In the figure, the area of the shaded segment is
A.
3\(\pi\)
B.
9\(\frac{\sqrt{3}}{4}\)
C.
3 \(\pi - 3 \frac{\sqrt{3}}{4}\)
D.
\(\frac{(\sqrt{3 - \pi)}}{4}\)
E.
\(\pi + \frac{9 \sqrt{3}}{4}\)
Correct answer is C
Area of sector = \(\frac{120}{360} \times \pi \times (3)^2 = 3 \pi\)
Area of triangle = \(\frac{1}{2} \times 3 \times 3 \times \sin 120^o\)
= \(\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9\sqrt {3}}{4}\)
Area of shaded portion = 3\(\pi - \frac{9\sqrt {3}}{4}\)
= 3 \(\pi - 3 \frac{\sqrt{3}}{4}\)