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

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Answer 1 · 2018-06-21T04:46:27

#color(purple)("Area of incircle " = A_i = pi * (12/13.12)^2 = 2.628 " sq units"#

#hat A = pi/12, hat B = pi/12, hat C = (5pi) / 6, A_t = 12#

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

#ab = (2 A_t) / sin C = 24 / sin ((5pi)/6) = 48#

#bc = A_t / sin A = 24/ sin (pi/12) = 92.73#

#ca = A_t / sin C = 26 / sin(pi/12) = 92.73#

#a = (abc) / (bc) = sqrt(48 * 92.73 * 92.73) / 92.73 = 6.93#

#b = (abc) / (ac) = sqrt(48 * 92.73 * 92.73) / 92.73 =6.93#

#c = (abc) / (ab) = sqrt(48 * 92.73 * 92.73) / 48 = 13.38#

#"Semiperimeter " = s = (a + b + c) / 2 = (6.93 + 6.93 + 13.38) / 2 = 13.12#

#"Incircle radius " = r = A_t / s = 12 / 13.12#

#color(purple)("Area of incircle " = A_i = pi * (12/13.12)^2 = 2.628 " sq units"#