Question

Please help! 50 points and i'll give brainliest!

Answer 1

Answer:

36\sqrt{3} -12\pi

Answer 2

Answer:

$36\sqrt{3} -12\pi$

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,  

$\tan 60 = \frac{AC}{AO} = \frac{x}{6}$

⇒ x = 6√3  

So, area of Δ AOC = $\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}$

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 $\frac{120}{360} \times \pi r^{2} = 12\pi$

Therefore, the area of the shaded portion is $36\sqrt{3} -12\pi$ (Answer)