6. In the figure, there are 7 equilateral triangles with a side length of $4 \sqrt{3} \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_d9f78d2eb5276b887ecdg-2.jpg?height=274&width=1311&top_left_y=543&top_left_x=378)

The triangles are arranged such that the vertex of the second triangle is at the midpoint of the side of the first triangle, the vertex of the third triangle is at the midpoint of the side of the second triangle, and so on, with the vertex of the seventh triangle at the midpoint of the side of the sixth triangle. What is the area of the gray part?
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_46538cfe21908c5128afg-4.jpg?height=303&width=1240&top_left_y=2096&top_left_x=405)

The area of the gray part is equal to the sum of the areas of 7 congruent equilateral triangles with a side length of \(4 \sqrt{3} \, \text{cm}\) minus the sum of \(2 \cdot 6 = 12\) areas (the "white areas" overlap and subtract parts of two triangles!) of equilateral triangles with a side length of \(2 \sqrt{3} \, \text{cm}\). \quad 3 POINTS By applying the Pythagorean theorem, the height of the larger triangle is calculated.

![](https://cdn.mathpix.com/cropped/2024_05_30_46538cfe21908c5128afg-5.jpg?height=292&width=291&top_left_y=248&top_left_x=380)

\[ v^2 = (4 \sqrt{3})^2 - (2 \sqrt{3})^2 = 48 - 12 = 36 \]

\[ v = 6 \, \text{cm} \]

The area of this triangle is \( P_1 = \frac{1}{2} \cdot 4 \sqrt{3} \cdot 6 = 12 \sqrt{3} \, \text{cm}^2 \).

1 POINT

By applying the Pythagorean theorem, the height of the smaller triangle is calculated.

![](https://cdn.mathpix.com/cropped/2024_05_30_46538cfe21908c5128afg-5.jpg?height=182&width=212&top_left_y=800&top_left_x=385)

\[ v_1^2 = (2 \sqrt{3})^2 - (\sqrt{3})^2 = 12 - 3 = 9 \]

\[ v_1 = 3 \, \text{cm} \]

The area of this triangle is \( P_2 = \frac{1}{2} \cdot 2 \sqrt{3} \cdot 3 = 3 \sqrt{3} \, \text{cm}^2 \).

The area of the gray part of the figure is \( 7 \cdot P_1 - 12 \cdot P_2 = 7 \cdot 12 \sqrt{3} - 12 \cdot 3 \sqrt{3} = 84 \sqrt{3} - 36 \sqrt{3} = 48 \sqrt{3} \, \text{cm}^2 \).