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?

1 Answer
Cri007
Mar 12, 2017

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 :).

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