A circle has a chord that goes from $( pi)/3$ to $(7 pi) / 8$ radians on the circle. If the area of the circle is $24 pi$ , what is the length of the chord?

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Answer 1 · 2016-06-10T09:01:15

It is $7.36$ .

The length of a chord is given by

$c=2rsin(\theta/2)$ where $theta$ is the angle under the chord and $r$ is the radius of the circle.

We start calculating the radius. We have the area that is $24\pi$ and we know that the area of the circle is

$A=pir^2$ , then the radius is

$r=sqrt(A/pi)=sqrt(24)\approx4.9$

Then we have to calculate the angle. It is simply the difference between the final angle and the initial angle

$\theta=7/8pi-pi/3=(21-8)/24pi=13/24pi$

So the length of the chord is

$c=2rsin(\theta/2)$

$=2*4.9*sin(1/2*13/24pi)$

$=9.8*sin(13/48pi)$

$=7.36$ .

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A circle has a chord that goes from ( pi)/3 ( pi)/3 to (7 pi) / 8 (7 pi) / 8 radians on the circle. If the area of the circle is 24 pi 24 pi , what is the length of the chord?