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?

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))#

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)#