I'm sure there's an easier way to do this, but this method does work:
First, AB = CD = CG + GF + FD, so FG = 2.
By the Pythagorean theorem, in triangle AFD we get
[tex]1^2+3^2=A F^2\implies A F=\sqrt{10}[/tex]
and in triangle BCG,
[tex]2^2+3^2=BG^2\implies BG=\sqrt{13}[/tex]
Angles AFD and EFG form a vertical pair, so they are congruent and have measure
[tex]m\angle EFG=\tan^{-1}3[/tex]
Similarly, angles BGC and FGE are congruent and have measure
[tex]m\angle FGE=\tan^{-1}\dfrac32[/tex]
Then the remaining angle in triangle EFG has measure
[tex]m\angle FEG=\pi-\tan^{-1}3-\tan^{-1}\dfrac32[/tex]
We can solve for the lengths of FE and GE exactly by applying the law of sines:
[tex]\dfrac{\sin\left(\pi-\tan^{-1}3-\tan^{-1}\frac 32\right)}2=\dfrac{\sin\left(\tan^{-1}3\right)}{GE}=\dfrac{\sin\left(\tan^{-1}\frac32\right)}{FE}[/tex]
[tex]\implies GE=\dfrac{2\sqrt{13}}3,FE=\dfrac{2\sqrt{10}}3[/tex]
Let [tex]s[/tex] be the semiperimeter of triangle ABE, so that
[tex]s=\dfrac{AB+BE+EA}2[/tex]
Then according to Heron's formula, the area of triangle ABE is
[tex]\sqrt{s(s-AB)(s-BE)(s-EA)}=\dfrac{25}2[/tex]
