Answer:
[tex]36\sqrt{3} -12\pi[/tex]
Step-by-step explanation:
The area of the shaded portion, we have to calculate.
Now, join C and O to form two right triangles Δ AOC and Δ BOC.
Now, from Δ AOC,
[tex]\tan 60 = \frac{AC}{AO} = \frac{x}{6}[/tex]
⇒ x = 6√3
So, area of Δ AOC = [tex]\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3} [/tex]
Due to symmetry area of Δ BOC will also be 18√3.
Now, area of quadrilateral OACB is 2 × 18√3 = 36√3
Now, area covered by the arc AB about center O will be [tex]\frac{120}{360} \times \pi r^{2} = 12\pi[/tex]
Therefore, the area of the shaded portion is [tex]36\sqrt{3} -12\pi[/tex] (Answer)
