A circle has a chord that goes from $( 5 pi)/3$ to $(17 pi) / 12$ radians on the circle. If the area of the circle is $18 pi$ , what is the length of the chord?

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Answer 1 · 2016-10-13T17:53:26

The length of the chord is $c ~~ 3.25$

We can compute the radius, given the area of the circle:

$A = pir^2$

$18pi = pir^2$

$18 = r^2$

$sqrt18 = r$

Compute the angle:

$theta = (17pi)/12 - (5pi)/3$

$theta = (17pi)/12 - (20pi)/12$

$theta = -(3pi)/12 = -pi/4$

We can compute the length of the chord, c, using the Law of Cosines:

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

where $a = b = r$ and $theta = -pi/4$ :

$c^2 = r^2 + r^2 - 2(r)(r)cos(-pi/4)$

$c^2 = 36(1 - sqrt(2)/2)$

$c ~~ 3.25$

A circle has a chord that goes from ( 5 pi)/3 ( 5 pi)/3 to (17 pi) / 12 (17 pi) / 12 radians on the circle. If the area of the circle is 18 pi 18 pi , what is the length of the chord?