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?

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Answer 1 · 2018-08-09T15:58:59

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

$hat A = pi / 8, hat B = (7pi)/12, hat C = (7pi)/24, A_t = 22$

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

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