Question

A chord with a length of `12` runs from `pi/12` to `pi/6` radians on a circle. What is the area of the circle?

Answer 1

Area of a circle is
`S = (36pi)/sin^2(pi/24)=(72pi)/(1-sqrt((2+sqrt(3))/4))`

Explanation:

Picture above reflects the conditions set in the problem. All angles (enlarged for better understanding) are in radians counting from the horizontal X-axis `OX` counterclockwise.
`AB=12`
`/_XOA=pi/12`
`/_XOB=pi/6`
`OA=OB=r`

We have to find a radius of a circle in order to determine its area.
We know that chord `AB` has length `12` and an angle between radiuses `OA` and `OB` (where `O` is a center of a circle) is
`alpha=/_AOB = pi/6 - pi/12 = pi/12`

Construct an altitude `OH` of a triangle `Delta AOB` from vertex `O` to side `AB`. Since `Delta AOB` is isosceles, `OH` is a median and an angle bisector:
`AH=HB=(AB)/2=6`
`/_AOH=/_BOH=(/_AOB)/2=pi/24`

Consider a right triangle `Delta AOH`.
We know that cathetus `AH=6` and angle `/_AOH=pi/24`.
Therefore, hypotenuse `OA`, which is a radius of our circle `r`, equals to
`r=OA=(AH)/sin(/_AOH)=6/sin(pi/24)`

Knowing radius, we can find an area:
`S = pi*r^2 = (36pi)/sin^2(pi/24)`

Let's express this without trigonometric functions.

Since
`sin^2(phi) = (1-cos(2phi))/2`
we can express the area as follows:
`S = (72pi)/(1-cos(pi/12))`

Another trigonometric identity:
`cos^2(phi) = (1+cos(2phi))/2`
`cos(phi) = sqrt[(1+cos(2phi))/2]`

Therefore,
`cos(pi/12) = sqrt[(1+cos(pi/6))/2] =`
`= sqrt[(1+sqrt(3)/2)/2] = sqrt((2+sqrt(3))/4)`

Now we can represent the area of a circle as
`S = (72pi)/(1-sqrt((2+sqrt(3))/4))`

Answer 2

Another approach same result

Explanation:

The chord AB of length 12 in the above figure runs from`pi/12` to `pi/6` in the circle of radius r and center O, taken as origin .
`/_AOX=pi/12` and `/_BOX=pi/6`
So polar coordinate of A `=(r,pi/12)` and that of B `=(r,pi/6)`
Applying distance formula for polar coordinate
the length of the chord AB,`12=sqrt(r^2+r^2-2*r^2*cos(/_BOX-/_AOX)`

`=>12^2=r^2+r^2-2*r^2*cos(pi/6-pi/12)`
`=>144=2r^2(1-cos(pi/12))`
`=>r^2=144/(2(1-cos(pi/12))`
`=>r^2=cancel144^72/(cancel2(1-cos(pi/12))`
`=>r^2=72/(1-cos(pi/12))`
`=>r^2=72/(1-sqrt(1/2(1+cos(2*pi/12))`
`=>r^2=72/(1-sqrt(1/2(1+cos(pi/6))`
`=>r^2=72/(1-sqrt(1/2(1+sqrt3/2)`

So area of the circle
`=pi*r^2`
`=(72pi)/(1-sqrt(1/2(1+sqrt3/2)`
`=(72pi)/(1-sqrt((2+sqrt3)/4)`