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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