Solution:
The given values in the question are
[tex]\begin{gathered} \angle A=45^0 \\ r=10in \end{gathered}[/tex]
Concept:
The area of a sector is given below as
[tex]A_{\text{sector}}=\frac{\theta}{360}\times\pi r^2[/tex]
By substituting the values, we have
[tex]\begin{gathered} A_{\text{sector}}=\frac{\theta}{360}\times\pi r^2 \\ A_{\text{sector}}=\frac{45}{360}\times\pi\times10^2 \\ A_{\text{sector}}=\frac{4500\pi}{360} \\ A_{\text{sector}}=\frac{25\pi}{2} \end{gathered}[/tex]
Hence,
The final answer is OPTION D
