Regular heptagon area formula Be a regular heptagon. Can anyone demonstrate this formula?

I know:
Let apothem = ap
side = l
$ S=\frac{p.ap}{2}\implies S = \frac{7l.ap}{2}\\
\triangle OBC: OC^2 = a^2+l^2\\
\triangle PAB: b^2+l^2 = PB^2\\
OC^2-a^2=PB^2-b^2$
but i can't go on
 A: 
Hints: You can use this formula to find area of regular polygon:
$$A=\frac {n\cdot l^2}{4 tan \frac {\pi}n}$$
where n is number of sides, $l$. is length of side. For $l=1$ and $n=7$ we get $A=3.63..$
Now you can find a as :
$\angle KBC=\angle OFH$
because their rays are perpendicular on each other. So we have:
$\triangle KBC\sim \triangle HOF$
$\frac{KB=a}{FH}\frac {BC=1}{OF=r}$
$r=\frac{IF=1/2}{\sin \frac{2\pi}{14}}$
You  finally find $ r\approx 1.2$ and $a=0.78$. For b we have:
$b=FL+a$
You have to show $\angle GFM=\angle HFE\approx 25.7^o$ and in triangle FGM you find:
$FL\approx 0.97\Rightarrow b=0.97+0.87\approx 1.76$
$A=\sqrt 7 a\cdot b\approx 3.63$
That is the formula is correct.
I you want to find formula, then we have:
Sum of angles of heptagon s is:
$s=(2\times 7-4)\frac{\pi}2=5\pi$
$\angle ABC=\frac{5\pi}7$
$\angle KBC=\frac{5\pi}7-\frac{\pi}2=\frac{3\pi}{14} $
$a=1\cdot \cos KBC=\cos\frac{3\pi}{14}\space\space\space\space (1)$
$b_1=FH= FE\cos EFH$
$\angle EFH=\frac {5\pi}7-\frac {\pi}2-\frac{\angle EFH}2$
$\Rightarrow \angle EFH=\frac {\pi}7$
$\Rightarrow b_1=FH=\frac{\pi}7$
$b=a+b_1=\cos \frac{\pi}7+\cos\frac {3\pi}{14}\space\space\space\space (2)$
$A=\frac{7\cdot \cos\frac{\pi}7}{4\cdot \sin \frac {\pi}7}\space\space\space\space (3)$
now you have to find $A=\sqrt 7 ab$ from relation (1) , (2) and (3).
A: Let $d$ be the diameter of the circumcircle of $ABCDEFG$. Note that $a = BC \sin \frac{2\pi}{7}=d\sin \frac{\pi}{7}\sin \frac{2\pi}{7}$ and $b=AD \sin \frac{2\pi}{7} = d \sin \frac{3\pi}{7} \sin \frac{2\pi}{7}$. On the other hand, the area of $ABCDEFG$ equals $7 \cdot \frac 12 \left(\frac{d}{2}\right) \sin \frac{2\pi}{7}$. The problem boils down to showing that
$$x=\sin \frac{\pi}{7}\sin \frac{2\pi}{7}\sin \frac{3\pi}{7} = \frac{\sqrt 7}{8}.$$
Using $2\sin^2 t = 1-\cos 2t$ we see that
\begin{align*}
8x^2 = 8\sin^2 \frac{\pi}{7}\sin^2\frac{2\pi}{7}\sin^2 \frac{3\pi}{7} &= \left(1-\cos \frac{2\pi}{7}\right)\left(1-\cos \frac{4\pi}{7}\right)\left(1-\cos \frac{6\pi}{7}\right) \newline
&= 1 - \left( \cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}\right) + \newline
& \quad +\left( \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}+\cos \frac{4\pi}{7}\cos \frac{6\pi}{7}+\cos \frac{6\pi}{7}\cos \frac{2\pi}{7}\right) - \newline
& \quad - \cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{6\pi}{7}.
\end{align*}
Using $\cos t \cos u = \dfrac{\cos(t+u)+\cos(t-u)}{2}$ we get
\begin{align*}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}+&\cos \frac{4\pi}{7}\cos \frac{6\pi}{7}+\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} \newline 
&=\frac{\cos \frac{6\pi}{7}+\cos \frac{2\pi}{7}}{2}+\frac{\cos \frac{10\pi}{7}+\cos \frac{2\pi}{7}}{2}+\frac{\cos \frac{8\pi}{7}+\cos \frac{4\pi}{7}}{2}\newline 
&=\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}
\end{align*}
since $\cos\frac{8\pi}{7}=\cos\frac{6\pi}{7}$ and $\cos\frac{10\pi}{7}=\cos\frac{4\pi}{7}$.
Moreover, using $\sin 2t = 2 \sin t \cos t$, we have
\begin{align*}
\cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} &= \cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{8\pi}{7} \newline 
&= \frac{\sin \frac{2\pi}{7}\cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{8\pi}{7}}{\sin\frac{2\pi}{7}} \newline 
&= \frac{\frac 12 \sin \frac{4\pi}{7}\cos \frac{4\pi}{7}\cos \frac{8\pi}{7}}{\sin\frac{2\pi}{7}} \newline &= \frac{\frac 14 \sin \frac{8\pi}{7}\cos \frac{8\pi}{7}}{\sin\frac{2\pi}{7}} \newline &= \frac{\frac 18 \sin \frac{16\pi}{7}}{\sin\frac{2\pi}{7}} \newline &= \frac 18.\end{align*}
It follows that $8x^2= 1-\frac 18 = \frac 78$, i.e. $x=\frac{\sqrt{7}}{8}$, as desired.
