2. Given a triangle $A B C$ with side $B C$ of length $22 \mathrm{~cm}$ and side $A C$ of length $19 \mathrm{~cm}$, whose medians $t_{a}, t_{b}$ are perpendicular to each other. Calculate the length of side $A B$.
2. Let $D$ be the midpoint of side $A C$, $E$ the midpoint of side $B C$, and $T$ the centroid of triangle $A B C$ (Fig. 1). If we denote $3 x$ and $3 y$ as the lengths of the medians $t_{a}$ and $t_{b}$, we have $|A T|=2 x,|E T|=x,|B T|=2 y$, $|D T|=y$. From the problem statement, it follows that triangles $A T D, B E T$, and $A B T$ are right-angled, so by the Pythagorean theorem we have

$$
\begin{aligned}
(2 x)^{2}+y^{2} & =\left(\frac{b}{2}\right)^{2}, \\
x^{2}+(2 y)^{2} & =\left(\frac{a}{2}\right)^{2} \\
(2 x)^{2}+(2 y)^{2} & =c^{2} .
\end{aligned}
$$

By adding the first two equations, we get $5\left(x^{2}+y^{2}\right)=$ $=\frac{1}{4}\left(a^{2}+b^{2}\right)$ and substituting into the third equation, we have $c^{2}=$ $=4\left(x^{2}+y^{2}\right)=\frac{1}{5}\left(a^{2}+b^{2}\right)$. Numerically, given that $\frac{1}{5}\left(22^{2}+19^{2}\right)=169$, we get $c=13 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_fb1bfc1f5e30d8251ccdg-2.jpg?height=717&width=528&top_left_y=1600&top_left_x=1324)

Fig. 1

For a complete solution, award 6 points. Award one point for using the fact that the centroid divides the median in the ratio $1: 2$, one point each for using the Pythagorean theorem for triangles $A T D, B E T, A B T$, and two points for calculating the length of side $A B$.