A circle has a chord that goes from $pi/3$ to $pi/8$ radians on the circle. If the area of the circle is $25 pi$ , what is the length of the chord?

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Answer 1 · 2017-01-22T18:51:15

$5 sqrt 2 sqrt {1 - cos frac{5 pi}{24}}$

Put $x$ and $y$ axes adequately so that

$x^2 + y^2 = R^2$

Area = $pi R^2 = 25 pi Rightarrow R = 5$

The chord is $AB$ , such that

$A = 5 (cos frac{pi}{8}, sin frac{pi}{8})$

$B = 5 (cos frac{pi}{3}, sin frac{pi}{3})$

$|AB|^2 = (x_A - x_B)^2 + (y_A - y_B)^2$

$= 25 (cos a - cos b)^2 + 25 (sin a - sin b)^2$

$= 25 (cos^2 a + cos^2 b - 2 cos a cos b + sin^2 a + sin^2 b - 2 sin a sin b)$

$= 25 [1 + 1 - 2 cos(a - b) ]$

$= 50 [1 - cos(frac{pi}{3} - frac{pi}{8}) ]$

$|AB| = 5 sqrt 2 sqrt {1 - cos (pi (8 - 3)/24)}$

A circle has a chord that goes from pi/3 pi/3 to pi/8 pi/8 radians on the circle. If the area of the circle is 25 pi 25 pi , what is the length of the chord?