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