Let ABC be an equilateral triangle, with each side of length . If a circle is drawn with diameter AB, then the area of the portion of the triangle lying inside the circle is

A

$$(3\sqrt{3} + \pi)\frac{k^2}{24}$$

B

$$(3\sqrt{3} + \pi)\frac{k^2}{6}$$

C

$$(3\sqrt{3} - \pi)\frac{k^2}{24}$$

D

$$(3\sqrt{3} - \pi)\frac{k^2}{6}$$