The figure shows a regular dodecagon decomposed into six equilateral triangles, six squares, and one regular hexagon, all with the same side length.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-11.jpg?height=337&width=343&top_left_y=2173&top_left_x=959)

a) If each triangle has an area of $1 \mathrm{~cm}^{2}$, what is the area of the hexagon?

b) The figure below was obtained by removing twelve equilateral triangles from a regular dodecagon whose side measures $1 \mathrm{~cm}$. What is the area of this figure?

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-12.jpg?height=414&width=394&top_left_y=227&top_left_x=731)

c) The figure below was obtained by removing two regular hexagons from a regular dodecagon whose side measures $1 \mathrm{~cm}$. What is the area of this figure?

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-12.jpg?height=325&width=325&top_left_y=820&top_left_x=771)
Solution

a) The figure below shows that the hexagon can be decomposed into six triangles equal to the triangles that make up the dodecagon. Since each of these triangles has an area of $1 \mathrm{~cm}^{2}$, it follows that the hexagon has an area of $6 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-34.jpg?height=323&width=331&top_left_y=2140&top_left_x=771)

b) $1^{st}$ solution: The figure from the previous item shows that the dodecagon can be decomposed into twelve equal equilateral triangles and six squares. Thus, by removing twelve triangles from the dodecagon, the star that remains has an area equal to the area of six squares. Since the side of the dodecagon measures $1 \mathrm{~cm}$, each square has an area of $1 \mathrm{~cm}^{2}$, and thus the area of the star is $6 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-35.jpg?height=366&width=354&top_left_y=248&top_left_x=954)

$2^{nd}$ solution: We can decompose the central hexagon of the star into six triangles and "fit" them as indicated in the figure below. The figure thus obtained has the same area as the star and consists of six squares with a side of $1 \mathrm{~cm}$; its area is then $6 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-35.jpg?height=354&width=357&top_left_y=888&top_left_x=952)

c) The figure below shows that the two hexagons removed have the same area as twelve equilateral triangles; as in item $b$), the gray region has the same area as six squares with a side of $1 \mathrm{~cm}$; its area is then $6 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-35.jpg?height=374&width=374&top_left_y=1503&top_left_x=938)