Answer:
The area is roughly 22.1106cm².
Step-by-step explanation:
The area of the entire circle would be:
[tex]\pi {(6)}^{2} = 36\pi[/tex]
However, we only have 120° of it, which out of 360° means we have 1/3 of the circle, and therefore the sector is:
[tex] \frac{36\pi}{3} = 12\pi[/tex]
Using the Law of Cosines, it is possible to find the missing side of the triangle:
[tex] {x}^{2} = {6}^{2} + {6}^{2} - 2(6)(6) \cos(120) \\ {x}^{2} = 36 + 36 - 2(36)( - \frac{1}{2} ) \\ {x}^{2} = 72 + 36 \\ {x}^{2} = 108 \\ x = \sqrt{3 \times 9 \times 4} \\ x = 3 \times 2 \times \sqrt{3} \\ x = 6 \sqrt{3} [/tex]
Then, to calculate the area of the triangle, we find its height using the Pythagorean Theorem.
[tex] { (\frac{6 \sqrt{3} }{2} )}^{2} + {x}^{2} = {6}^{2} \\ {(3 \sqrt{3} )}^{2} + {x}^{2} = {6}^{2} \\ 27 + {x}^{2} = 36 \\ {x}^{2} = 36 - 27 \\ {x}^{2} = 9 \\ x = 3[/tex]
Finally, we calculate the area of the triangle and subtract it from the sector.
[tex]a = 12\pi - ( \frac{3 \times 6 \sqrt{3} }{2} ) \\ = 12\pi - 9 \sqrt{3} = 22.1106[/tex]
