2006 AMC 12B Problem 16

The distance is $AC = \sqrt{7^2 + 1^2} = \sqrt{50}.$ In a regular hexagon with side $s,$ the distance between vertices two apart is $s\sqrt3,$ so $s^2 \cdot 3 = 50,$ giving $s^2 = \dfrac{50}{3}.$ The hexagon’s area is $$\frac{3\sqrt3}{2}s^2 = \frac{3\sqrt3}{2} \cdot \frac{50}{3} = 25\sqrt3.$$ Thus, the correct answer is C.