Pentagon to Decagon

Let $x$ be the distance between a vertex of the decagon and the closest vertex of the pentagon, and $y$ be the side length of the decagon, as shown below on the left.

From the right triangle shown on the right, we get $\dfrac{y}{2} = x \cos 36^\circ$ , i.e. $y = 2x \cos 36^\circ$ . Then looking at the side length of the pentagon,

$\begin{array}{rl} 2x + y &= \ \ 100 \\ \\ 2x + 2x \cos 36^\circ &= \ \ 100 \\ \\ x \ (1 + \cos 36^\circ) &= \ \ 50 \\ \\ x &= \ \ \dfrac{50}{1 + \cos 36^\circ} \end{array}$

and then the side length of the decagon is

$\begin{array}{rl} y &= \ \ 2x \cos 36^\circ \\ \\ &= \ \ \dfrac{100 \cos 36^\circ}{1 + \cos 36^\circ} \\ \\ &\approx \ \ \boxed{44.72} \end{array}$

By law of cosines,

$x^2=\left(50-\dfrac{x}{2}\right)^2+\left(50-\dfrac{x}{2}\right)^2-2\left(50-\dfrac{x}{2}\right)\left(50-\dfrac{x}{2}\right)(\cos 108^\circ)$

$x^2=2500-50x+\dfrac{x^2}{4}+2500-50x+\dfrac{x^2}{4}-2 \cos 108^\circ \left(2500-50x+\dfrac{x^2}{4}\right)$

$0.3455x^2+130.9x-6545.08=0$

Use the quadratic formula, to solve for $x$ , we have

$x\dfrac{-130.9\pm\sqrt{(-130.9)^2-4(0.3455)(-6545.08)}}{2(0.3455)}$

$x=-189.44\pm 123.16$

get the positive value

$\boxed{x\approx 44.72}$