a) $${\left( {mn} \right)^{\frac{1}{2}}}$$
b) $$m{n^{\frac{1}{2}}}$$
c) $${m^{\frac{1}{2}}}n$$
d) $$mn$$

Answer: d
Explanation:

$$\eqalign{ & {\text{tan }}\theta = \frac{H}{{{m^2}}} \cr & \tan \left( {90^ \circ – \theta } \right) = \frac{H}{{{n^2}}} \cr & \cot \theta = \frac{H}{{{n^2}}} \cr & \tan \theta .\cot \theta = \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} \cr & \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} = 1 \cr & {H^2} = {m^2}{n^2} \cr & H = mn \cr} $$