Given:
[tex]3)\theta=120^0,radius(r)=9cm[/tex]
To Determine: The length of the minor and the major arc and the area of the sector
[tex]\begin{gathered} Length\text{ of an arc} \\ =\frac{\theta}{360^0}\times2\pi r \\ 3)minorarclength=\frac{120^0}{360^0}\times2\pi\times9cm=6\pi cm \\ Area\text{ of sector=}\frac{\theta}{360^0}\times\pi r^2=\frac{120^0}{360^0}\times\pi(9cm)^2=27\pi cm^2 \\ majorarclength=2\pi r-6\pi=2\pi\times9-6\pi=18\pi-6\pi=12\pi cm \end{gathered}[/tex]
