In the figure, three identical, overlapping equilateral triangles are depicted. Determine the area of each of them, given that the following conditions are simultaneously satisfied:

$\triangleright$ The intersection of triangle $T_{1}$ and triangle $T_{2}$ is an equilateral triangle with an area of $\sqrt{3} \mathrm{~cm}^{2}$.

$\triangleright$ The intersection of triangle $T_{2}$ and triangle $T_{3}$ is an equilateral triangle with an area of $\frac{9}{4} \sqrt{3} \mathrm{~cm}^{2}$.

$\triangleright$ The intersection of triangle $T_{1}$ and triangle $T_{3}$ is an equilateral triangle with an area of $\frac{1}{4} \sqrt{3} \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_9ae8228539612b8057feg-3.jpg?height=522&width=688&top_left_y=236&top_left_x=747)
Let $U T$ (fig.) denote the side of the equilateral triangle which is the intersection of $T_{1}$ and $T_{3}$. For the area of this triangle, we have

$$
\frac{|U T|^{2} \cdot \sqrt{3}}{4}=\frac{\sqrt{3}}{4}
$$

thus $|U T|=1(\mathrm{~cm})$. Next, let $V T$ denote the side of the equilateral triangle which is the intersection of $T_{1}$ and $T_{2}$. For the area of this triangle, we have

$$
\frac{|V T|^{2} \cdot \sqrt{3}}{4}=\sqrt{3}
$$

thus $|V T|=2(\mathrm{~cm})$.

Finally, let $Y Z$ denote the side of the equilateral triangle which is the intersection of $T_{2}$ and $T_{3}$, with $Y Z \| V T$. For the area of this triangle, we have

$$
\frac{|Y Z|^{2} \cdot \sqrt{3}}{4}=\frac{9}{4} \sqrt{3}
$$

thus $|Y Z|=3(\mathrm{~cm})$. Let $X Z$ denote the side of triangle $T_{2}$ which is parallel to $V T$.

$$
|X Z|=|X Y|+|Y Z| \text {. }
$$

Quadrilateral $X Y U V$ is a parallelogram, so $|X Y|=|V U|$. At the same time, $|V U|=|V T|-|U T|$. Therefore,

$$
|X Z|=|V U|+|Y Z|=|V T|-|U T|+|Y Z|=2-1+3=4(\mathrm{~cm}) \text {. }
$$

The area of triangle $T_{2}$ is

$$
\frac{|X Z|^{2} \cdot \sqrt{3}}{4}=\frac{4^{2} \sqrt{3}}{4}=4 \sqrt{3}\left(\mathrm{~cm}^{2}\right) .
$$

[Calculation of $U T, V T, Y Z-3 \times 1$ b., calculation of $X Z-1$ b., area of triangle $T_{2}$ - 2 b.]