A circle has a chord that goes from $( pi)/4$ to $(13 pi) / 8$ radians on the circle. If the area of the circle is $32 pi$ , what is the length of the chord?

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Answer 1 · 2017-04-27T20:13:58

$c ~~ 9.4" "larr$ the length of the chord.

We can use the area of the circle, $32pi$ , to compute the radius:

$"Area" = pir^2$

$32pi = pir^2$

$r = sqrt(32) = 4sqrt2$

Two radii and the chord form an isosceles triangle where "c" is the unknown length of the chord.

The lengths of the other two sides are:

$a = b = r = 4sqrt(2)$

The angle between the radii is:

$theta = (13pi)/8-pi/4 = (11pi)/8$

We can use the Law of Cosines to find the length of the chord:

$c = sqrt(a^2+b^2-2(a)(b)cos(theta)$

$c = sqrt(32+32-64cos((11pi)/8)$

$c ~~ 9.4" "larr$ the length of the chord.

A circle has a chord that goes from ( pi)/4 ( pi)/4 to (13 pi) / 8 (13 pi) / 8 radians on the circle. If the area of the circle is 32 pi 32 pi , what is the length of the chord?