Correct Answer: Option B

Explanation

A_sector = \(\frac{60}{360} \times \pi r^2\)

= \(\frac{1}{6} \pi r^2\)

A_{\(\bigtriangleup\)} = \(\frac{1}{2}r^2 \sin 60^o\)

\(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\)

A_{\(\text{shaded portion}\)} = A_sector -
A_{\(\bigtriangleup\)}

= (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\)

= \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\)

= \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)

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