Let $s$ be the edge length of a regular dodecahedron. As a function of $s$, what is the dodecahedron's minimum and maximum diagonal (i.e. cross-section)?
2026-03-25 12:24:49.1774441489
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Minimum and maximum diagonal of a dodecahedron
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For unit edge length, the longest diagonal is 2.802517076888147. If this is not what you
mean (as joriki's comment question indicates), my apologies.
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In this post of mine I posted this link to a small notebook (see section 15.3. Dodecaedro, in Portuguese), where I compute the relation
$$\frac{l}{d}=\frac{1}{2\sqrt{3}\cos \left( \frac{\pi }{5}\right) }=\frac{2}{% \sqrt{3}\left( 1+\sqrt{5}\right) }\approx 0.356\,82,$$
where $l$ is the edge length (denoted $s$ by you) and $d$ is the diagonal length connecting two opposite vertices.
So
$$\frac{d}{l}=2\sqrt{3}\cos \left( \frac{\pi }{5}\right) =\frac{\sqrt{3}% \left( 1+\sqrt{5}\right) }{2}\approx 2.802\,5.$$
Top figure: edges (black), diagonal (red), axes of faces (magenta). Bottom figure: cross-section rotated with respect to the top figure.
It you want the cross-section area, then I think the bottom figure may help. One would have to compute the height (brown), the radius of the circumscribed circle to the faces, and its apothem.
As far as the other "diagonals" I have not computed them.
Wikipedia gives the radius of the circumscribed sphere as $\frac{\sqrt{15}+\sqrt{3}}{4}s\approx 1.401s$. Your maximum diameter is twice this. It also gives the radius of the inscribed sphere as $\frac{\sqrt{250+110\sqrt{5}}}{20}s\approx 1.1135s$. Is this whatyou menant by minimum diagonal-it is between two face centers.