A circle has a chord that goes from $\frac{π}{6}$ $( pi)/6$ to $\frac{5 π}{6}$ $(5 pi) / 6$ radians on the circle. If the area of the circle is $18 π$ $18 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{π}{6}$ $( pi)/6$ to $\frac{5 π}{6}$ $(5 pi) / 6$ radians on the circle. If the area of the circle is $18 π$ $18 pi$ , what is the length of the chord?

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Answer 1 · 2017-04-20T17:28:10

$c = \sqrt{54}$ $c = sqrt(54)$

Explanation:

We can compute the radius from the area of the circle:

$π r^{2} = 18 π$ $pir^2=18pi$

$r = \sqrt{18}$ $r = sqrt18$

To radii and the chord form a triangle. The angle between the two radii is:

$θ = \frac{5 π}{6} - \frac{π}{6} = \frac{2 π}{3}$ $theta = (5pi)/6-pi/6= (2pi)/3$

If we use the angle and the length of the two radii, we can use the Law of Cosines:

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

where $a = b = r = \sqrt{18} and θ = \frac{2 π}{3}$ $a = b = r = sqrt18 and theta = (2pi)/3$

$c = \sqrt{{(\sqrt{18})}^{2} + {(\sqrt{18})}^{2} - 2 (\sqrt{18}) (\sqrt{18}) cos (\frac{2 π}{3})}$ $c = sqrt((sqrt18)^2+(sqrt18)^2-2(sqrt18)(sqrt18)cos((2pi)/3)$

$c = \sqrt{54}$ $c = sqrt(54)$

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A circle has a chord that goes from $\frac{π}{6}$ $( pi)/6$ to $\frac{5 π}{6}$ $(5 pi) / 6$ radians on the circle. If the area of the circle is $18 π$ $18 pi$ , what is the length of the chord?

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Explanation:

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A circle has a chord that goes from π6( pi)/6 to 5π6(5 pi) / 6 radians on the circle. If the area of the circle is 18π18 pi , what is the length of the chord?

1 Answer

Explanation:

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A circle has a chord that goes from $\frac{π}{6}$ $( pi)/6$ to $\frac{5 π}{6}$ $(5 pi) / 6$ radians on the circle. If the area of the circle is $18 π$ $18 pi$ , what is the length of the chord?