A circle has a chord that goes from $( pi)/6$ to $(11 pi) / 8$ radians on the circle. If the area of the circle is $64 pi$ , what is the length of the chord?

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Answer 1 · 2017-06-24T08:08:42

The length of the chord is $=15.2$

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

$theta=11/8pi-1/6pi=66/48pi-8/48pi=58/48pi=29/24pi$

Let the radius of the circle be $=r$

The area of the circle is $A=pir^2$

Here, $A=64pi$

So,

$pir^2=64pi$

$r^2=64$

$r=8$

The length of the chord is

$l=2*rsin(theta/2)$

$=2*8*sin(29/48pi)$

$=15.2$

A circle has a chord that goes from ( pi)/6 ( pi)/6 to (11 pi) / 8 (11 pi) / 8 radians on the circle. If the area of the circle is 64 pi 64 pi , what is the length of the chord?