Problem 4. Given is a square $ABCD$ with side length $9 \, \text{cm}$. The square is folded as shown in the diagram on the right, so that vertex $B$ coincides with point $E$, which lies on side $\overline{CD}$. In this way, three triangles are formed: $CEF$, $EDG$, and $GAH$. What is the sum of the perimeters of these three triangles?

![](https://cdn.mathpix.com/cropped/2024_06_05_e235c78af50361de0ec1g-04.jpg?height=272&width=348&top_left_y=1859&top_left_x=1114)
Solution. Let the side of the square be denoted by $a$. The perimeter of triangle $C E F$ is $L_{1}=\overline{C F}+\overline{C E}+\overline{E F}$, of $E D G$ is $L_{2}=\overline{D G}+\overline{D E}+\overline{E G}$, and of $G A H$ is $L_{3}=\overline{A G}+\overline{G H}+\overline{A H}$.

The sum of the perimeters of the three triangles is then:

$$
\begin{aligned}
L_{1}+L_{2}+L_{3} & =\overline{C F}+\overline{C E}+\overline{E F}+\overline{D G}+\overline{D E}+\overline{E G}+\overline{A G}+\overline{G H}+\overline{A H} \\
& =(\overline{C F}+\overline{E F})+(\overline{C E}+\overline{D E})+(\overline{D G}+\overline{G H}+\overline{A H})+(\overline{A G}+\overline{E G}) \\
& =a+a+a+a=4 a=4 \cdot 9=36 \text{~cm}
\end{aligned}
$$