Answer:
[tex]\text{Area of segment =}13.102 in^2[/tex]
Step-by-step explanation:
Given a segment of a circle has a 120 arc and a chord of 8 in. We have to find the area of segment.
As, radius from the centre perpendicularly bisect the chord ∴ 120 degree arc is also bisected.
Hence, we get a right triangle with the radius of circle the hypotenuse and 4 opposite side and the 60 degree angle i.e
In ΔOAB
AB=4 in
∠AOB=60°
[tex]\sin60=\frac{AB}{OB}[/tex]
⇒ [tex]Radius=OB=4 \times \csc(60) = 4 \times \frac{2}{\sqrt3} = \frac{8}{sqrt3}in[/tex]
[tex]\text{Then},\text{ the area of the sector is }=\frac{1}{2}r^2\theta=\frac{1}{2}(\frac{8}{\sqrt3})^2 120(\frac{\pi}{180})=\frac{64\pi}{9}in^2[/tex]
Now, we have to find the area of triangle ODB
[tex]Base=DB=8 in[/tex]
[tex]Height=OA=\frac{AB}{\tan60}=\frac{4}{\sqrt3}in[/tex]
[tex]\text{Area of triangle ODB=}\frac{1}{2}\times DB\times OA=\frac{1}{2}\times8\times \frac{4}{\sqrt3}=\frac{16}{\sqrt3}in^2[/tex]
[tex]\text{Area of segment =} \frac{64\pi}{9} - \frac{16}{\sqrt3}=13.102 in^2[/tex]
