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,

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

1. Find the area of the sector. Since $\alpha =120^{\circ},$ then

$A_{sector}=\dfrac{\pi r^2}{3}=\dfrac{64\pi}{3}\ un^2.$

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

$A_{triangle}=\dfrac{1}{2}\cdot r\cdot r\cdot \sin \alpha=\dfrac{64}{2}\cdot \dfrac{\sqrt{3}}{2}=16\sqrt{3}\ un^2.$

3. The area of the segment is

$A_{segment}=A_{sector}-A_{triangle}=\dfrac{64\pi}{3}-16\sqrt{3}=\dfrac{64\pi-48\sqrt{3}}{3}\ un^2.$

Answer: $\dfrac{64\pi-48\sqrt{3}}{3}\ un^2.$