A circle has a chord that goes from $( pi)/3$ to $(7 pi) / 12$ radians on the circle. If the area of the circle is $36 pi$ , what is the length of the chord?

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Answer 1 · 2016-11-25T18:09:48

The length of the chord, $c ~~ 4.59$

Given: $Area = 36pi$

Use $Area = pir^2$ to solve for r:

$36pi = pir^2$

$r^2 = 36$

$r = 6$

Two radii, one each drawn to its respective end of the chord, form a triangle. The angle, C, between the radii is:

$C = (7pi)/12 - pi/3 = pi/4$

To find the length of the chord, c, we can use the Law of Cosines:

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

where $a = b = r = 6 and C = pi/4$

$c = sqrt(6^2 + 6^2 - 2(6)(6)cos(pi/4)$

$c ~~ 4.59$

A circle has a chord that goes from ( pi)/3 ( pi)/3 to (7 pi) / 12 (7 pi) / 12 radians on the circle. If the area of the circle is 36 pi 36 pi , what is the length of the chord?