Question

A circle has a chord that goes from `pi/12` to `pi/4` radians on the circle. If the area of the circle is `8 pi`, what is the length of the chord?

Answer 1

`2sqrt(3)-2`

Explanation:

The area of a circle is given by the formula:

`"area" = pi r^2`

where `r` is the radius.

So in our example:

`8pi = pi r^2`

Hence:

`r = sqrt(8) = 2sqrt(2)`

So the chord is the base of an isoceles triangle with the other two sides of length `2sqrt(2)`.

Bisecting the chord with a line joining the origin to the midpoint at `pi/6` we obtain two right angled triangles, each with hypotenuse of length `2sqrt(2)` and acute angle `pi/12`.

Hence the length of half of the chord is the length of the smaller leg of the right angled triangle:

`2sqrt(2)*sin(pi/12) = 2sqrt(2)*1/4(sqrt(6)-sqrt(2)) = sqrt(3)-1`

So the length of the chord is `2sqrt(3)-2`

graph{(x^2+y^2-8)(y-x)(y - (1/4(sqrt(6)-sqrt(2)))x)(y-1/(sqrt(3))x)((y-2)+tan(pi/3)(x-2)) = 0 [-1.055, 3.945, -0.26, 2.24]}