c) $AB=AC=2R\Rightarrow \Delta ABC$ đều $\Rightarrow S_{ABC}=\frac{{\left(2R\right)}^2\sqrt{3}}{4}=R^2\sqrt{3}$

Khi $\Delta ABC$ đều mà $CE\perp AB,BF\perp AC\Rightarrow CE,BF$ là hai dường trung tuyến

$\Rightarrow \Delta AE\text{}F$ đều có $AE=\frac{AB}{2}=R\Rightarrow S_{AE\text{}F}=\frac{R^2\sqrt{3}}{4}$

$\Delta \text{EOF}$ cũng đều $\Rightarrow S_{q\left(EOF\right)}=\frac{\pi R^2{.60}^0}{{360}^0}=\frac{\pi R^2}{6}$

$S_{EOF}=\frac{R^2\sqrt{3}}{4}\Rightarrow S$ viên phân cung EF = $\frac{R^2\sqrt{3}}{4}-\frac{\pi R^2}{6}=\frac{R^2\left(3\sqrt{3}-2\pi \right)}{12}$

Diện tích cần tìm $=S_{ABC}-S_{AEF}+S_{vp\stackrel{⏜}{E\text{}F}}=R^2\sqrt{3}-\frac{R^2\sqrt{3}}{4}+\frac{R^2\left(3\sqrt{3}-2\pi \right)}{12}=\frac{R^2\left(6\sqrt{3}-\pi \right)}{6}\left(dvdt\right)$