5. Let's return the square $A B C D$ to its original form and mark all the vertices of the observed triangles on it. Mark point $G$ twice, once on side $\overline{A D}(G)$, and once on side $\overline{A B}\left(G^{\prime}\right)$.

Do the same with point $E$.

![](https://cdn.mathpix.com/cropped/2024_05_30_10d6775da96883f7c302g-1.jpg?height=345&width=348&top_left_y=2072&top_left_x=908)

Notice that each side of the three triangles $C E F, E D G$, and $G A H$ is contained in one of the sides of the square $A B C D$, and that these sides together cover all the sides of the square.

Therefore, the sum of the perimeters of the triangles $C E F, E D G$, and $G A H$ is equal to the perimeter of the square, which is $4 \cdot 9=36 \mathrm{~cm}$.
5. The square $A B C D$ has a side length of $9 \, \text{cm}$. The square is folded, as shown in the figure, so that vertex $B$ falls onto point $E$, which is the midpoint of side $\overline{C D}$. In this way, three triangles are determined: $C E F$, $E D G$, and $G A H$. What is the sum of the perimeters of these three triangles?

![](https://cdn.mathpix.com/cropped/2024_06_03_f9b99612d02b8cc7c925g-1.jpg?height=388&width=457&top_left_y=2099&top_left_x=845)

Each problem is scored out of 10 points.

The use of a pocket calculator or any reference materials is not allowed.