Area Of A Quadrilateral!

I filled in the angles in the diagram above using properties of an equilateral triangle, isosceles triangle and the measure of a straight angle.

Let $m\angle{FBC} = \theta, \:\ m\angle{FCB} = \lambda, \:\ FE = a$ and $BE = x$ .

Using $\triangle{DBC} \implies \lambda = 60 - \theta$

Using the law of sines on $\triangle{FBE} \implies x = \dfrac{a}{\sin(\theta)}$

Using the law of sines on $\triangle{FEC} \implies 2 = \dfrac{a}{\sin(60 - \theta)} \implies a = 2\sin(60 - \theta) \implies$

$x = \dfrac{2(\sin(60^{\circ})\cos(\theta) - \cos(60^{\circ})\sin(\theta))}{\sin(\theta)}$ $= \sqrt{3}\cot(\theta) - 1$ .

Using $\triangle{FBE} \implies \cot(\theta) = \dfrac{1}{a}$ and $x^2 = a^2 + 1 \implies a = \sqrt{x^2 - 1} \implies$

$\cot(\theta) = \dfrac{1}{\sqrt{x^2 - 1}} \implies x = \dfrac{\sqrt{3}}{\sqrt{x^2 - 1}} - 1 \implies$ $(x + 1)\sqrt{x^2 - 1} = \sqrt{3}$

$\implies (x^2 + 2x + 1)(x^2 - 1) = 3 \implies x^4 + 2x^3 - 2x - 4 = 0$

$\implies x^3(x + 2) - 2(x + 2) = 0 \implies (x^3 - 2)(x + 2) = 0$

$x > 0 \implies x \neq -2 \implies x = 2^{\frac{1}{3}} \implies a = \sqrt{2^{\frac{2}{3}} - 1}$ .

Let $FC = l$

$m\angle{BEF} = 90 - \theta \implies m\angle{FEC} = 90 + \theta$ and $\sin(\theta) = \dfrac{a}{x} \implies$

Using the law of cosines on $\triangle{FEC} \implies$ $l^2 = a^2 + 1 - 2a\cos(90 + \theta) =$

$a^2 + 1 + 2a\sin(\theta) =$ $a^2 + 1 + \dfrac{2a^2}{x} = 2^{\frac{2}{3}} - 1+ 1 + (2^{\frac{2}{3}})(2^{\frac{2}{3}} - 1) =$

$2^ {\frac{2}{3}}(1 + 2^{\frac{2}{3}} - 1) = 2^{\frac{4}{3}} \implies l = 2^{\frac{2}{3}}$

$\implies A_{\triangle{FEC}} = \dfrac{1}{2}2^{\frac{2}{3}}(a\sin(30^{\circ}) =$ $\dfrac{1}{2}2^{\frac{2}{3}}\sqrt{2^{\frac{2}{3}} - 1} (\dfrac{1}{2}) =$ $\dfrac{\sqrt{2^{\frac{2}{3}} - 1}}{2^{\frac{4}{3}}}$

$A_{\triangle{FBE}} = \dfrac{1}{2}(a)(1) = \dfrac{\sqrt{2^{\frac{2}{3}} - 1}}{2}$

$A_{\triangle{FDB}} = \dfrac{1}{2}(1)(\dfrac{\sqrt{3}}{2}) = \dfrac{\sqrt{3}}{4}$

Let $AF = m$

Using the law of cosines on $\triangle{ADF} \implies$

$m^2 = 2 - 2\cos(120^{\circ}) = 2 + 2(\dfrac{1}{2}) = 3 \implies m = \sqrt{3} \implies$

$A_{\triangle{ADF}} = \dfrac{1}{2}\sqrt{3}(\sin(30^{\circ}) = \dfrac{\sqrt{3}}{4}$

$\implies A = \dfrac{\sqrt{2^{\frac{2}{3}} - 1}}{2^{\frac{4}{3}}} + \dfrac{\sqrt{2^{\frac{2}{3}} - 1}}{2} +$ $\dfrac{\sqrt{3}}{2} \approx \boxed{1.55339022}$