to the risk of sounding redundant.
[tex]\bf \textit{area of a segment of a circle}\\\\ A=\cfrac{r^2}{2}\left(\cfrac{\pi \theta }{180}-sin(\theta ) \right)~~ \begin{cases} r=&radius\\ \theta =&angle~in\\ °rees\\ \cline{1-2} r=&6\\ \theta =&60 \end{cases}\implies A=\cfrac{6^2}{2}\left(\cfrac{\pi 60}{180}-sin(60^o ) \right) \\\\\\ A=18\left( \cfrac{\pi }{3}-\cfrac{\sqrt{3}}{2} \right)\implies \boxed{A=6\pi -9\sqrt{3}}\implies \implies A\approx 3.26[/tex]
