A circle has a chord that goes from $\frac{2 π}{3}$ $(2 pi)/3$ to $\frac{5 π}{8}$ $(5 pi) / 8$ radians on the circle. If the area of the circle is $6 π$ $6 pi$ , what is the length of the chord?

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Answer 1 · 2017-06-06T08:10:21

The length of the chord is $= 0.32$ $=0.32$

The area of the circle is $= 6 π$ $=6pi$

Let the radius of the circle be $= r$ $=r$

Then,

$π r^{2} = 6 π$ $pir^2=6pi$ , $\Rightarrow$ $=>$ , $r^{2} = 6$ $r^2=6$ , $r = \sqrt{6}$ $r=sqrt6$

The angle subtended by the chord at the centre of the circle is

$θ = \frac{2}{3} π - \frac{5}{8} π = \frac{16}{24} π - \frac{15}{24} π = \frac{1}{24} π$ $theta=2/3pi-5/8pi=16/24pi-15/24pi=1/24pi$

The length of the chord is

$= 2 r sin (\frac{θ}{2}) = 2 \cdot \sqrt{6} sin (\frac{1}{2} \cdot \frac{1}{24} π)$ $=2rsin(theta/2)=2*sqrt6sin(1/2*1/24pi)$

$= 0.32$ $=0.32$

A circle has a chord that goes from 2π3(2 pi)/3 to 5π8(5 pi) / 8 radians on the circle. If the area of the circle is 6π6 pi , what is the length of the chord?