Calculate volume of a uniform tetradecahederon

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The solid is shown with all sides a = 20 in green and those on the 'equator' with length (R = circumscribing radius) in white. Take the central plane of symmetry yielding a hexagonal cross section shown.

$R \cos \alpha = 20$ and $2R \sin \frac{\alpha}{2} = 20$ equating the two gives a quadratic in $\sin \frac{\alpha}{2}$ solving which gives $\sin \frac{\alpha}{2} = \frac{\sqrt{3}-1}{2}$ that is $\alpha = 42.94$ and R = 27.321

Area of top hexagon $A_T = 1039.23$ and area of equatorial hexagon $A_E = 1939.23$ Height of the two frustra of hexagonal pyramids = $R \sin \alpha = 18.612$

Volume of the solid = 2 x volume of each frustrum = $\frac{2h}{3}(A_T+\sqrt{A_T \times A_E}+A_E)$ = $\boxed{54572}$

@Niranjan Khanderia