A d10, in roleplayers' lingo, is a 10-sided dice.
Several different shapes for this item exist, but the most common is that of two 5-sided pyramids with their equilateral bases lying on the same plane and offset by 36° (1/10, or half a side of a pentagon), as shown in the lower part of the image:
Both pyramids have their apex on the vertical of A, and the upper part of the drawing (shown at a different zoom below, so that all letters can be read) is built upon points B and C for the upper pyramid (projected: H and J) and I and F for the lower one (projectet: K and G).
It is trivial to see that the x component of AC is ABcos(36°)
Now, the planes defined by the 10 side faces of the pyramids meet on a jagged line that goes up and down the plane where the original bases lie. I suppose the meeting point between a face and the edge of an opposite pyramid are N and P but I'm not sure I'm right.
Suppose that ∠HAJ (and therefore also ∠KLG) is 90° (then, because of simmetry, ∠LPM and ∠LNM are 90° too), what's the LN/LP ratio?
I'm sure I can calculate ∠JHL and ∠HJL from the ratio between the sides (which is the same as the ratio between the x component of AC and and ABcos(36°)) and that I could get JL and HL from there but I can't think about a way to find NJ and PH.
Images created with geogebra
Not a very geometric solution but it gets the job done. Take three consecutive points on the jagged equator: $$(\cos\frac{\pi}5, -\sin\frac{\pi}5, x), (1,0,-x), (\cos\frac{\pi}5, \sin\frac{\pi}5, x)$$ where $x$ is some still to be determined deviation. The plane through these points determines the apex at $(0,0,(5+2\sqrt 5) x)$. Now $x$ needs to be taken such that the three points $$(-1, 0, x), (0,0,(5+2\sqrt 5) x), (1, 0, -x)$$ in the $XZ$ plane (vertex apex vertex) form a right angle. This gives $$(4+2\sqrt 5)(6+2\sqrt 5)x^2 = (44 + 20 \sqrt 5)x^2=1.$$ For this value of $x$ the length of the long edge is $\sqrt{1+(4+2\sqrt 5)^2x^2}=\tfrac12\sqrt{5+\sqrt5}$ and of the long diagonal $\sqrt{1+(6+2\sqrt 5)^2x^2} = \sqrt{\sqrt 5}$. The ratio long diagonal to long edge is therefore $\sqrt{\sqrt 5 -1}$.