The formula for the area (A) of a sector is,
[tex]A=\frac{\theta}{360^0}\times\pi r^2[/tex]
Given
[tex]\begin{gathered} \angle NMP=120^0 \\ r=10cm \end{gathered}[/tex]
Therefore,
[tex]A_s=\frac{120^0}{360^0}\times\pi(10)^2=104.71975\approx104.72cm^2[/tex]
Let us now solve for the area of the larger sector
[tex]\begin{gathered} \theta=360^0-120^0=240^0 \\ r=10cm \end{gathered}[/tex]
Therefore,
[tex]A_l=\frac{240^0}{360^0}\times\pi(10)^2=209.43951\approx209.44cm^2[/tex]
Hence, the answers are
[tex]\begin{gathered} A_s=104.72cm^2 \\ A_l=209.44cm^2 \end{gathered}[/tex]
