A circle has a chord that goes from $\frac{π}{2}$ $( pi)/2$ to $\frac{3 π}{4}$ $(3 pi) / 4$ radians on the circle. If the area of the circle is $96 π$ $96 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{π}{2}$ $( pi)/2$ to $\frac{3 π}{4}$ $(3 pi) / 4$ radians on the circle. If the area of the circle is $96 π$ $96 pi$ , what is the length of the chord?

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Answer 1 · 2017-04-07T06:34:37

length of chord $\approx 7.499$ $~~ 7.499$

Explanation:

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Area of a circle $A = π r^{2}$ $A=pir^2$ ,
Given $A = 96 π, \Rightarrow r = \sqrt{96}$ $A=96pi, => r=sqrt96$

As shown in the figure, the angle $θ$ $theta$ subtended by the chord at the centre is :
$θ = \frac{3 π}{4} - \frac{π}{2} = \frac{π}{4}$ $theta=(3pi)/4-pi/2=pi/4$
$\Rightarrow \frac{θ}{2} = \frac{π}{8}$ $=> theta/2=pi/8$

$\Rightarrow A M = r sin (\frac{θ}{2})$ $=> AM=rsin(theta/2)$

$\Rightarrow$ $=>$ length of chord $A B = 2 A M = 2 \cdot r \cdot sin (\frac{θ}{2})$ $AB=2AM=2*r*sin(theta/2)$
$= 2 \cdot \sqrt{96} \cdot sin (\frac{π}{8}) \approx 7.499$ $= 2*sqrt96*sin(pi/8)~~7.499$

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A circle has a chord that goes from $\frac{π}{2}$ $( pi)/2$ to $\frac{3 π}{4}$ $(3 pi) / 4$ radians on the circle. If the area of the circle is $96 π$ $96 pi$ , what is the length of the chord?

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