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?

1 Answer
Douglas K.
Nov 16, 2016

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

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