A circle has a chord that goes from $\frac{2 π}{3}$ $( 2 pi)/3$ to $\frac{17 π}{12}$ $(17 pi) / 12$ radians on the circle. If the area of the circle is $12 π$ $12 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{2 π}{3}$ $( 2 pi)/3$ to $\frac{17 π}{12}$ $(17 pi) / 12$ radians on the circle. If the area of the circle is $12 π$ $12 pi$ , what is the length of the chord?

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Answer 1 · 2016-12-23T19:10:31

The length of the chord is $6.4$ $6.4$

Explanation:

Given: The area of the circle is $12 π$ $12pi$

The area of a circle is:

$A r e a = π r^{2}$ $Area = pir^2$

$12 π = π r^{2}$ $12pi = pir^2$

$r = \sqrt{12}$ $r = sqrt(12)$

If you draw a radius from the center to one end of the chord and a radius from the center to the other end of the chord, you have a triangle. The angle, $θ$ $theta$ , between the two radii is:

$θ = \frac{17 π}{12} - \frac{2 π}{3} = \frac{3 π}{4}$ $theta = (17pi)/12 - (2pi)/3 = (3pi)/4$

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

$c = \sqrt{a^{2} + b^{2} - 2 (a) (b) cos (θ)}$ $c = sqrt(a^2 + b^2 - 2(a)(b)cos(theta))$

where $a = b = r = \sqrt{12} and θ = \frac{3 π}{4}$ $a = b = r = sqrt(12) and theta = (3pi)/4$

$c = \sqrt{12 + 12 - 2 (12) cos (\frac{3 π}{4})}$ $c = sqrt(12 + 12 - 2(12)cos((3pi)/4))$

$c = 6.4$ $c = 6.4$

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A circle has a chord that goes from $\frac{2 π}{3}$ $( 2 pi)/3$ to $\frac{17 π}{12}$ $(17 pi) / 12$ radians on the circle. If the area of the circle is $12 π$ $12 pi$ , what is the length of the chord?

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Explanation:

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A circle has a chord that goes from 2π3( 2 pi)/3 to 17π12(17 pi) / 12 radians on the circle. If the area of the circle is 12π12 pi , what is the length of the chord?

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Explanation:

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A circle has a chord that goes from $\frac{2 π}{3}$ $( 2 pi)/3$ to $\frac{17 π}{12}$ $(17 pi) / 12$ radians on the circle. If the area of the circle is $12 π$ $12 pi$ , what is the length of the chord?