A circle has a chord that goes from ( 5 pi)/4  to (4 pi) / 3  radians on the circle. If the area of the circle is 64 pi , what is the length of the chord?

Apr 15, 2018

The length of the chord is $= 4 \sqrt{8 - 2 \left(\sqrt{6} + \sqrt{2}\right)} \approx 2.088$

Explanation:

Calculate the radius of the circle from the area of the circle.

area of circle = $A = \pi {r}^{2}$

$r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{64 \pi}{\pi}} = 8$

Now calculate the coordinates of the endpoints of the chord.

Let's represent the coordinates for the endpoint at $5 \frac{\pi}{4}$ by

$\left({x}_{1} , {y}_{1}\right) = \left(8 \cos \left(\frac{5 \pi}{4}\right) , 8 \sin \left(\frac{5 \pi}{4}\right)\right) = \left(- 4 \sqrt{2} , - 4 \sqrt{2}\right)$

, the coordinates for the endpoint at $\frac{4 \pi}{3}$ by

$\left({x}_{2} , {y}_{2}\right) = \left(8 \cos \left(\frac{4 \pi}{3}\right) , 8 \sin \left(\frac{4 \pi}{3}\right)\right) = \left(- 4 , - 4 \sqrt{3}\right)$

The length, $l$, of the chord can be calculated by the distance formula.

$l = \sqrt{{\left({y}_{1} - {y}_{2}\right)}^{2} + {\left({x}_{1} - {x}_{2}\right)}^{2}}$

$= \sqrt{{\left(- 4 \sqrt{2} - \left(- 4 \sqrt{3}\right)\right)}^{2} + {\left(- 4 \sqrt{2} - \left(- 4\right)\right)}^{2}}$

$= 4 \sqrt{{\left(\sqrt{3} - \sqrt{2}\right)}^{2} + {\left(1 - \sqrt{2}\right)}^{2}}$

$= 4 \sqrt{5 - 2 \sqrt{6} + 3 - 2 \sqrt{2}}$

$= 4 \sqrt{8 - 2 \left(\sqrt{6} + \sqrt{2}\right)} \approx 2.088$