Question

Medians $AD$ and $BE$ of the $△ABC$ with vertices $A(0,b),B(0,0)$ and $C(a,0)$ are perpendicular. If $b$ is the geometric mean between $7$ and $17$ then $\frac{a^2}{119}$ is equal to

Answer 1

$b=$ geometric mean of $7$ and $17=\sqrt{119}$
$AD$ and $BE$ are perpendicular to each other. So, The product of slope is $-1$
Mid-point of $BC=(\frac{a}{2},0)$ and that of $AC=(\frac{a}{2},\frac{b}{2})$
Product of slope$=\frac{b-0}{0-\frac{a}{2}}\times \frac{\frac{b}{2}-0}{\frac{a}{2}-0}=-1$
$\Rightarrow \frac{2b^2}{a^2}=1\Rightarrow a^2=238$