A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/8 #, and the triangle's area is #14 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-06-27T02:20:45

#color(maroon)("Area of inscribed circle " = A_i = pi r^2 = 3.4165#

#"Area of "Delta = A_t = (1/2) a b sin C = (1/2) b c sin A = (1/2) c a sin B#

#"Given " hat A = pi/12, hat B = pi/8, hat C = (19pi)/24, A_t = 14#

#a b = (2 A_t) / sin C = 28 / sin ((19pi)/24) = 46#

#b c = (2 A_t) / sin A = 28 / sin (pi/12) = 108.18#

#c a = (2 A_t) / sin B = 28 / sin (pi/8) = 73.17#

#a = (a b c) / (b c) = sqrt(46 * 108.18 * 73.17) / 108.18 = 5.58#

#b = (a b c) / (c a) = sqrt(46 * 108.18 * 73.17) / 73.17 = 8.15#

#c = (a b c) / (a b) = sqrt(46 * 108.18 * 73.17) / 46 = 13.12#

#"Semi-perimeter " = s = (a + b + c) / 2 = 26.85 / 2 = 13.425#

#"Radius of inscribed circle " = r = A_t / s = 14 / 13.425 = 1.0428#

#"Area of inscribed circle " = A_i = pi r^2 = pi * 1.0428^2 = 3.4165#