Question

A chord with a length of `4` runs from `pi/8` to `pi/6` radians on a circle. What is the area of the circle?

Answer 1

`Area = 935pi`

Explanation:

The chord and two radii, each drawn from the center to its respective end of the chord form an isosceles triangle.

The angle, `theta` between the two radii is:

`theta = pi/6 - pi/8`

`theta = pi/24`

let side `c = 4`
Let sides `a = b = r`

Use the Law of Cosines

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

`4^2 = r^2 + r^2 - 2(r)(r)cos(theta)`

`4^2 = r^2(1 + 1 - 2cos(theta))`

`r^2 = 4^2/(2 - 2cos(pi/24)`

`r^2 ~~ 935`

The area of a circle is:

`Area = pir^2`

`Area = 935pi`