A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #(7pi)/12 #, and the triangle's area is #22 #. What is the area of the triangle's incircle?

1 Answer
Aug 9, 2018

Area of Incircle #A_r = pi * (22/13.13) ~~ 541.6 # sq.units

Explanation:

#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