A triangle has vertices A, B, and C. Vertex A has an angle of #pi/3 #, vertex B has an angle of #(5 pi)/12 #, and the triangle's area is #16 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-01-09T10:56:33

Area of I’m circle #color(green)(A_i = 9.19) sq. Units#

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Given #A = pi /3, B = (5pi) / 12, C = pi - A - B = pi / 4#

Area of triangle #A_t = 16#

#a*b = A_t / ((1/2) sin C) = 16 / (0.5 sin (pi/3)) ~~ 36.95#

Similarly,
#b*c = 16 / (0.5 sin A) = 16 / (0.5 * sin ((5pi)/12)) ~~ 33.13#

#c*a = 16 / (0.5 sin B) = 16 / (0.5 * sin (pi/4)) ~~ 45.25#

#(ab * bc * ca) = (abc)^2 = 36.95 * 33.13 * 45.25 = 55392.95#

#abc = sqrt(55392.95) ~~ 235 36#

#a = (abc) / (bc) = 235.36 / 33.13 ~~ 7.1#

#b = (abc) / (ca) = 235.36 / 45.25 ~~ 5.2#

#c = (abc) / (ab) = 235.36 / 36.95 ~~ 6.37#

Semi perimeter #S/ 2 = (a + b + c) / 2 = (7.1 + 5.2 + 6.37) / 2 = 9.34#

In radius #r = A_t / (S/2) = 16 / 9.34 = 1.71#

Area of in circle #A_i = pi r^2 = pi * (1.71)^2 = color(green)(9.19)#