An equilateral triangle with a side length of $4 \mathrm{~cm}$ is given (see figure). Determine the area of the dark part and also what percentage of the area of the original triangle is occupied by the dark part? (P. Tlustý)

![](https://cdn.mathpix.com/cropped/2024_04_17_5e753ad895f65083a512g-1.jpg?height=465&width=491&top_left_y=664&top_left_x=817)
Let $v$ denote the height of the given triangle as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_5e753ad895f65083a512g-1.jpg?height=468&width=491&top_left_y=1291&top_left_x=817)

It is given that $v=2 \sqrt{3} \mathrm{~cm}$, so the area of the equilateral triangle is $S=4 \sqrt{3} \mathrm{~cm}^{2}$.

Considering that the base of the dark triangle is $\frac{3}{4}$ the length of the base of the large triangle and the height $v^{\prime}$ of the dark triangle is half the height of the large triangle, the area of the dark triangle is $\frac{3}{8}$ of the area of the entire triangle, i.e., 37.5%. The area of the dark triangle is $\frac{3}{8} \times 4 \sqrt{3} = \frac{3}{2} \sqrt{3} \mathrm{~cm}^{2}$.