Find the area of ​the shaded region. in the triangle $HGE$ below

Question

Calculate the area of ​​the shaded region. if $AD = 12; DC = 5$.

My progress

$\triangle ACD:\\ AC^2 = \sqrt{12^2-5^2 }=13\\ 2r+13=CD+AD\implies 2r = 4 \therefore r = 2\\ SABCD = 12,6 = 60\\ SABC = \frac{12.5}{2}=30 $

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Answer 1

$EL \perp CD(L \in CD)\implies CL=CI = 5-2 = 3\\ \triangle ABC \sim \triangle GIC: \frac{GI}{AB}=\frac{CI}{BC}=\frac{GC}{AC}\implies \\ \frac{GI}{5}=\frac{3}{12}=\frac{GC}{13}\\ \therefore GI = \frac{15}{12} ~e~CG=\frac{39}{12}\\EG = EI+GI = 2+\frac{15}{12} =\frac{39}{12} \\ BG = 12 - CG=\frac{105}{12}\\ \triangle BGH \sim \triangle ICG: \frac{GH}{CG}=\frac{BG}{CI}\implies\frac{12.GH}{39}=\frac{105}{12.3}\therefore GH = \frac{455}{48}\\ S_{HGE} = \frac{1}{2}.\frac{455}{48}.\frac{39}{12}=\frac{5915}{384}$

Answer 2

HINT.-The incenter is $E=(10,2)$; the $\perp$ from $E$ to line $AC$ is $y-2=-\dfrac{12}{5}(x-10)$ which cuts line $BC$ in $G=(8.75,5)$.

The $\perp$ in $G$ to line $y-2=-\dfrac{12}{5}(x-10)$ is $y-5=\dfrac{5}{12}(x-8.75)$ which determine the point $H=(0,1.354)$.

You do have now the vertex of the right triangle $\triangle{EGH}$ so the required area (which is approximately equal to $15.40375$ )