A circle has a chord that goes from $\frac{5 π}{8}$ $( 5 pi)/8$ to $\frac{4 π}{3}$ $(4 pi) / 3$ radians on the circle. If the area of the circle is $27 π$ $27 pi$ , what is the length of the chord?

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Answer 1 · 2018-01-01T14:46:39

The length of a chord is $9.32$ $9.32$ unit.

Formula for the length of a chord is $L_{c} = 2 r sin (\frac{θ}{2})$ $L_c= 2r sin (theta/2)$

Where $r$ $r$ is the radius of the circle and $θ$ $theta$ is the angle

subtended at the center by the chord. Area of circle is

$cancelpi * r^2 = 27 cancelpi :. r^2 = 27 or r = sqrt27 =3 sqrt3$

$theta= (4pi)/3-(5pi)/8 = (32pi-15pi)/24=(17pi)/24$

$:. L_c= 2 *3sqrt3 *sin (((17pi)/24)/2)$ or

$L_c= 6sqrt3 * sin ((17pi)/48) = 6 sqrt3 * sin 63.75$

$((17pi)/48)=63.75^0$ or

$L_c= 9.32$ unit. The length of a chord is $9.32$ unit.[Ans]

A circle has a chord that goes from ( 5 pi)/8 5π8( 5 pi)/8 to (4 pi) / 3 4π3(4 pi) / 3 radians on the circle. If the area of the circle is 27 pi 27π27 pi , what is the length of the chord?