<s>[INST] 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
 [/INST] 
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).
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