A circle has a chord that goes from $( 3 pi)/8$ to $(4 pi) / 3$ radians on the circle. If the area of the circle is $48 pi$ , what is the length of the chord?

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Answer 1 · 2017-10-02T14:27:12

$13.83$

To start, we know that the area of a circle is equal it the radius square times pi.

$A=r^2xxpi$

We also know the area of the circle is $48pi$ , so using this we know that

$48pi=r^2xxpi$

We can divide through by pi.

$48=r^2$

And square root.

$4sqrt3=r$

We have calculated the radius of the circle.

Now to find the angle across our chord we subtract the two angles we have been given.

$theta=(4pi)/3-(3pi)/8=(23pi)/24$

From the image we can see the angle has been bisected, also bisecting the chord creating two right-angled triangles.

Using trigonometry we can calculate half the length of the chord.

We have the radius/hypotenuse, the angle $theta/2$ / $(23pi)/48$ and we are looking for the opposite, so we are using sin.

$sintheta=o/h$

$hsintheta=o$

$4sqrt3sin((23pi)/48)=6.91=o$

This is the length of half the chord, so the chord length is

$13.83$ to 2 s.f

A circle has a chord that goes from ( 3 pi)/8 ( 3 pi)/8 to (4 pi) / 3 (4 pi) / 3 radians on the circle. If the area of the circle is 48 pi 48 pi , what is the length of the chord?