What's the actual volume of a Great Stellated Dodecahedron? (Im getting seemingly different formulas) So I was trying to derive the formula for the Great Stellated Dodecahedron starting from a unitary side length Dodecahedron. I managed only to get so far.
So when I went to check the actual formula, I found that Mathworld and WolframAlpha were giving me seemingly different formulas, but both assuming unit edge length.
The WolframAlpha formula:

$V = \frac{1}{4}(7 \sqrt5 - 15) $

The Mathworld formula:

$V = \frac{5}{4}(3 + \sqrt5 ) $

I could be misreading the entry on Mathworld (I don't speak english natively)
Now im curious about this difference, both are clearly formulas for something but they can't be both the volume of the Great Stellated Dodecahedron. What am I missing?
 A: Suppose the pentagram faces have unit edge length. Then the $20$ spikes of the great stellated dodecahedron have edge lengths $\varphi^{-3}$ (base) and $\varphi^{-2}$ (from tip) where $\varphi$ is the golden ratio. The base circumradius is $\frac1{\sqrt3}\varphi^{-3}$, yielding a height of $\varphi^{-3}\sqrt{\varphi^2-\frac13}$ and a volume for each spike of
$$\frac13×\frac{\sqrt3}4\varphi^{-6}×\varphi^{-3}\sqrt{\varphi^2-\frac13}=\frac1{12}\varphi^{-9}\sqrt{3\varphi^2-1}=\frac1{12}\varphi^{-7}$$
The whole solid has $20$ spikes, so we multiply this by $20$ to get the spike volume as $\frac53\varphi^{-7}$. To this we add the hidden icosahedron's volume of
$$\frac5{12}(3+\sqrt5)\varphi^{-9}=\frac56\varphi^{-7}$$
and obtain the volume of the great stellated dodecahedron as
$$\frac52\varphi^{-7}=0.086104\dots$$
We can rewrite the MathWorld $V$ as $\frac52\varphi^2$. This means the MathWorld formula is correct if the edge length of the hidden icosahedron (length of each side of the spike base) is $1$. The Wolfram Alpha formula seems completely wrong.

MathWorld says

Cumulating a unit dodecahedron to produce a great stellated dodecahedron…

"Cumulation" is another word for augmentation, adding pyramids to each face of the original polyhedron.
This is rubbish. You need to augment an icosahedron to produce a great stellated dodecahedron.
