$\triangle ABC$ is a right triangle $(\measuredangle ACB=90^\circ)$. The medians are $AA_1=4\sqrt{13}$ and $BB_1=2\sqrt{73}.$ Find the sides of $\triangle ABC$.
By the Pythagorean theorem: $\triangle AA_1C\rightarrow AC^2+CA_1^2=AC^2+(\dfrac12 BC)^2=AA_1^2=16\times 13=208$
and
$\triangle BB_1C \rightarrow BC^2+CB_1^2=BC^2+(\dfrac12 AC)^2=BB_1^2=4\times 73=292$.
So $\begin{cases} 4AC^2+BC^2=832 \\ 4BC^2+AC^2=1168 \end{cases} \Leftrightarrow AC=12; BC=16.$ The answer given in my book is $4;6;2\sqrt{13}$. Is there a problem with my solution? Is this the fastest way to find the medians? (we have not studied trigonometry)