A circle has a chord that goes from $\frac{π}{12}$ $pi/12$ to $\frac{π}{4}$ $pi/4$ radians on the circle. If the area of the circle is $32 π$ $32 pi$ , what is the length of the chord?

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Answer 1 · 2017-03-27T08:53:55

length of chord $= 4 (\sqrt{3} - 1) \approx 2.928$ $=4(sqrt3-1)~~2.928$

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As area of a circle is given by $π r^{2}$ $pir^2$ , and it is $32 π$ $32pi$ , we have $r = \sqrt{32}$ $r=sqrt32$ .

As shown in the figure, the angle $Theta$ subtended by the chord at the centre is :
$Theta=pi/4-pi/12=pi/6$
$=> Theta/2=pi/12$

$=> AM=rsin(Theta/2)$

$=>$ length of chord $AB=2AM=2*r*sin(Theta/2)$
$= 2*sqrt32*sin(pi/12)~~2.928$

exact value:
$AB=2*sqrt32*((sqrt6-sqrt2)/4)$
$=1/2(8sqrt3-8)$
= $4(sqrt3-1)$

A circle has a chord that goes from pi/12 π12pi/12 to pi/4 π4pi/4 radians on the circle. If the area of the circle is 32 pi 32π32 pi , what is the length of the chord?