Show that the area of the face of the coin is $\frac{a^2}{2}(\pi-7\tan\frac{\pi}{14})$ The diagram shows a British 50 pence coin. $$ How can i prove it?

Taking O to be the centre, let $AO=OE=r$ and we have $\angle AOB=\frac {2\pi}{7}, \angle AOE=\frac {6\pi}{7}, \angle AEO=\frac{\pi}{14}$

Using the Sine Rule in $\triangle AOE$, we have$$\frac{a}{\sin\left(\frac{6\pi}{7}\right)}=\frac{r}{\sin\left(\frac{\pi}{14}\right)}$$ $$\Rightarrow r=\frac{a}{2\cos\left(\frac{\pi}{14}\right)}$$

The area of $\triangle AOE$ is $$\frac 12 r^2\sin\left(\frac{6\pi}{7}\right)$$

This simplifies to $$\frac{a^2}{4}\tan\left(\frac{\pi}{14}\right)$$

Hence the required area is$$7\times\left[\frac{a^2}{2}\times\frac {\pi}{7}-2\times\frac{a^2}{4}\tan\left(\frac{\pi}{14}\right)\right]$$

Hence the answer.