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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