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

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

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Answer 1 · 2017-01-22T18:51:15

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

Explanation:

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

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

Area = $π R^{2} = 25 π \Rightarrow R = 5$ $pi R^2 = 25 pi Rightarrow R = 5$

The chord is $A B$ $AB$ , such that

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

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

${| A B |}^{2} = {(x_{A} - x_{B})}_{A}^{2} + {(y_{A} - y_{B})}_{A}^{2}$ $|AB|^2 = (x_A - x_B)^2 + (y_A - y_B)^2$

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

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

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

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

$| A B | = 5 \sqrt{2} \sqrt{1 - cos (π \frac{8 - 3}{24})}$ $|AB| = 5 sqrt 2 sqrt {1 - cos (pi (8 - 3)/24)}$

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

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