Let's calculate the area of the triangle if we know the lengths of the medians $\left(s_{a}, s_{b}, s_{c}\right)$.
The figure shows the construction.

![](https://cdn.mathpix.com/cropped/2024_05_02_895f4982d4507b5b129ag-1.jpg?height=610&width=507&top_left_y=218&top_left_x=798)

The area of $\triangle SAB$ is one-third of the area of the sought $\triangle ABC$, because they share the side $AB$, and the ratio of the corresponding heights (which is the same as the ratio of the segments of the medians) is $1:3$.

The area of $\triangle SAB$ can be calculated from the given medians $s_a, s_b, s_c$ as follows. If we reflect point $S$ over the midpoint $F$, the resulting $\triangle S'FA$ is the mirror image of $\triangle SFB$, and thus $\triangle SAS'$ has the same area as $\triangle SAB$, with sides $\frac{2}{3}s_a, \frac{2}{3}s_b, \frac{2}{3}s_c$. Therefore, the area of $\triangle SAS'$ can be calculated using Heron's formula:

$$
\begin{aligned}
t_{\triangle SAS'} & =\sqrt{\frac{s_a+s_b+s_c}{3}\left(\frac{s_a+s_b+s_c}{3}-\frac{2s_a}{3}\right)\left(\frac{s_a+s_b+s_c}{3}-\frac{2s_b}{3}\right)\left(\frac{s_a+s_b+s_c}{3}-\frac{2s_c}{3}\right)}= \\
& =\frac{1}{9} \sqrt{\left(s_a+s_b+s_c\right)\left(s_b+s_c-s_a\right)\left(s_a+s_c-s_b\right)\left(s_a+s_b-s_c\right)}
\end{aligned}
$$

The area of $\triangle ABC$ is three times this:

$$
t_{\triangle ABC}=\frac{1}{3} \sqrt{\left(s_a+s_b+s_c\right)\left(s_b+s_c-s_a\right)\left(s_a+s_c-s_b\right)\left(s_a+s_b-s_c\right)}
$$

Klara Raisz (Miskolc, Vamos I. lg. I. o. t.)