Question

A chord with a length of `13` runs from `pi/8` to `pi/2` radians on a circle. What is the area of the circle?

Answer 1

`pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)`

Explanation:

The angle, A, is:

`A = pi/2 - pi/8`

`A = (4pi)/8 - pi/8`

`A = 3pi/8`

Two radii and the chord form an isosceles triangle with sides, a = 13, b = r and c = r.

Using the law of cosines, `a² = b² + c² - 2(b)(c)cos(A)`, we make the substitutions:

`13² = r^2 + r^2 - 2(r)(r)cos((3pi)/8)`

`13² = 2r^2 - 2(r^2)cos((3pi)/8)`

`13² = r^2(2 - 2cos((3pi)/8))`

`r^2 = (13^2)/(2 - 2cos((3pi)/8)`

To obtain the area of the circle, multiply both sides by `pi`:

`pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)`

Answer 2

`pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)`

Explanation:

The angle, A, is:

`A = pi/2 - pi/8`

`A = (4pi)/8 - pi/8`

`A = 3pi/8`

Two radii and the chord form an isosceles triangle with sides, a = 13, b = r and c = r.

Using the law of cosines, `a² = b² + c² - 2(b)(c)cos(A)`, we make the substitutions:

`13² = r^2 + r^2 - 2(r)(r)cos((3pi)/8)`

`13² = 2r^2 - 2(r^2)cos((3pi)/8)`

`13² = r^2(2 - 2cos((3pi)/8))`

`r^2 = (13^2)/(2 - 2cos((3pi)/8)`

To obtain the area of the circle, multiply both sides by `pi`:

`pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)`