Question

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

Answer 1

`2103pi`

Explanation:

From diagram. chord = c = 6.

Angle C subtended at the centre is:

`pi/6-pi/8=pi/24`

The sum of the angles of triangle ABC = `pi`

Triangle ABC is isosceles so angles A and B are equal.

Angles A and B are:

`pi-pi/24=1/2*(23pi)/24=(23pi)/48`

a and b are radii, so we can solve for either of these.

Solving for a using the Sine Rule:

`sinA/a=sinB/b=sinC/c`

We know angle C and side c

`:.`

`sin((23pi)/48)/a=sin(pi/24)/6=>a=(6sin((23pi)/48))/(sin(pi/24))~~45.869`

So radius is `~~45.869`

Area of a circle:

`pir^2`

`pi(45.869)^2=2103pi`