Circular arcs in a pentagon

Let $O$ be the center point, $C$ be the center of the top right circular arc, and $A$ , $B$ , $D$ , and $E$ be labelled as follows:

By symmetry, $\angle AOB = \frac{360°}{5} = 72°$ , and since a central angle is twice the angle that subtends the same arc, $\angle ACB = 2 \cdot 72° = 144°$ , which means $\angle BCD = 72°$ and $\angle BCO = 108°$ . Using trigonometry on right triangle $\triangle BCD$ and using the given information that $AB = 2$ , $BC = \frac{1}{\sin 72°}$ , and since $BC$ is a radius, $BC = CE = CO = \frac{1}{\sin 72°}$ .

Also by symmetry, $\angle COE = \frac{360°}{5} = 72°$ , and since $\triangle COE$ is an isosceles triangle, $\angle ECO = 36°$ .

The area of region between the line $BO$ and arc $BO$ is the difference between the area of the sector $BCO$ and the triangle $\triangle BCO$ , which is $A_{BO} = (\frac{108°}{360°}\pi - \frac{1}{2} \sin 108°)(\frac{1}{\sin 72°})^2$ , and similarly, $A_{EO} = (\frac{36°}{360°}\pi - \frac{1}{2} \sin 36°)(\frac{1}{\sin 72°})^2$ .

The desired area is then $A_P = 2A_{BO} - 4A_{EO} = 2(\frac{108°}{360°}\pi - \frac{1}{2} \sin 108°)(\frac{1}{\sin 72°})^2 - 4(\frac{36°}{360°}\pi - \frac{1}{2} \sin 36°)(\frac{1}{\sin 72°})^2$ which simplifies to $A_P = (2 \sin 36° - \sin 108° + \frac{1}{5}\pi)(\frac{1}{\sin 72°})^2$ . Substituting $\sin 36° = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$ and $\sin 72° = \sin 108° = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$ further simplifies it to $A_P = (\frac{\sqrt{10 - 2\sqrt{5}}}{2} - \frac{\sqrt{10 + 2\sqrt{5}}}{4} + \frac{1}{5}\pi) \cdot \frac{16}{10 + 2\sqrt{5}} \approx \boxed{0.942868444}$ .