Divide the side

Refer to the figure above. By Pythagorean theorem : $AM^2 = AB^2-BM^2$ , $\implies AM = 9\sqrt 3$ .

We note that $\triangle ACM$ and $\triangle DCG$ are similar, therefore, $\dfrac {DG}{CG} = \dfrac {AM}{CM}$ , $\implies DG = \dfrac {CG}{CM}AM = \dfrac 23 \times 9 \sqrt 3 = 6\sqrt 3$ .

We note that $\triangle DFG$ and $\triangle EFM$ are similar, therefore, $\dfrac {EM}{FM} = \dfrac {DG}{FG}$ , $\implies EM = \dfrac {FM}{FG}DG = \dfrac 23 \times 6 \sqrt 3 = 4\sqrt 3$ .

The area of the blue triangle $[EFH] = \dfrac 12 \times FH \times EM = \dfrac 12 \times \dfrac 46 \times 18 \times 4 \sqrt 3 = 24 \sqrt 3$

$\implies n = \boxed{24}$