Answer: $\mathbf B$
The mid-points of $BC$ and $AC$ are: $D\Big(\frac{a}{2}, 0\Big)$ and $E\Big(\frac{a}{2}\frac{b}{2}\Big)$
$$\;\text{Slope of AD} =m_1= \frac{y_2-y_1}{x_2-x_1} =\Bigg(\frac{0-b}{\frac{a}{2}-0}\Bigg)$$
Similarly, $$\text{Slope of BE } =m_2=\frac{y_2-y_1}{x_2-x_1} = \Bigg(\frac{ \frac{-b}{2}} {\frac{-a}{2}}\Bigg)$$
It is given that the medians are mutually perpendicular to each other.
$$\therefore m_1\times m_2 = -1$$
$$\implies \Bigg(\frac{0-b}{\frac{a}{2}-0}\Bigg)*\Bigg(\frac{ \frac{-b}{2}} {\frac{-a}{2}}\Bigg) = -1$$
$$\implies a = \pm \sqrt{2b}$$
$\therefore$ $\mathbf B$ is the correct answer.