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 $48 π$ $48 pi$ , 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 $48 π$ $48 pi$ , what is the length of the chord?

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Answer 1 · 2016-11-05T16:32:36

The chord, $c \approx 4.45$ $c ~~ 4.45$

Explanation:

Given the Area, A, of the circle is $48 π$ $48pi$ , allows us to solve for the radius, using the equation:

$A = π r^{2}$ $A = pir^2$

$48 π = π r^{2}$ $48pi = pir^2$

$r^{2} = 48$ $r^2 = 48$

Two radii and the chord form a triangle with the angle, $θ = \frac{π}{8} - \frac{π}{3}$ $theta = pi/8 - pi/3$ , between the two radii. This allows us to use the Law of Cosines to find the length of the chord, c:

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

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

$c^{2} = 2 r^{2} (1 - cos (θ))$ $c^2 = 2r^2(1 - cos(theta))$

$c^{2} = 96 (1 - cos (\frac{π}{8} - \frac{π}{3}))$ $c^2 = 96(1 - cos(pi/8 - pi/3))$

$c = \sqrt{96 (1 - cos (\frac{π}{8} - \frac{π}{3}))}$ $c = sqrt(96(1 - cos(pi/8 - pi/3)))$

$c \approx 4.45$ $c ~~ 4.45$

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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 $48 π$ $48 pi$ , what is the length of the chord?

1 Answer

Explanation:

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A circle has a chord that goes from $\frac{π}{3}$ $pi/3$ to $\frac{π}{8}$ $pi/8$ radians on the circle. If the area of the circle is $48 π$ $48 pi$ , what is the length of the chord?