Question

Geometry Question attached below

Answer 1

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

$1^2+3^2=A F^2\implies A F=\sqrt{10}$

and in triangle BCG,

$2^2+3^2=BG^2\implies BG=\sqrt{13}$

Angles AFD and EFG form a vertical pair, so they are congruent and have measure

$m\angle EFG=\tan^{-1}3$

Similarly, angles BGC and FGE are congruent and have measure

$m\angle FGE=\tan^{-1}\dfrac32$

Then the remaining angle in triangle EFG has measure

$m\angle FEG=\pi-\tan^{-1}3-\tan^{-1}\dfrac32$

We can solve for the lengths of FE and GE exactly by applying the law of sines:

$\dfrac{\sin\left(\pi-\tan^{-1}3-\tan^{-1}\frac 32\right)}2=\dfrac{\sin\left(\tan^{-1}3\right)}{GE}=\dfrac{\sin\left(\tan^{-1}\frac32\right)}{FE}$

$\implies GE=\dfrac{2\sqrt{13}}3,FE=\dfrac{2\sqrt{10}}3$

Let $s$ be the semiperimeter of triangle ABE, so that

$s=\dfrac{AB+BE+EA}2$

Then according to Heron's formula, the area of triangle ABE is

$\sqrt{s(s-AB)(s-BE)(s-EA)}=\dfrac{25}2$