A circle has a chord that goes from #( 3 pi)/4 # to #(5 pi) / 4 # radians on the circle. If the area of the circle is #81 pi #, what is the length of the chord?

First, let's obtain the radius from the area:

#A = pi r^2 = 81 pi#

Here, #pi# cancels out:

#r^2 = 81#

And taking the square root gets us:

#r = 9#.

Here's what the circle looks like:

Just like any other circle. Let's just say its radius is #9#.

Anyway, we now need a chord from #(3 pi)/4# to #(5 pi)/4#.

#pi# itself is halfway around the circle. Since we're taking quarters of halfs of the circle, let's split it into eighths:

Let's start rotating! We know to start at the rightmost point of the circle, which is in this case #(9, 0)#:

Then we can start rotating it by #(3 pi)/4#, which is #3# quarters of half of the circle, or equivalently, #3# eigths of the circle:

How much left do we need to go?

#(5 pi)/4 - (3 pi)/4 = (2 pi)/4 = pi/2#

Ah, half-#pi# radians left! That's half of a half of the circle, or, a quarter of the circle, which is #2# eights:

We could now draw the chord between these two points, and lift off slices we no longer need:

Those familiar with the unit circle might immediately get that the length of the chord must be double the sine of the angle inside that triangle, times the radius.

Why? The sine is defined as the ratio of the side of the triangle opposite of the angle, to the hypotenuse:

#sin(theta) = "Opposite"/"Hypotenuse"#

Multiplying by the hypotenuse, which is in this case the radius, #r = 9#, should get us the length of the opposite side:

#r * sin(theta) = r * "Opposite"/r = "Opposite"#

But, what is the angle inside the triangle?

Well, we could immediately see that it is a quarter of the semicircle, which is #pi/4# radians.

The length of the side opposite to that angle would be

#L_"opposite" = r * sin(pi/4) = 9 * (sqrt(2))/2 = (9 sqrt(2))/2#

And the chord is double that length:

#L_"chord" = 2 * L_"opposite" = 2 * (9 sqrt(2))/2 = 9 sqrt(2)#