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

Question

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.

Answer 1

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, $$|AB|^2=|AG|^2+|BG|^2=(2x)^2+(2y)^2=4(x^2+y^2)=\frac{4}{5}(170)=136.$$

Answer 2

Let $a=|BC|=22$, $b=|AC|=14$, $c=|AB|$.

The medians from sides of lengths $a$ and $b$ are perpendicular if and only if

\begin{align} a^2 + b^2 = 5c^2 , \end{align}

hence

\begin{align} c=|AB|&=\sqrt{\tfrac15\,(a^2 + b^2)} =2\,\sqrt{34} . \end{align}