As we can see in the image the segment that was removed is 90°, which implies 174 of the circle, therefore the area should be 3/4 of the total area. So:
[tex]A_T=\pi r^2[/tex]
So:
[tex]\begin{gathered} A_T=\pi\times(7mi)^2 \\ A_T=49\pi\text{ }mi^2 \end{gathered}[/tex]
Now, the sector area has to be:
[tex]A_S=\frac{3}{4}A_T=\frac{3}{4}49\pi=\frac{147\pi}{4}mi^2[/tex]
The answer is A. 147pi/4 mi^2.
