Three differently shaped quadrilaterals I, II, III can be obtained:
![](https://cdn.mathpix.com/cropped/2024_04_17_1c3f74535a8c6f3ba064g-4.jpg?height=340&width=1606&top_left_y=914&top_left_x=240)

Quadrilateral I is half of the hexagon, so its area is $78 \mathrm{~cm}^{2}$.

Quadrilaterals II and III are obtained by removing two congruent triangles from the hexagon, so they have the same areas. First, we determine the areas of the triangles to be removed; see, for example, triangle $A B C$:

![](https://cdn.mathpix.com/cropped/2024_04_17_1c3f74535a8c6f3ba064g-4.jpg?height=502&width=571&top_left_y=1551&top_left_x=774)

A regular hexagon can be divided into 6 congruent triangles ( $\triangle A B S, \triangle B C S$, $\triangle C D S, \ldots)$, each with an area of $156: 6=26\left(\mathrm{~cm}^{2}\right)$. Quadrilateral $A B C S$ is composed of two such triangles, so it has an area of $52 \mathrm{~cm}^{2}$. The same quadrilateral can be divided into two other congruent triangles $A B C$ and $A C S$, which also have areas of $26 \mathrm{~cm}^{2}$. Therefore, the area of triangle $A B C$ is $26 \mathrm{~cm}^{2}$.

The area of quadrilaterals II and III is thus $156-2 \cdot 26=104\left(\mathrm{~cm}^{2}\right)$.
Sketch all possible differently shaped quadrilaterals that have vertices at the vertices of a given regular hexagon.

Determine what their areas would be if the hexagon had an area of $156 \mathrm{~cm}^{2}$.