A circle has a chord that goes from $\frac{π}{3}$ $pi/3$ to $\frac{π}{2}$ $pi/2$ radians on the circle. If the area of the circle is $9 π$ $9 pi$ , 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 $\frac{π}{2}$ $pi/2$ radians on the circle. If the area of the circle is $9 π$ $9 pi$ , what is the length of the chord?

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Answer 1 · 2016-11-07T12:11:13

Chord length is $1.5528$ $1.5528$

Explanation:

As area of a circle is given by $π r^{2}$ $pir^2$ and it is $9 π$ $9pi$ , we have $r = \sqrt{9} = 3$ $r=sqrt9=3$ .

Now see the given figure. The angle subtended by the chord at the center is $\frac{π}{2} - \frac{π}{3} = \frac{π}{6}$ $pi/2-pi/3=pi/6$ .
enter image source here
If the chord length is $a$ $a$ and angle subtended by the chord at the center is $θ$ $theta$ , the relation we have is

$sin (\frac{θ}{2}) = \frac{\frac{a}{2}}{r}$ $sin(theta/2)=(a/2)/r$ or $a = 2 r sin (\frac{θ}{2})$ $a=2rsin(theta/2)$

Hence, chord length is $2 \times 3 \times sin (\frac{π}{12}) = 6 \times 0.2588 = 1.5528$ $2xx3xxsin(pi/12)=6xx0.2588=1.5528$

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A circle has a chord that goes from $\frac{π}{3}$ $pi/3$ to $\frac{π}{2}$ $pi/2$ radians on the circle. If the area of the circle is $9 π$ $9 pi$ , what is the length of the chord?

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