4. Inside an equilateral triangle, a point $T$ is chosen which is at distances of $1 \mathrm{~cm}, 2 \mathrm{~cm}$, and $3 \mathrm{~cm}$ from the sides of the triangle, respectively. What is the area of this triangle?
4. Let $D, E, F$ be the feet of the perpendiculars from point $T$ to the sides $\overline{A B}, \overline{B C}, \overline{C A}$ as shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_30_a9a8cb41faa417e2420fg-2.jpg?height=343&width=357&top_left_y=431&top_left_x=907)

According to the problem, $|T D|=1, |T E|=2$, and $|T F|=3$. We can now express the area of triangle $A B C$ as the sum of the areas of three triangles:

$$
P(A B C)=P(A B T)+P(B C T)+P(C A T)
$$

Since the area of a triangle is the product of the length of a side and the height, we have

$$
P(A B C)=\frac{a \cdot|T D|}{2}+\frac{a \cdot|T E|}{2}+\frac{a \cdot|T F|}{2}=\frac{a}{2}(|T D|+|T E|+|T F|)=\frac{a}{2}(1+2+3)=3 a
$$

On the other hand, the area of an equilateral triangle is $P(A B C)=\frac{a^{2} \sqrt{3}}{4}$, so

by equating the two expressions for the area, we get $\frac{a^{2} \sqrt{3}}{4}=3 a$, from which $a=\frac{12}{\sqrt{3}}=4 \sqrt{3}$.

Finally, using the formula for the area of the triangle, we get $P(A B C)=\frac{a^{2} \sqrt{3}}{4}=\frac{48 \sqrt{3}}{4}=12 \sqrt{3} \mathrm{~cm}^{2}$.