Given a sector with radius, r,
[tex]\begin{gathered} \text{ and subtends an angle }\theta\text{ at the center, then the area, A, of the } \\ \text{ sector is given by} \end{gathered}[/tex][tex]A=\frac{\theta}{360^0}\times\pi r^2[/tex]In this case,
[tex]\theta=86^0,r=4units[/tex]then
[tex]A=\frac{86}{360}\times\pi\times4^2=\frac{86\times16\pi}{360}[/tex]Take
[tex]\pi\approx3.142[/tex]Then,
[tex]A=\frac{86\times16\times3.142}{360}\approx12.01\text{units}^2[/tex]Therefore the area of the sector is 12.01 square units
