Big Pentagon Small Pentagon - 2

Using cosine rule as @Fletcher Mattox has used in solving the problem Big Hexagon Small Hexagon - 2 . Let the side length of the big pentagon be $1$ . The the side length of the small pentagon is given by:

$\begin{aligned} s^2 & = \left(\frac 13\right)^2 + \left(\frac 23\right)^2 - 2 \cdot \frac 13 \cdot \frac 23 \blue{\cos 108^\circ} & \small \blue{\text{Note that }\cos (180^\circ - \theta) = - \cos \theta} \\ & = \frac 59 + \frac 49 \blue{\cos 72^\circ} \\ & = \frac 59 + \frac 49 (2\cos^2 36^\circ - 1) & \small \blue{\text{and } \cos 36^\circ = \frac {1+\sqrt 5}4 = \frac \varphi 2} \\ & = \frac 19 + \frac 29 \varphi^2 \\ & = \frac {1+2(\varphi + 1)}9 \\ & = \frac {3+2\varphi}9 \end{aligned}$

The ratio of areas $\dfrac {A_s}{A_B} = \dfrac {s^2}{1^2} = \boxed{\dfrac {3+2\varphi}9}$ .