On Area percentage -1

In the Discussion I will generalise a case one of which is in this question .

In $\parallel_{gm} JOGC$ assuming $\overline{JC}=\overline{CG}$ the $\red{red}$ area is $\frac{100}{\pi}(\angle COG - \frac{\sin^2(\angle COG)}{\sin(2\angle COG)})\%$ of the whole circle(all the angles are measured in radians only)

Proof:

Let us denote $\angle COG=\alpha$ overall the proof (to make things easier to write for me)

And let $\overline{JC}=\overline{CG}=a;\overline{OC}=r$

Now as $JOCG$ is a parallelogram $\therefore$ $\overline{JO} || \overline{CG} \Rightarrow \angle JOG=\angle CGD$

Also as the diagonal of a parallelogram bisects its vertex angles ${2\alpha=\angle CGD}$

In $\triangle CGD$ : $\sin(\angle CGD)=\sin(2\alpha)=\frac{\overline{CD}}{a}$ $\Rightarrow \overline{CD}=a\sin(2\alpha)$ In $\triangle COD$ : $\sin(\alpha)=\frac{\overline{CD}}{r}$ $\overline{CD}=r\sin(\alpha)$ $\Rightarrow r\sin(\alpha)=a\sin(2\alpha)$ $\Rightarrow r=a\frac{\sin(2\alpha)}{\sin(\alpha)}$ Area of $||_{gm}JOGC=A_1=\overline{OG}\times\overline{CD}=a\times a\sin(2\alpha)=a^2\sin(2\alpha)$

Area of circle $=A_2=\pi r^2=\pi a^2\frac{\sin^2(2\alpha)}{\sin^2(\alpha)}$

Area of the sector $\frac{\angle JOG}{2\pi}A=\frac{\cancel{2}\alpha}{\cancel{2}\pi}A=\frac{\alpha}{\pi}A=\frac{\alpha}{\cancel{\pi}}(\cancel{\pi} a^2\frac{\sin^2(2\alpha)}{\sin^2(\alpha)})=a^2\alpha \frac{\sin^2(2\alpha)}{\sin^2(\alpha)}$

Area of the $\red{red}$ region $=A-A_1$

Percentage of the area to the circles area$=\frac{A-A_1}{A_2}\times 100=(a^2\alpha \frac{\sin^2(2\alpha)}{\sin^2(\alpha)}-a^2\sin(2\alpha))\times \frac{100\sin^2(\alpha)}{\pi a^2\sin^2(2\alpha)}=\boxed{\frac{100}{\pi}({\alpha}-\frac{\sin^2(\alpha)}{\sin(2\alpha)})}$

Bonus:I have proved it for the case when all sides of the parallelogram are equal, you can find a formula for general case quadrilateral