In order to find the interior angles of the green triangle, first let's find the angle CAB in the upper blue triangle, using its tangent relation:
[tex]\begin{gathered} \tan (\text{CAB})=\frac{2}{4}=\frac{1}{2} \\ \text{CAB}=\tan ^{-1}(\frac{1}{2}) \\ \text{CAB}=26.565\degree \end{gathered}[/tex]Since the figure is symmetrical, angle FAE is also 26.565° and triangle AEC is isosceles.
Now, we can find angle CAE:
[tex]\begin{gathered} CAE+\text{CAB}+\text{FAE}=90 \\ CAE+26.565+26.565=90 \\ CAE+53.13=90 \\ CAE=36.87\degree \end{gathered}[/tex]Finding angles ECA and AEC:
[tex]\begin{gathered} ACE+\text{AEC}+CAE=180 \\ ACE+ACE+36.87=180 \\ 2ACE=143.13 \\ ACE=71.565\degree \\ \\ \text{AEC}=ACE \\ \text{AEC}=71.565\degree \end{gathered}[/tex]In order to calculate the area of green triangle, we can first calculate the length of all three sides using Pythagorean theorem:
[tex]\begin{gathered} AC^2=AB^2+BC^2 \\ AC^2=16+4 \\ AC^2=20 \\ AC=4.47 \\ \\ AE=AC=4.47 \\ \\ EC^2=ED^2+DC^2 \\ EC^2=4+4 \\ EC^2=8 \\ EC=2.83 \end{gathered}[/tex]So the area of the triangle using Heron's formula is:
[tex]\begin{gathered} A=\sqrt[]{p(p-a)(p-b)(p-c)} \\ p=\frac{4.47+4.47+2.83}{2}=5.885 \\ A=\sqrt[]{5.885(1.415)\mleft(1.415\mright)\mleft(3.055\mright)} \\ A=6 \end{gathered}[/tex]So the area of the triangle 6 mm².
