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.rt. 3in. Find the area of the segment.

Answer 1

In order to better understand, let's draw a figure and include a triangle created by 2 radius and the given chord:

Since the triangle is isosceles, we can calculate the radius dividing the triangle in two with the height relative to its base:

Using the sine relation of the 60° angle, we have:

`\begin{gathered} \sin (60\degree)=\frac{4\sqrt[]{3}}{r} \\ \frac{\sqrt[]{3}}{2}=\frac{4\sqrt[]{3}}{r} \\ (1)/(2)=(4)/(r) \\ r=8 \end{gathered}`

Now, let's find the height and then calculate the triangle area:

`\begin{gathered} \cos (60\degree)=(h)/(r) \\ (1)/(2)=(h)/(8) \\ h=4 \\ \\ A=(b\cdot h)/(2) \\ A=\frac{8\sqrt[]{3}\cdot4}{2}=16\sqrt[]{3} \end{gathered}`

Now, calculating the area of the segment, we have:

`\begin{gathered} A=(\theta)/(360)\cdot\pi r^2-A_(triangle) \\ A=(120)/(360)\cdot\pi\cdot8^2-16\sqrt[]{3} \\ A=(1)/(3)\cdot64\pi-16\sqrt[]{3} \\ A=(64\pi)/(3)-16\sqrt[]{3} \end{gathered}`