In the dodecagon $A B C D E F G H I J K L$, every two adjacent sides are perpendicular, and all sides except for sides $A L$ and $G F$ are congruent. Sides $A L$ and $G F$ are twice as long as the other sides. Segments $B G$ and $E L$ intersect at point $M$ and divide the dodecagon into six figures (three triangles, two quadrilaterals, and one pentagon). The quadrilateral $E F G M$ has an area of $7 \mathrm{~cm}^{2}$.

Determine the areas of the other five figures.

(E. Semerádová)

![](https://cdn.mathpix.com/cropped/2024_04_17_e836f4ee23919875e85ag-5.jpg?height=502&width=474&top_left_y=1485&top_left_x=791)

Hint. Divide the image into smaller congruent figures.
From the problem statement, we know that the dodecagon $A B C D E F G H I J K L$ consists of six identical squares. Each of these squares can be further divided by their diagonals into four identical triangles. Therefore, the dodecagon consists of 24 identical triangles, and each of the above-mentioned figures is formed by several such triangles:

![](https://cdn.mathpix.com/cropped/2024_04_17_e836f4ee23919875e85ag-6.jpg?height=508&width=485&top_left_y=183&top_left_x=791)

The quadrilateral $E F G M$ consists of seven of these triangles and has an area of $7 \mathrm{~cm}^{2}$. Therefore, one triangle has an area of $1 \mathrm{~cm}^{2}$, and the areas of all the figures are

$$
\begin{aligned}
S_{I J M} & =1 \mathrm{~cm}^{2}, \quad S_{I G H}=S_{J K L}=2 \mathrm{~cm}^{2}, \\
S_{C D E M B} & =5 \mathrm{~cm}^{2}, \quad S_{A B M L}=S_{E F G M}=7 \mathrm{~cm}^{2} .
\end{aligned}
$$