A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/4 #, and the triangle's area is #12 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-03-01T04:28:32

Area of Incircle #A_i = pi r_i^2 = color (purple)(6.4683)# sq units

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#hat A = pi/2, hat B = pi/4, hat C = pi - pi/2 -pi/4 = pi/4, A_t = 12#

#It’s an isosceles right triangle with sides b and c equal

#A_t = (1/2)ab sin C = (1/2) ca sin B#

#ab = ca = (2 * A_t) / sin C = (2*12)/ sin (pi/4) = 24sqrt2#

#bc = (2 * 12) / sin (pi/2) = 24#

#abc = sqrt(ab* bc* ca) = sqrt (24sqrt2 * 24sqrt2 * 24) = 96 sqrt3#

#a = (96 sqrt3) / 24 = 4 sqrt3#

#b = c = (96 sqrt3) / (24 sqrt2 )= 2 sqrt(6)#

Radius of Incircle #r_i = (b + c - a)/2 = (4 sqrt6 - 4 sqrt3) /2 = 1.4349#

Area of Incircle #A_i = pi r_i^2 = pi * 1.4349^2 = color (purple)(6.4683)# sq units