It's All Tetrahedrons

$\triangle{AEC}$ is an isosceles triangle $\implies EM$ is the perpendicular bisector of base $AC$ $\implies \triangle{MEC}$ is a right triangle and $AM = MC$ .

Since $\angle{ACB}$ is a common angle to both right triangle $ABC$ and $MEC \implies \triangle{ABC} \sim \triangle{MEC} \implies \dfrac{2m}{m} = \dfrac{x + a}{a} \implies x = a$ and the pythagorean theorem in $\triangle{ABC} \implies y = 2m = \sqrt{3}a$ .

The surface area $S = \dfrac{a^2}{2}(\sqrt{3} + \sqrt{18} + \sqrt{6}) + A_{4}$

For $A_{4}$ :

Using Heron's formula $\implies A_{4} = \dfrac{a^2}{4}\sqrt{(5 + \sqrt{7})(5 - \sqrt{7})(\sqrt{7} - 1)(\sqrt{7} + 1)} = \dfrac{a^2}{4}\sqrt{18*6} = \dfrac{3\sqrt{3}}{2}a^2$

$\implies S = \dfrac{a^2}{2}(4\sqrt{3} + \sqrt{6} + 3\sqrt{2}) = \dfrac{1}{2} \implies$

$a = \dfrac{1}{\sqrt{4\sqrt{3} + \sqrt{6} + 3\sqrt{2}}} \approx \boxed{0.27096}$ .