What's the area of the regular hexagon?

Using the law of cosines on $\triangle{CBO} \implies$

$r^2 = \dfrac{r^2}{4} + 1 - r\cos(150^{\circ}) = \dfrac{r^2}{4} + 1 + \dfrac{\sqrt{3}}{2}r \implies$ $4r^2 = r^2 + 4 + 2\sqrt{3}$

$\implies 3r^2 - 2\sqrt{3}r - 4 = 0 \implies r = \dfrac{\sqrt{3} \pm \sqrt{15}}{3} =$ $\dfrac{1 \pm \sqrt{5}}{\sqrt{3}}$

dropping the negative root we have $r = \dfrac{1 + \sqrt{5}}{2}(\dfrac{2}{\sqrt{3}}) =\dfrac{2}{\sqrt{3}}\phi \implies$

$\overline{AB} = \dfrac{r}{2} = \dfrac{\phi}{\sqrt{3}} \implies h_{\triangle_{AOB}} = \dfrac{\phi}{2} \implies$

$A_{\triangle{AOB}} = \dfrac{1}{2}\overline{AB} * h_{\triangle{AOB}} = \dfrac{\phi^2}{4\sqrt{3}} \implies A_{Hexagon} = 6 * A_{\triangle{AOB}} = \dfrac{\sqrt{3}}{2}\phi^2 =$

$\dfrac{\sqrt{a}}{b}\phi^2 \implies a + b = \boxed{5}$ .