a) $$\sqrt {2a} \,{\text{metres}}$$
b) $$\frac{a}{{2\sqrt 2 }}\,{\text{metres}}$$
c) $$\frac{a}{{\sqrt 2 }}\,{\text{metres}}$$
d) $$2a\,{\text{metres}}$$

Answer: b
Explanation: Let height of pole CD = h
and AB = 2h, BD = a
M is mid-point of BD

$$ DM = MB = \frac{a}{2}$$
$${\text{Let }}\angle CMD = \theta ,$$    $${\text{then }}\angle AMB = $$     $${90^ \circ } – \theta $$
$$\eqalign{ & \tan \theta = \frac{{CD}}{{DM}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a}\,……({\text{i}}) \cr & {\text{and}} \cr & tan\left( {{{90}^ \circ } – \theta } \right) = \frac{{AB}}{{MB}} = \frac{{2h}}{{\frac{a}{2}}} = \frac{{4h}}{a} \cr & \Rightarrow \cot \theta = \frac{{4h}}{a}\,……….({\text{ii}}) \cr & {\text{Multiplying (i) and (ii)}} \cr & {\text{tan}}\theta \times {\text{cot}}\theta = \frac{{2h}}{a} \times \frac{{4h}}{a} \cr & 1 = \frac{{8{h^2}}}{{{a^2}}} = {h^2} = \frac{{{a^2}}}{8}\,m \cr & h = \sqrt {\frac{{{a^2}}}{8}} = \frac{a}{{\sqrt 8 }} = \frac{a}{{2\sqrt 2 }}\,m \cr} $$