A circle has a chord that goes from #( pi)/6 # to #(5 pi) / 6 # radians on the circle. If the area of the circle is #18 pi #, what is the length of the chord?

Question

A circle has a chord that goes from #(  pi)/6  # to #(5  pi) /  6  # radians on the circle.  If the area of the circle is #18   pi  #, what is the length of the chord?

Aaliyah Anderson · Accepted Answer

#c = sqrt(54)#

We can compute the radius from the area of the circle:

#pir^2=18pi#

#r = sqrt18#

To radii and the chord form a triangle. The angle between the two radii is:

#theta = (5pi)/6-pi/6= (2pi)/3#

If we use the angle and the length of the two radii, we can use the Law of Cosines:

#c^2=a^2+b^2-2(a)(b)cos(theta)#

where #a = b = r = sqrt18 and theta = (2pi)/3#

#c = sqrt((sqrt18)^2+(sqrt18)^2-2(sqrt18)(sqrt18)cos((2pi)/3)#

#c = sqrt(54)#

Autumn Armstrong · Answer

undefined The length of the chord can be calculated using the formula $ L = 2r\sin\left(\frac{\theta}{2}\right) $, where $ L $ is the length of the chord, $ r $ is the radius of the circle, and $ \theta $ is the angle subtended by the chord at the center of the circle.

Given that the area of the circle is $ 18\pi $, we can find the radius using the formula for the area of a circle, $ A = \pi r^2 $. So, $ r^2 = \frac{A}{\pi} = \frac{18\pi}{\pi} = 18 $. Therefore, $ r = \sqrt{18} = 3\sqrt{2} $.

The angle $ \theta $ subtended by the chord is $ \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} $.

Substituting $ r $ and $ \theta $ into the formula for the length of the chord:

$$ L = 2(3\sqrt{2})\sin\left(\frac{2\pi}{3}\right) $$

Using the sine of $ \frac{2\pi}{3} $, which is $ \frac{\sqrt{3}}{2} $:

$$ L = 2(3\sqrt{2})\left(\frac{\sqrt{3}}{2}\right) $$

$$ L = 6\sqrt{2} \cdot \frac{\sqrt{3}}{2} $$

$$ L = 3\sqrt{6} $$

So, the length of the chord is $ 3\sqrt{6} $ units.