The length l of the arc is calculated below as
[tex]\begin{gathered} l=\frac{\theta}{360}\times2\pi r \\ Substitute\text{ }\theta=60^0\text{ and }l=12\text{ into the equation:} \\ 12=\frac{60}{360}\times2\pi\times r \end{gathered}[/tex]
Therefore, the value of r:
[tex]r=11.46[/tex]
To find the area, A, of a sector, the formula is
[tex]\begin{gathered} A=\frac{\theta}{360}\times\pi r^2 \\ Where\text{ } \\ r=11.46\text{ cm \lparen two decimal places\rparen} \end{gathered}[/tex]
Substitute for r
[tex]\begin{gathered} A=\frac{60}{360}\times\pi\times(11.46)^2 \\ A=68.76\text{ cm}^2\text{ } \end{gathered}[/tex]
Hence,
The area of the sector is approximately is 68.76cm²
The perimeter of the sector is given by:
[tex]length\text{ of arc}+r+r[/tex]
Therefore, the perimeter is given by:
[tex]12+2\times11.4615\approx34.92cm[/tex]
Hence,
The perimeter of the sector is approximately 34.92cm
