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

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Answer 1 · 2016-06-10T09:01:15

It is $7.36$ $7.36$ .

The length of a chord is given by

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

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

$A = π r^{2}$ $A=pir^2$ , then the radius is

$r = \sqrt{\frac{A}{π}} = \sqrt{24} \approx 4.9$ $r=sqrt(A/pi)=sqrt(24)\approx4.9$

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

$θ = \frac{7}{8} π - \frac{π}{3} = \frac{21 - 8}{24} π = \frac{13}{24} π$ $\theta=7/8pi-pi/3=(21-8)/24pi=13/24pi$

So the length of the chord is

$c = 2 r sin (\frac{θ}{2})$ $c=2rsin(\theta/2)$

$= 2 \cdot 4.9 \cdot sin (\frac{1}{2} \cdot \frac{13}{24} π)$ $=2*4.9*sin(1/2*13/24pi)$

$= 9.8 \cdot sin (\frac{13}{48} π)$ $=9.8*sin(13/48pi)$

$= 7.36$ $=7.36$ .

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A circle has a chord that goes from π3( pi)/3 to 7π8(7 pi) / 8 radians on the circle. If the area of the circle is 24π24 pi , what is the length of the chord?