Inscribed Circles in a Unit Hexagon!

Right $\triangle{BOA}$ with $m\angle{OBA} = 60^{\circ} \implies \dfrac{1 - 2r}{2} = \dfrac{\overline{OA}}{2} \implies \overline{OA} = 1 - 2r$

and $r = \dfrac{\sqrt{3}}{2}\overline{OA} \implies \overline{OA} = \dfrac{2r}{\sqrt{3}} \implies \dfrac{2r}{\sqrt{3}} = 1 - 2r \implies \boxed{r = \dfrac{3 - \sqrt{3}}{4}}$

$\implies r + R = \dfrac{4R + 3 - \sqrt{3}}{4}$ and $\sqrt{3} - (R + r) = \dfrac{(5\sqrt{3} - 3) - 4R}{4}$

Right $\triangle{OO'E} \implies \dfrac{(3 - \sqrt{3})^2}{16} + \dfrac{((5\sqrt{3} - 3) - 4R)^2}{16} = \dfrac{(4R + (3 - \sqrt{3}))^2}{16} \implies$

$32\sqrt{3}R = 2(42 - 15\sqrt{3}) \implies \boxed{R = \dfrac{14\sqrt{3} - 15}{16}}$

$\implies A_{R} = \dfrac{813 - 420\sqrt{3}}{256}\pi$ and $A_{r} = \dfrac{12 - 6\sqrt{3}}{16}\pi \implies$

$A_{T} = A_{R} + 2A_{r} = \dfrac{9(133 - 68\sqrt{3})}{256}\pi = \dfrac{a(b - c\sqrt{d})}{e} \pi \implies a + b + c + e = \boxed{469}$ .