Let $T$ be the center of mass of the triangle $\Delta ABC$ and $\overline{AA_1} = t_a$, $\overline{BB_1} = t_b$ are the two corresponding medians to the sides $BC$ and $AC$. From the right-angled triangle $\Delta BA_1T$, using that $\overline{A_1T} = \frac{1}{3}t_a$ and $\overline{TB} = \frac{2}{3}t_b$, we obtain
$$\left(\frac{a}{2}\right)^2 = \left(\frac{1}{3}t_a\right)^2 + \left(\frac{2}{3}t_b\right)^2.$$
Similarly from the right-angled triangle $\Delta B_1AT$, using that $\overline{AT} = \frac{2}{3}t_a$ and $\overline{TB_1} = \frac{1}{3}t_b$, we have
$$\left(\frac{b}{2}\right)^2 = \left(\frac{2}{3}t_a\right)^2 + \left(\frac{1}{3}t_b\right)^2.$$
Summing the two equalities we get
$$\frac{a^2 + b^2}{4} = \frac{5}{9}(t_a^2 + t_b^2)$$
or
$$t_a^2 + t_b^2 = \frac{9}{20}(a^2 + b^2).$$
Finally from the right-angled triangle $\Delta ABT$ we obtain
$$c^2 = \left(\frac{2}{3}t_a\right)^2 + \left(\frac{2}{3}t_b\right)^2 = \frac{4}{9}(t_a^2 + t_b^2) = \frac{a^2 + b^2}{5},$$
from where it follows
$$c = \sqrt{\frac{a^2 + b^2}{5}}.$$