Ten in a Circle!

$\triangle OAC$ is a right triangle $\Rightarrow AC=\sqrt{OC^2 - OA^2} = \sqrt{(R-r)^2 - r^2}$

$= \sqrt{R^2 - 2R}$ .

$\triangle CBH$ is a right triangle $\Rightarrow BH = \sqrt{CH^2 - BC^2} = \sqrt{2^2 - \left( \frac{1}{2} \sqrt{R^2-2R} \right)^2}$

$= \frac{1}{2}\sqrt{16 + 2R- R^2}$ .

$\triangle OEH$ is a right triangle $\Rightarrow OH^2 = OE^2 + EH^2 \rightarrow (OG- GH )^2 = \left( \frac{1}{2} AC \right)^2 + (BH+BE)^2$

$R^2 - 2R + 1 = \frac{1}{4}(R^2-2R)+ \frac{1}{4}(16+2R-R^2) + \sqrt{16+2R-R^2} +1$

Simplify, we have

$R^2 - 2R -4 = \sqrt{16 + 2R - R^2} \rightarrow \text{Squaring both sides and simplify, we have} \quad R^{4}-4R^3 - 3R^2 + 14R = 0$

$R(R-2)(R^2-2R-7) =0$

$R=0 \quad cm \Rightarrow \color{#D61F06} \text{Not Accepted} \rightarrow \color{#3D99F6} \text{Since R>0}$

$R=2 \quad cm \Rightarrow \color{#D61F06} \text{Not Accepted} \rightarrow \color{#3D99F6} \text{Since there are 5 smaller circles in a half of bigger circle, then R must be >} \frac{5}{2}$

$R^2 - 2R -7 = 0 \Rightarrow \text{By Al-Khawarizimi Formula, we have} \quad R_{1,2} =1 \pm 2\sqrt{2}$ .

Since $R>0$ , we have $R = 1 + 2\sqrt{2} \quad cm$ , which is approximately equal to $\boxed{3.82}$

I disagree with $ADC$ being a straight line. For some reason, we have exactly the same answer though!