Question

A circle has a chord that goes from `( 5 pi)/6` to `(5 pi) / 4` radians on the circle. If the area of the circle is `21 pi`, what is the length of the chord?

Answer 1

`c ~~ 5.579`

Explanation:

Given: `"Area" = 21pi = pir^2`

Therefore, `r^2 = 21`

Given: The angle between two the radii on either end of the chord, `theta = (5pi)/4 - (5pi)/6 = (5pi)/12`

Let c = the length of the chord; its length can found using a variant of the Law of Cosines:

`c = sqrt(r^2 + r^2 - 2(r)(r)cos(theta))`

`c = sqrt(2r^2 - 2(r^2)cos(theta))`

`c = sqrt(2r^2(1 - cos(theta)))`

`c = sqrt(42(1 -cos((5pi)/12)))`

`c ~~ 5.579`