Find the area $ S_ {OGBH}$ in the figure below

For reference:

In the figure, calculate area $ S_ {OGBH} $ if the triangle area $ABC=9\ \mathrm{m^2}$ and $AB=3\ \mathrm m$ and $AO = 2\ \mathrm m$. (Answer: $5\ \mathrm{m^2}$)

My progress:

Draw $BO.$ Let $AG = x$ and $HO=GO=R.$

$BG = BH.$

$S_{BGO} = S_{BOH}$

$\displaystyle \frac{9}{S_{CHO}} = \frac{BC\cdot CH}{AC\cdot CO}$

$\displaystyle \frac{S_{BGO}}{S_{CHO}}=\frac{R\cdot BG}{R\cdot CH}=\frac{BG}{CH}$

$\displaystyle \frac{S_{AGO}}{S_{BGO}}=\frac{Rx}{R\cdot BG}=\frac{x}{BG}$

$\displaystyle S_{OAG} = \frac{Rx}{2}$

$\displaystyle \frac{S_{ABC}}{S_{ABO}} = \frac{3AC}{3\cdot2}\implies \frac{9}{S_{ABO}} = \frac{3AC}{6}\implies S_{ABO}=\frac{18}{AC}$

....???

Say radius of the semicircle is $r$, Then $OG = OH = r$. Now drop a perp from $B$ to $AC$. Say the foot of perp is $E$.

$\triangle ABE \sim \triangle AOG$ with hypotenuse $3$ and $2$ respectively.

So, $S_{AOG} = \frac 49 S_{ABE}$ and $BE = \dfrac{3r}{2}$

Next, using similarity of $\triangle COH$ and $\triangle CBE$,

$S_{COH} = \frac 49 \cdot S_{BCE}$

But also, $S_{ABC} = S_{ABE} + S_{CBE} = 9$. That leads to $S_{BGOH} = 5$.