Answer:
[tex]=(\frac{64}{3}\pi-16\sqrt{3}) in^2[/tex]
Step-by-step explanation:
Area of segment equals area of sector minus area of isosceles triangle.
[tex]=\frac{\theta}{360}\times \pi r^2 -\frac{1}{2} r^2 \sin(\theta)[/tex]
Given; the length of chord, [tex]d=8\sqrt{3} in.[/tex]
and the angle of the sector, [tex]\theta=120\degree[/tex].
We can use the formula for calculating the length of a chord to find the radius of the circle.
[tex]d=2r\sin(\frac{\theta}{2})[/tex]
[tex]8\sqrt{3}=2r\sin(60\degree)[/tex]
[tex]8\sqrt{3}=2r(\frac{\sqrt{3}}{2})[/tex]
[tex]\Rightarrow r=8in.[/tex]
Area of segment[tex]=\frac{120}{360}\times \pi \times 8^2-\frac{1}{2}\times 8^2\sin(120\degree)[/tex]
[tex]=(\frac{64}{3}\pi-16\sqrt{3}) in^2[/tex]
