Explanation

(d)
Let the vertices be $\mathcal{z}_{0},\ \mathcal{z}_{1},\ldots,\mathcal{z}_{5}$ w.r.t. centre O at origin and $\left| \mathcal{z}_{0} \right| = \sqrt{5}$
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$\Rightarrow \ \ A_{0}A_{1} = \left| \mathcal{z}_{1} - \mathcal{z}_{0} \right|$
$= \left| \mathcal{z}_{0}e^{i\theta} - \ \mathcal{z}_{0} \right|$
$= \left| \mathcal{z}_{0} \right||\cos{\theta + i\sin{\theta - 1|}}$
$= \sqrt{5}\sqrt{({\cos{\theta - 1)}}^{2} + \sin^{2}\theta}$
$= \sqrt{5}\sqrt{2(1 - \cos{\theta)}}$
$= \sqrt{5}.2\sin\frac{\theta}{2}$
$\Rightarrow \ A_{0}A_{1} = \sqrt{5}\ 2\sin{\left( \frac{\pi}{6} \right) = \sqrt{5}\ \ \left( \because\theta = \frac{2\pi}{6} = \frac{\pi}{3} \right)\ldots}$(i)
Similarly, $A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{0} = \sqrt{5}$
Hence, the perimeter of regular hexagon
$= A_{0}A_{1} + A_{1}A_{2} + A_{2}A_{3} + A_{3}A_{4} + A_{4}A_{5} + A_{5}A_{0}$
$= 6\sqrt{5}$