A circle has a chord that goes from $\frac{π}{3}$ $pi/3$ to $\frac{π}{4}$ $pi/4$ radians on the circle. If the area of the circle is $49 π$ $49 pi$ , what is the length of the chord?

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Answer 1 · 2017-12-20T13:50:22

The length of the chord is $= 1.83 u$ $=1.83u$

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The angle subtended at the center of the circle is

$ˆ A O B = θ = \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$ $hat(AOB)=theta=pi/3-pi/4=pi/12$

The area of the circle is

$A = π r^{2} = 49 π$ $A=pir^2=49pi$

Therefore,

The radius of the circle is

$r = \sqrt{49} = 7$ $r=sqrt49=7$

The length of the chord is

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

$= 2 \cdot 7 \cdot sin (\frac{π}{24})$ $=2*7*sin(pi/24)$

$= 14 \cdot 0.13$ $=14*0.13$

$= 1.83 u$ $=1.83u$

A circle has a chord that goes from π3pi/3 to π4pi/4 radians on the circle. If the area of the circle is 49π49 pi , what is the length of the chord?