Cutting A Hexagon

$AC$ and $FB$ are perpendicular bisectors of $\triangle ABO \Rightarrow$ the areas $b$ and $e$ are equals. Then $b+c+d$ is equal to $\frac{1}{2}$ of $\triangle ABO$ area. Also $b+c+d=a$ because $OA$ is perpendicular bisector of $OA \therefore$ $a+b+c+d$ is equal to $\triangle ABO$ area.

$\triangle ABO$ area is equal to $\boxed{\frac{\sqrt{3}}{4}}$

Area $g$ is two times area $f$ . Area $g$ is equal to a sixth part of the circle's area minus $\triangle ABO$ area. This is $\frac{\pi}{6}-\frac{\sqrt{3}}{4}=\frac{2\pi-3\sqrt{3}}{12}\Rightarrow g+f=1.5 \left(\frac{2\pi-3\sqrt{3}}{12}\right) =\boxed{\frac{2\pi-3\sqrt{3}}{8}}$

$a+b+c+d+e+f+g=\frac{2\pi-3\sqrt{3}}{8}+\frac{\sqrt{3}}{4}=\boxed{\frac{2\pi-\sqrt{3}}{8}}$

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