Isosceles Triangles

First we need to obtain $\theta$ to get the measure of the desired angles to obtain the area and perimeter.

$m\angle{ECD} = 180^{\circ} - 2\theta, m\angle{ECB} = 2\theta - 60^{\circ}, m\angle{BEC} = 300^{\circ} - 4\theta$

$m\angle{ABE} = 240^{\circ} - 2\theta, m\angle{AEB} = 3\theta - 120^{\circ}$ and $m\angle{A} = 60^{\circ} - \theta \implies$

$60^{\circ} - \theta = 3\theta - 120^{\circ} \implies \theta = 45^{\circ} \implies m\angle{ECD} = 90^{\circ}$

and $m\angle{A} = m\angle{AEB} = 15^{\circ} \implies m\angle{ABE} = 150^{\circ}$

$\triangle{ECD}$ is a $(45^{\circ},45^{\circ},90^{\circ})$ triangle $\implies \overline{ED} = \sqrt{2}a$

and in $\triangle{ABE}$ :

$\overline{AE}^2 = 2a^2(1 - \cos(150^{\circ}) = 4a^2\sin^2(75^{\circ}) \implies \overline{AE} = 2a\sin(75^{\circ}) = \dfrac{\sqrt{3} + 1}{\sqrt{2}}a$

$\implies \overline{AD} = \overline{AE} + \overline{ED} = \dfrac{3 + \sqrt{3}}{\sqrt{2}}a$ and $h = a\sin(45^{\circ}) = \dfrac{a}{\sqrt{2}} \implies$

$\boxed{A_{\triangle{ACD}} = \dfrac{3 + \sqrt{3}}{4}a^2}$

For $P_{\triangle{EBC}}$ :

$\overline{BC}^2 = 2a^2(1 - \cos(120^{\circ}) = 4a^2\sin^2(60^{\circ}) \implies \overline{BC} = 2a\sin(60^{\circ}) = \sqrt{3}a$

and $\overline{BE} = \overline{EC} = a \implies \boxed{P_{\triangle{EBC}} = (\sqrt{3} + 2)a}$

$A_{\triangle{ACD}} + P_{\triangle{EBC}} = 1 \implies (3 + \sqrt{3})a^2 + 4(\sqrt{3} + 2)a - 4 = 0 \implies$

dropping the negative root we have:

$a = \dfrac{2(-(\sqrt{3} + 2) + \sqrt{5(\sqrt{3} + 2)})}{3 + \sqrt{3}} =$ $\dfrac{2(-3 - \sqrt(3) + \sqrt{30(2 - \sqrt{3})(2 + \sqrt{3})})}{6}$

$= \dfrac{\sqrt{30} - \sqrt{3} - 3}{3} = \dfrac{\sqrt{3 * 10} - \sqrt{3} - 3}{3} = \dfrac{\sqrt{\alpha * \beta} - \sqrt{\alpha} - \alpha}{\alpha} \implies$

$\alpha + \beta = \boxed{13}$ .