I found this question from another user account. Here angle A, B,C = 120. If i construct AD and AC. Then AD and AC are equal? If equal then how? If we add EB then EAB are isosceles triangle. And the area of these isosceles triangle became 2sin120. Anybody please tell me where did i mistake or how to progress. Thank you.
On
Equality of $AD$ and $AC$ following from the facts $AE=AB$ and $ED=BC$ (use Pythagoras' Theorem)
Hints to compute the area:
Let $\angle ADE$ be $\theta$. Also let $BC=CD=DE=x$.
Using sine rule in $\triangle AED$, we get $\frac{\sin\theta}{2}=\frac{\cos\theta}{x}$ (or simply, $\tan\theta=\frac{2}{x}$).
After marking some angles and using sine rule in $\triangle ADC$, we get $$\frac{\sin(2\theta-60)}{x}=\frac{\sin(120-\theta)}{\sqrt{2^2+x^2}}$$
which is same as $\sin(2\theta-60)=\sin(120-\theta)\cos\theta$.
Solving this and using the answers would give the area.
By considering the angles of the polygon and the position of the midpoint of $BE$ we have that such pentagon can be decomposed into an equilateral triangle with side length $2$ and three equilateral triangles with side length $\sqrt{3}$, so its area is $\frac{\sqrt{3}}{4}\left(2^2+3\cdot \sqrt{3}^2\right)=\frac{13}{4}\sqrt{3}$.