top_left_x=755)

The axes of the sides in the equilateral triangle $A B C$ are axes of symmetry of this triangle. It follows that the quadrilaterals $O E C D, O D B F$, and $O F A E$ are congruent to each other and each has an area of $12 \mathrm{~cm}^{2}$ ( $F$ denotes the midpoint of the segment $A B$ ). The area of the triangle $A B C$ is equal to the sum of the areas of these three quadrilaterals:

$$
S_{A B C}=3 \cdot 12=36\left(\mathrm{~cm}^{2}\right) .
$$

Evaluation. 2 points for the correct interpretation of the first fold (triangle $B E C$); 2 points for the correct interpretation of the second fold (quadrilateral $O E C D$); 2 points for expressing the area of the triangle $A B C$.