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

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Answer 1 · 2017-06-06T08:10:21

The length of the chord is $=0.32$

The area of the circle is $=6pi$

Let the radius of the circle be $=r$

Then,

$pir^2=6pi$ , $=>$ , $r^2=6$ , $r=sqrt6$

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

$theta=2/3pi-5/8pi=16/24pi-15/24pi=1/24pi$

The length of the chord is

$=2rsin(theta/2)=2*sqrt6sin(1/2*1/24pi)$

$=0.32$

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