Hexagon 's Segment

$Isosceles ~\Delta~BCD, ~~~vertex~ \angle~120,~~\therefore~ base, ~BD~=~\sqrt3. \\ The~median~BH~of ~\Delta~BCD=\frac 1 2 *\sqrt{2(BD^2+BC^2)-DC^2}=\frac 1 2 *\sqrt{2(3+1)-1}=\dfrac{\sqrt7} 2 =~\dfrac{\sqrt a} b.\\ a+b=\Huge 9.$

Consider $\triangle CJG:$

$sin~30=\dfrac{CJ}{\frac{1}{2}}$

$\dfrac{1}{2}=\dfrac{CJ}{\frac{1}{2}}$

$CJ=\dfrac{1}{2}(\dfrac{1}{2})=\dfrac{1}{4}$

By Pythagorean Theorem

$GJ=\sqrt{(\dfrac{1}{2})^2-(\dfrac{1}{4})^2}=\sqrt{\dfrac{3}{16}}=\dfrac{\sqrt{3}}{4}$

It follows that

$BJ=1+CJ=1+\dfrac{1}{4}=\dfrac{5}{4}$

Consider $\triangle BJG$ , by pythagorean theorem

$BG=\sqrt{(\dfrac{5}{4})^2+(\dfrac{\sqrt{3}}{4})^2}=\sqrt{\dfrac{28}{16}}=\sqrt{\dfrac{7}{4}}=\dfrac{\sqrt{7}}{2}$

Finally,

$a=7$ and $b=2$

$a+b=7+2=$ $\boxed{9}$

Here is a solution without cosine law

By cosine law:

$\large \displaystyle \overline{BG}^2 = \overline{BC}^2 + \overline{CG}^2 - 2\cdot \overline{BC} \cdot \overline{CG} \cdot \cos(\angle BCG)$

$\large \displaystyle \overline{BG}^2 = 1^2 + \frac12 ^2 - 2\cdot 1 \cdot \frac12 \cdot \left ( - \frac12 \right)$

$\large \displaystyle \overline{BG}^2 = \frac74$

$\color{#20A900} \boxed{\large \displaystyle \overline{BG} = \frac{\sqrt{7}}{2}}$

So:

$\color{#3D99F6} \large \displaystyle a = 7, b = 2, \boxed{\large \displaystyle a+b=9}$

From law of cosines in $\triangle BCG$ we can get $BG=\sqrt{\frac{1}{4}+1-\cos120^\circ}=\frac{\sqrt{7}}{2}$

$7+2=\boxed{9}$