#hat A = pi / 8, hat B = (7pi)/12, hat C = (7pi)/24, A_t = 22#
Incircle radius #r = A_t / s# where s is the semi perimeter of the triangle.
Area of triangle #A_t = (1/2) bc sin A = (1/2) ac sin B = (1/2) ab sin C#
#bc = (2 A_t) / sin A = (2*22)/sin (pi/8) = 114.98#
#ca = (2 A_t) / sin A = (2*22)/sin ((7pi)/12) = 45.55#
#ab = (2 A_t) / sin A = (2*22)/sin ((7pi)/24) = 55.46#
#sqrt(ab * bc * ca) = sqrt(114.98 * 45.55 * 55.46) = 538.95#
#b = sqrt(ab * bc * ca) / (ca) = 538.95 / 45.45 ~~ 11.86#
#c = sqrt(ab * bc * ca) / (ab) = 538.95 / 55.46 ~~ 9.72#
#a = sqrt(ab * bc * ca) / (bc) = 538.95 / 114.98 ~~ 4.69#
#s = (a + b + c) / 2 = 13.13#
Incircle radius #r = A_t / s = 22 / 13.13 #
Area of Incircle #A_r = pi * (22/13.13) ~~ 541.6 # sq.units
