Question

Express answer in exact form.

A segment of a circle has a 120 arc and a chord of 8sqrt3 in. Find the area of the segment.

i had gotten 44/3pi-16sqrt3 i forgot the "sqrt of 3"

Answer 1

Ok.
You got an equal sided triangle with central angle 120 degrees.
If you draw the median to chord, you got 2 right-angled triangles with 60 and 30 degree angles and \[a=4\sqrt3\] one side or a.
so you can find a radius using sin60(it would be easier to understand if you would post a picture):
\[\sin60^0=\frac{4\sqrt3}{radius}\]
\[radius=\frac{4\sqrt3 \times 2}{\sqrt3}=8\]
The area of segment:\[A_{segment}=A_{sector}-A_{\triangle}\]
\[A_{sector}=\frac{n \pi r^2}{360}=\frac{120 \pi 8^2}{360}=\frac{64 \pi}{3}\]
Let's find second side of right-angled triangle:
\[\cos60^0=\frac{b}{r}\]
\[b=\cos60^0 \times 8=4\]
So\[A_{\triangle of120^0}=2\times \frac{a \times b}{2}=4\sqrt3 \times 4=16\sqrt3\]
\[A_{segment}=A_{sector}-A_{\triangle of 120^0}=\frac{64 \pi}{3}-16\sqrt3\approx 39.31\]

Answer 2

yep = i got same answer