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 $A_t = sqrt(s (s-a) (s-b) (s-c)), " using Heron's formula"$

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

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

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