Cutting a regular dodecahedron in half - Part 2

One side of the hexagon is the midsegment of the isosceles trapezoid formed by three sides of a pentagon and one diagonal.

Since the side of the pentagon is $1$ and the diagonal of the pentagon is $\frac{1+\sqrt{5}}{2}$ , the midsegment (and the side of the hexagon) is the average of these two sides, which is $\frac{3 + \sqrt{5}}{4} \approx \boxed{1.309}$ .

The sides of the hexagon connect the midpoints of two non-adjacent sides of a pentagon, a face of the dodecahedron. As such, their length is $1+\sin(18^{o})=\frac{1}{4}(3+\sqrt{5})$ $\approx \boxed{1.309}$ , as illustrated in the attached, rather primitive graphic. Note that the angle between the green lines and the sides of the pentagon is $18^{o}$ .