Circumscribing a hexagonal prism

Label the diagram as follows:

Since $BC = \frac{1}{2}A$ and $\angle CBO = 30°$ , $BO = \frac{\sqrt{3}}{3}A$ and $CO = \frac{\sqrt{3}}{6}A$ .

Since $BH = A$ and $BO = \frac{\sqrt{3}}{3}A$ , by the Pythagorean Theorem on $\triangle BHO$ , $HO = \frac{\sqrt{6}}{3}A$ .

Since $\triangle GEC \sim \triangle HOC$ by AA similarity, $\frac{CE}{EG} = \frac{CO}{HO}$ . Substituting $EG = 30$ , $CO = \frac{\sqrt{3}}{6}A$ , and $HO = \frac{\sqrt{6}}{3}A$ and solving gives $CE = \frac{15\sqrt{2}}{2}$ .

$EO$ is the height of equilateral $\triangle DFO$ with a side length of $DF = 10$ , so $EO = 5\sqrt{3}$ .

By segment addition, $CO = CE + EO$ . Substituting $CO = \frac{\sqrt{3}}{6}A$ , $CE = \frac{15\sqrt{2}}{2}$ , and $EO = 5\sqrt{3}$ and solving gives $A = 15\sqrt{6} + 30 \approx \boxed{66.74}$ .

Extending the lines of the Hexagon will create a triangle of side length $3a$

Fomula for Tetrahedron's heigt: $H = \frac{A\sqrt6}{3}$

Wich makes $A = \frac{3H}{\sqrt6}$

The total height is the prism's height + the smaller tetrahedron's height (on top of the prism)

$H = h + \frac{3a \sqrt6}{3}$

$A = \frac{3 ( h + \frac{3a \sqrt6}{3} ) }{\sqrt6} = \frac{3h + \frac{9a \sqrt6}{3} ) }{\sqrt6} = 66.74$