A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/8 #, and the triangle's area is #16 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-06-26T15:12:29

Area of in circle # A_i = 6.7# sq units

#A_t = (1/2) ab sin C =(1/2)b c sin A = (1/2) c a sin B#

#hat A = pi/2, hat B = pi/8, hat C = (3pi)/8#

#a b = (2 * 16) / sin ((3pi)/8) = 34.64#

#c a = (2*16)/ sin (pi/8) = 83.62#

#b c= (2*16)/ sin(pi/2) = 32#

#a = (abc) / (b c) = sqrt(34.64*83.62*32) / 32 = 9.51#

#b = (abc) / (c a) = sqrt(34.64* 83.62*32) / 83.62 = 3.64#

#c = sqrt(34.64*83.62*32)/34.64 = 8.79#

Semi perimeter #s = (a + b + c) / 2#

#s = (9.51 + 3.64 + 8.79) / 2 = 10.97#

Radius of in circle # r = A_t / s = 16 / 10.97 = 1.46#

Area of in circle # A_i = pi r^2 = pi * 1.46^2 = 6.7#