A circle has a chord that goes from $\frac{5 π}{6}$ $( 5 pi)/6$ to $\frac{5 π}{4}$ $(5 pi) / 4$ radians on the circle. If the area of the circle is $39 π$ $39 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{5 π}{6}$ $( 5 pi)/6$ to $\frac{5 π}{4}$ $(5 pi) / 4$ radians on the circle. If the area of the circle is $39 π$ $39 pi$ , what is the length of the chord?

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Answer 1 · 2017-10-19T22:27:15

Chord length $= * \frac{39}{2} or 19.5 *$ $= **39/2 or 19.5**$

Explanation:

Angle sub tended at the center $θ$ $theta$ by the chord is
$\frac{5 π}{4} - \frac{5 π}{6} = \frac{15 π - 10 π}{6} = \frac{10 π}{6} = \frac{5 π}{3}$ $(5pi)/4 - (5pi) /6 = (15pi - 10pi) / 6 = (10pi)/6 = (5pi)/3$
$\frac{θ}{2} = \frac{5 π}{6}$ $theta / 2 = (5pi)/6$

Circumference $= 2 π r = 39 π$ $= 2pir = 39pi$
$r = \frac{39 π}{2 π} = \frac{39}{2}$ $r = (39pi) / (2pi) = 39/2$

If chord length is c,
$sin (\frac{θ}{2}) = \frac{\frac{c}{2}}{r}$ $sin (theta/2) = (c/2)/r$

$c = 2 r sin (\frac{θ}{2}) = 2 \cdot (\frac{39}{2}) \cdot sin (\frac{5 π}{6})$ $c = 2r sin (theta/2) = 2* (39/2) * sin ((5pi)/6)$
$sin (\frac{5 π}{6}) = sin (π - (\frac{5 π}{6}) = sin (\frac{π}{6}))$ $color(red)(sin ((5pi)/6) = sin (pi - ((5pi)/6) = sin (pi/6)))$

$c = 39 \cdot sin (\frac{π}{6}) = 39 \cdot (\frac{1}{2}) = \frac{39}{2} = 19.5$ $c = 39 *sin(pi/6) = 39 * (1/2) = 39/2 = 19.5$

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A circle has a chord that goes from $\frac{5 π}{6}$ $( 5 pi)/6$ to $\frac{5 π}{4}$ $(5 pi) / 4$ radians on the circle. If the area of the circle is $39 π$ $39 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{5 π}{6}$ $( 5 pi)/6$ to $\frac{5 π}{4}$ $(5 pi) / 4$ radians on the circle. If the area of the circle is $39 π$ $39 pi$ , what is the length of the chord?