Solution:
Given:
A circle with the sector details;
[tex]\begin{gathered} r=19\text{ }in \\ \theta=135^0 \end{gathered}[/tex]
The area of a sector is given by;
[tex]\begin{gathered} A=\frac{\theta}{360}\times\pi r^2 \\ A=\frac{135}{360}\times\pi\times19^2 \\ A=\frac{1083\pi}{8}\text{ }in^2 \end{gathered}[/tex]
Therefore, the area of the sector is;
[tex]\frac{1083\pi}{8}\text{ }in^2[/tex]
