Question

A circle has a chord that goes from `( 11 pi)/6` to `(7 pi) / 4` radians on the circle. If the area of the circle is `128 pi`, what is the length of the chord?

Answer 1

See below.

Explanation:

Let us call the center of the circle `O` and the chord `AB`. Then, `angleAOB=frac(11pi)(6)-frac(7pi)(4)=frac(pi)(12)=15` degrees. Let us call the radius of the circle `r`. Then, `DeltaAOB` is isoceles. By law of cosines, we find that:
`AB^2= AO^2+BO^2-2*AO*BO*cos(O)`
`=r^2+r^2-2*r*rcos(15)`
`=2r^2-2r^2cos(15)`
`=2r^2(1-cos(15))`

However, we know that `pir^2=128pi` and `r^2=128`.
Replacing, we get that `AB^2=256(1-cos(15))`. Using either half-angle or the cosine difference formula or a calculator, we can find that `cos(15)=frac(sqrt(6)+sqrt(2))(4)`.

Substituting,
`AB^2=256(1-frac(sqrt(6)+sqrt(2))(4))=256-64sqrt(6)-64sqrt(2)`
`AB=sqrt(256-64sqrt(6)-64sqrt(2))` or `2.95347`.

Hopefully this helps :).