A triangle has corners at #(6 , 6 )#, #(4 ,4 )#, and #(1 ,2 )#. What is the radius of the triangle's inscribed circle?

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Answer 1 · 2018-07-29T12:34:59

#color(blue)("Radius of in-circle "= r = A_t / s ~~ 0.1558#

#A(6, 6), B(4, 4), C(1, 2)#

#c = sqrt((6-4)^2 + (6-4)^2) ~~ sqrt 8#

#a= sqrt ((4-1)^2 + (4-2)^2) ~~ sqrt 13#

#b = sqrt((6-1)^2 + (6-2)^2) ~~ sqrt 41#

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

#s = (sqrt 8 + sqrt 13 + sqrt 41 ) / 2 = 6.4186#

Area of triangle #A_t = sqrt(s (s-a) (s-b) (s-c)), " using Heron's formula"#

#A_t = sqrt(6.4186 (6.4186- sqrt 8) (6.4186-sqrt 13) (6.4186-sqrt 41)) ~~ 1#

#color(blue)("Radius of in-circle "= r = A_t / s = 1 / 6.4186 ~~ 0.1558#