In $\triangle ABC$, $AD$ and $BE$ are medians, and $AD \perp BE$. If $AC=14$ and $BC=22$, find $AB$. 
In $\triangle ABC$, $AD$ and $BE$ are medians, and $AD \perp BE$. If $AC=14$ and $BC=22$, find $AB$.

I'm using Apollonius's theorem to find the medians. Instead, I found
$$BE^2-AD^2=72$$
Not sure how to proceed.

Let $G$ be the centroid of $\triangle ABC$.  Suppose that $3x=|AD|$ and $3y=|BE|$.  Then by Pythagoras
$$7^2=\left(\frac{1}{2}|AC|\right)^2=|AD|^2=|AG|^2+|GD|^2=(2x)^2+y^2$$
and 
$$11^2=\left(\frac{1}{2}|BC|\right)^2=|BE|^2=|BG|^2+|GE|^2=(2y)^2+x^2.$$
Thus 
$$170=7^2+11^2=(4x^2+y^2)+(4y^2+x^2)=5(x^2+y^2).$$
Therefore,
$$|AB|^2=|AG|^2+|BG|^2=(2x)^2+(2y)^2=4(x^2+y^2)=\frac{4}{5}(170)=136.$$
