Truncated Right Hexagonal Prism

The average length of the lateral edges is $\dfrac{5}{3}$ .

Let $O':(0,0,0)$ be center of the hexagonal base and erect a normal at $O'$ so that the top surface is centered at $O: (0,0,\dfrac{5}{3})$ .

Using coordinates for the surface we have:

$F: (\dfrac{1}{2},-\dfrac{\sqrt{3}}{2},2), A:(1,0,2), B:(\dfrac{1}{2},\dfrac{\sqrt{3}}{2},2)$ $C:(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2},2), D:(-1,0,1)$

and $E:(-\dfrac{1}{2},-\dfrac{\sqrt{3}}{2},1)$

$\implies FA = AB = BC = DE = 1$ and $CD = EF = \sqrt{2}$

and the radii $OF =OA = OB = OC = \dfrac{\sqrt{10}}{3}$ and $OD = OE = \dfrac{\sqrt{13}}{3}$

$h = \dfrac{\sqrt{31}}{6} \implies A_{\triangle{FOA}} = \dfrac{\sqrt{31}}{12} = A_{\triangle{AOB}} =$ $A_{\triangle{BOC}}$ .

$h^{*} = \dfrac{\sqrt{43}}{6} \implies A_{\triangle{DOE}} = \dfrac{\sqrt{43}}{12}$ .

$|\vec{U} X \vec{V}| = |\vec{U}| |\vec{V}|\sin(\theta) = |\vec{U}| d \implies$ $d = \dfrac{|\vec{U} X \vec{V}|}{ |\vec{U}|}$

$\vec{U} = \vec{i} + 0\vec{j} + \vec{k}$

$\vec{V} = \dfrac{1}{2}\vec{i} + \dfrac{\sqrt{3}}{2}\vec{j} + \dfrac{2}{3}\vec{k}$

$\implies$

$\vec{U} X \vec{V} = -\dfrac{\sqrt{3}}{2}\vec{i} - \dfrac{1}{6}\vec{j} + \dfrac{\sqrt{3}}{2}\vec{k}$

$\implies (|\vec{U} X \vec{V}| = \dfrac{\sqrt{55}}{6}$ and $|\vec{U}| = \sqrt{2} \implies$ $d = \dfrac{1}{6}\sqrt{\dfrac{55}{2}} \implies$ $A_{\triangle{COD}} = A_{\triangle{EOF}} =$

$\dfrac{1}{2}(\sqrt{2})(\dfrac{1}{6})(\sqrt{\dfrac{55}{2}}) = \dfrac{\sqrt{55}}{12}$

$\implies A_{surface} = \dfrac{2\sqrt{55} + 3\sqrt{31} + \sqrt{43}}{12}$

$A_{hexagonal base} = 6(\dfrac{1}{2})(1)(\dfrac{\sqrt{3}}{2}) = \dfrac{3\sqrt{3}}{2}$

For the area of the trapezoids we have:

$A_{FAF'A'} = A_{ABA'B'} = A_{BCB'C'} = \dfrac{1}{2}(4)(1) = 2$

and

$A_{CDC'D'} = A_{EFE'F'} = \dfrac{1}{2}(3)(1) = \dfrac{3}{2}$

and $A_{DED'E'} = \dfrac{1}{2}(2)(1) = 1$

$\implies$ The total area of the trapezoids $A_{T} = 3(2) + 2(\dfrac{3}{2}) + 1 = 10$ .

$\implies A_{s} = \dfrac{2\sqrt{55} + 3\sqrt{31} + \sqrt{43}}{12} + \dfrac{3\sqrt{3}}{2} + 10 =$

$\dfrac{120 + 18\sqrt{3} + 2\sqrt{55} + 3\sqrt{31} + \sqrt{43}}{12} =$

$\dfrac{a + b\sqrt{c} + d\sqrt{e} + c\sqrt{f} + \sqrt{g}}{h} \implies$

$a + b + c + d + e + f + g + h = \boxed{284}$ .