Find the area of ​the shaded region in the triangle below? For reference: In figure $G$ is the centroid of the triangle $ABC$; if the area of ​​the $FGC$ triangle is $9m^2$, the area of ​​the FGB triangle is $16m^2$
Calculate the area of ​​the shaded region. (Answer:$7m^2$)
If possible by geometry
My progress:

$S_{FGC} = \frac{b.h_1}{2} = \frac{FG.h_1}{2}\implies FG = \frac{18}{h_1}\\
S_{FGB}=\frac{b.h_2}{2} = \frac{FG.h_2}{2} \implies FG = \frac{32}{h_2}\\
\therefore \frac{18}{h_1} = \frac{32}{h_2}\implies \frac{h_1}{h_2} = \frac{32}{18}=\frac{16}{9}\\
S_{ABG} = S_{BCG} = S_{ACG}$
...??? I'm not able to develop this

$\begin{array}{} A=(p,q) & B=(a,0) & C=(0,0) & F=(x,y) \end{array}$
$S_{FGC}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ 0 & 0 & i \\ \end{array} \right| |=9 \\ S_{FGB}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ a & 0 & i \\ \end{array} \right| |=16$
$\begin{array}{} |x·y_{G}-x_{G}·y|=18 &  & |x·y_{G}-x_{G}·y+a·y-a·y_{G}|=32  \end{array}$
Solving the modular equations we get four solutions. Replace $x_{G}=\frac{a+p}{3}$ and $y_{G}=\frac{q}{3}$
$\begin{array}{} \left( x=\frac{aq(a+p)±6(2a-25p)}{3a},y=\frac{aq∓150}{3a}  \right)  & ⇒ & S_{AFG}=7 \\ \left( x=\frac{aq(a+p)±6(34a+7p)}{3a},y=\frac{aq±42}{3a}  \right)  & ⇒ & S_{AFG}=25 \\ \end{array}$

