We can draw the diagram.
DIAGRAM
The triangle $\triangle \text{ABC}$ is a right angle triangle, so we can apply the Pythagorean theorem.
$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}$
$\Rightarrow \text{(AC)}^{2} = \text{(AB)}^{2} + \text{(BC)}^{2}$
$\Rightarrow \text{(AC)}^{2} = \text{(2)}^{2} + \text{(2)}^{2}$
$\Rightarrow \text{(AC)}^{2} = 4 +4$
$\Rightarrow \text{(AC)}^{2} = 8$
$\Rightarrow \boxed{\text{AC} = \sqrt{8}}$
Similarly $\boxed{\text{BD} = \sqrt{8}}$
$\therefore$ The sum of the perimeters of the triangles = sum of all sides of square + $2 \times $ sum of diagonal lengths
$= 4 \times 2 + 2 \times \sqrt{8}$
$= 8 + 2 \times \sqrt{4 \times 2} = 8 + 4 \sqrt{2} = 4(2 + \sqrt{2}).$
Correct Answer $: \text{D}$