Area of the sector is given by
[tex]A\text{ = }\frac{\emptyset}{360}\text{ x }\pi r^2[/tex]r = diameter /2 = 10/2 inches = 5 inches
[tex]\emptyset=45^0[/tex][tex]\begin{gathered} A\text{ =}\frac{45}{360}\text{ x }\pi\text{ x 5 x 5} \\ \\ A\text{ = }\frac{1125\text{ }\pi}{360} \\ \\ A\text{ = }3.125\text{ }\pi \end{gathered}[/tex]Option A is correct
