2. In triangle $\mathrm{ABC}$ with sides $\overline{\mathrm{AB}}=32 \mathrm{~cm}$ and $\overline{\mathrm{BC}}=24 \mathrm{~cm}$, the medians AA and CC intersect at a right angle. Calculate the length of side $\mathrm{AC}$ of triangle $\mathrm{ABC}$.
Solution. Let T be the centroid of triangle $\mathrm{ABC}$ and let $\overline{\mathrm{TC}_{1}}=x, \overline{\mathrm{TA}_{1}}=\mathrm{y}$

![](https://cdn.mathpix.com/cropped/2024_06_05_b25f91ce23f40fe870e6g-5.jpg?height=288&width=367&top_left_y=673&top_left_x=249)

fig. 11

(fig. 11). Then from the right triangle CTA, we have $(2 \mathrm{x})^{2}+\mathrm{y}^{2}=\left(\frac{1}{2} \overline{\mathrm{CB}}\right)^{2}=144$, and from the right triangle $\operatorname{ATC}_{1}(2 \mathrm{y})^{2}+\mathrm{x}^{2}=\left(\frac{1}{2} \overline{\mathrm{AB}}\right)^{2}=256$. By adding the last two equations, we get $400=5\left(x^{2}+y^{2}\right)$, or $80=x^{2}+y^{2}$. Now, from the right triangle $\mathrm{A}_{1} \mathrm{TC}$, it follows that $\mathrm{x}^{2}+\mathrm{y}^{2}={\overline{\mathrm{A}_{1} \mathrm{C}_{1}}}^{2}=\frac{\overline{\mathrm{AC}}^{2}}{4}$. Therefore, we have $\overline{\mathrm{AC}}^{2}=4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=320$, or $\overline{\mathrm{AC}}=8 \sqrt{5} \mathrm{~cm}$.