Explanation
We are given the following:
[tex]Minor\text{ }Sector\begin{cases}MP(radius)={12\text{ }cm} \\ \angle NMP=\text{ }{40\degree}\end{cases}[/tex]
We are required to determine the area of the smaller sector.
We know that the area of a sector is given as:
[tex]\begin{gathered} Area=\frac{\pi\theta r^2}{360} \\ where \\ \pi=3.14 \\ \theta=40\degree \\ r=12\text{ }cm \end{gathered}[/tex]
Therefore, we have:
[tex]\begin{gathered} Area=\frac{3.14\times40\degree\times12^2}{360\degree} \\ Area=\frac{3.14\times12^2}{9} \\ Area=3.14\times16 \\ Area=50.24cm^2 \end{gathered}[/tex]
Hence, the answer is:
[tex]Area=50.24cm^{2}[/tex]
