Answer: $\mathbf B$
Explanation:
$\mathrm {AD}$ and $\mathrm {BE}$ are $\perp$s
$\mathrm D$ and $\mathrm E$ are the mid-points of $\mathrm {BC}$ and $\mathrm {AC}$ respectively.
So, coordinates of $\mathrm {D = (0+a/2, 0+0/2)$ = $(a/2,0)}$
$\mathrm {AD \perp BE}$
$\therefore\; \text{Slope of AD} \times \;\text{Slope of BE} = -1$ tex
$\mathrm {= \frac{(0-b)}{(a/2-0)} *\frac{ (b/2-0)}{(a/2-0)} = -1}$
$\mathrm{= \frac{b^2}{2} = \frac{a^2}{4} \\\Rightarrow a^2 = 2b^2}$
$\mathrm {\Rightarrow a = \pm\sqrt{(2b)}}$
$\therefore \mathbf B$ is the correct option.