Hexagon's Segment (2)

Relevant wiki: Cosine Rule (Law of Cosines)

The measure of an internal angle of a regular polygon of $n$ sides is given by $\dfrac {(n-2)180^\circ}n$ . Therefore, $\angle BCD =$ $\angle CDE =$ $\dfrac {(6-2)180^\circ}6 =$ $120^\circ$ . Since $\triangle BCD$ is isosceles, then $\angle BDC =$ $\dfrac {180^\circ - 120^\circ}2 =$ $30^\circ$ and $\angle BDE =$ $\angle CDE - \angle BDC =$ $120^\circ - 30^\circ =$ $90^\circ$ . And by Pythagorean theorem on $\triangle BDH$ , we have:

$\begin{aligned} BH^2 & = DH^2 + \color{#3D99F6} BD^2 & \small \color{#3D99F6} \text{By cosine rule on }\triangle BCD \\ & = \left(\frac 12\right)^2 + \color{#3D99F6} BC^2+CD^2-2(BC)(CD)\cos \angle BCD \\ & = \frac 14 + 1^2+1^2 - 2(1)(1) \cos 120^\circ \\ & = \frac 14 + 1+1 - 2\left(-\frac 12\right) \\ & = \frac {13}4 \\ \implies BH & = \frac {\sqrt{13}}2 \end{aligned}$

$\implies a+b = 13+2 = \boxed{15}$