Given the octagon $ABCDEFGH$, with $\angle GHA=\angle ABC=\angle CDE=\angle EFG=120^o$, $AE\perp GC$ and $AO=OE=OG=OC$ find the area of the octagon.
We have that $\triangle AGH, \triangle GFE, \triangle DEC, \triangle CBA$ isosceles. $ACEG$ is a square with side $2$. $\angle HGA=\angle HAG=\angle BAC=\angle BCA=\angle DCE=\angle DEC=\angle GEF=\angle EGF$.
Also we have that $HBDF$ is a square. Basically all you need to work out is $GH$, but I can't work it out. Could you please explain to me how to solve the question?
Best Answer
With this symmetrical figure, focus on $\triangle AOK$ , $\triangle ABK$,$\triangle AOB$.
For the simpliticy of calulation, first let us assume BK=1 $\to$AB=2,AK=OK=$\sqrt 3$:
\begin{align} \frac{S_{ABCDEFGH}}{S_{ACEG}} =\frac{S_{\triangle AOB}}{S_{\triangle AOK}}=\frac{S_{\triangle AOK}-S_{\triangle ABK}}{S_{\triangle AOK}}=1-\frac{S_{\triangle ABK}}{S_{\triangle AOK}}=1-\frac{BK}{OK}=1-\frac{1}{\sqrt{3}}\end{align}
Now scale back with $S_{ACEG}=4$, $S_{ABCDEFGH}$= 4$(1-\frac{1}{\sqrt{3}})$