A circle has a chord that goes from $\frac{π}{4}$ $pi/4$ 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 · 2017-07-18T06:16:16

The length of the chord is $= 2.3 u$ $=2.3u$

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The angle subtented at the center of the circle is

$θ = \frac{1}{2} π - \frac{1}{4} π = \frac{1}{4} π$ $theta=1/2pi-1/4pi=1/4pi$

The area of the circle is

$A = π r^{2}$ $A=pir^2$

So, the radius is

$r = \sqrt{\frac{A}{π}} = \sqrt{9 \frac{π}{π}} = 3$ $r=sqrt(A/pi)=sqrt(9pi/pi)=3$

The length of the chord is

$A B = 2 \cdot A C = 2 \cdot r \cdot sin (\frac{θ}{2})$ $AB=2*AC=2*r*sin(theta/2)$

$= 2 \cdot 3 \cdot sin (\frac{1}{8} π)$ $=2*3*sin(1/8pi)$

$= 2.3$ $=2.3$

A circle has a chord that goes from π4pi/4 to π2pi/2 radians on the circle. If the area of the circle is 9π9 pi , what is the length of the chord?