What's wrong with my calculation for the sphericity of the Disdyakis Triacontahedron?

Question

I was looking at the Wikipedia page for Sphericity, and it lists that of the Disdyakis Triacontahedron at $0.9857$. This makes complete sense. However, I checked the formula for sphericity that they use, which is $$\Psi=\frac{\pi^{\frac{1}{3}}(6V)^{\frac{2}{3}}}{A}$$ and plugged it into the formulas they use for the volume, which is $$\frac{180}{11}\sqrt{179-24\sqrt{5}}$$ and the surface area, which is $$\frac{180}{11}(5+4\sqrt{5})$$ and got $0.6836$ instead. Where did my calculation go wrong?

Answer 1

I get an answer of about $0.9857$ (with the same exact formula as Wikipedia) if I assume that the two constants in area and volume are swapped: if the disdyakis triacontahedron should actually have \begin{align} V &= \frac{180}{11} (5 + 4\sqrt 5) s^3 \\ A &= \frac{180}{11} \sqrt{179 - 24 \sqrt 5} s^2 \end{align} This seems like a plausible mistake to make.

I'm not entirely sure what $s$ is in these formulas, so I can't actually confirm that either of them is correct. Mathematica's PolyhedronData command gives the following values for a disdyakis triacontahedron whose shortest edge length is $1$: \begin{align} V &= \frac{1}{5} \sqrt{39612 \sqrt{5}+88590} \\ A &= \sqrt{\frac{22626}{5}+\frac{9738}{\sqrt{5}}} \end{align} Using these in the formula for sphericity gives the same result close to $0.9857$ again.