Question

express answer in exact form. Show all work for full credit. A segment of a circle has a 120 arc and a chord of 8 sq root 3. Find the area of the segment.

Answer 1

Let
`\alpha` be the angle of an arc,
`\alpha =120^(\circ),` and r be the radius of the circle.

Then by the cosine rule,

`\text{chord}^2=r^2+r^2-2\cdot r\cdot r\cdot \cos \alpha,\\ \\(8√(3))^2=2r^2-2r^2\cdot \cos 120^(\circ)=2r^2-2r^2\cdot \left(-(1)/(2)\right)=3r^2,\\ \\192=3r^2,\\ \\r^2=64,\\ \\r=8\ un.`

1. Find the area of the sector. Since
`\alpha =120^(\circ),` then

`A_(sector)=(\pi r^2)/(3)=(64\pi)/(3)\ un^2.`

2. Find the area of the triangle formed by two radii and chord:

`A_(triangle)=(1)/(2)\cdot r\cdot r\cdot \sin \alpha=(64)/(2)\cdot (√(3))/(2)=16√(3)\ un^2.`

3. The area of the segment is

`A_(segment)=A_(sector)-A_(triangle)=(64\pi)/(3)-16√(3)=(64\pi-48√(3))/(3)\ un^2.`

Answer:
`(64\pi-48√(3))/(3)\ un^2.`