A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(pi)/2 #, and the triangle's area is #24 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-06-21T04:57:43

#color(purple)("Area of incircle " = A_i = pi * (24/15.645)^2 = 7.393 " sq units"#

#hat A = pi/12, hat B = pi/2, hat C = (5pi) / 12, A_t = 24#

#A_t = (1/2) a b sin C = (1/2) bc sin A = (1/2) ca sin B#

#ab = (2 A_t) / sin C = 48 / sin ((5pi)/12) = 49.69#

#bc = A_t / sin A = 48/ sin (pi/12) = 185.46#

#ca = A_t / sin C = 48 / sin(pi/2) = 48#

#a = (abc) / (bc) = sqrt(185.46 * 48 * 49.69) / 185.46 = 3.59#

#b = (abc) / (ac) = sqrt(185.46 * 48 * 49.69) / 48 =13.86#

#c = (abc) / (ab) = sqrt(185.46 * 48 * 49.69) / 49.69 = 13.84#

#"Semiperimeter " = s = (a + b + c) / 2 = (3.59 + 13.86 + 13.84) / 2 = 15.645#

#"Incircle radius " = r = A_t / s = 24 / 15.645#

#color(purple)("Area of incircle " = A_i = pi * (24/15.645)^2 = 7.393 " sq units"#