Question

A triangle has corners at `(3 , 2 )`, `(5 ,7 )`, and `(9 ,5 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`r = 1.44879`

Explanation:

A triangle given by their vertices `p_1,p_2,p_3` has area given by

`A=1/2abs((y_2+y_1)(x_2-x_1)+(y_3+y_2)(x_3-x_2)+(y_1+y_3)(x_1-x_3))`

or by the famous Heron's formula (circa 60 AD). Given their sides lenght

`a = norm(p_2-p_1)`
`b = norm(p_3-p_2)`
`c = norm(p_1-p_3)`

and calling `s = (a+b+c)/2` semiperimeter,
then

`A = sqrt(s(s-a)(s-b)(s-c))`

Calling now `o` the triangle's orthocenter, which is the intersection of bissectrices from each vertice's angle, the distance from `o` to each side is given by `r` which is the internal circle radius. So we can state.

`1/2(a xx r + b xx r +c xx r) = r(a+b+c)/2 = r xx s = A`

solving for `r`

`r = A/s= sqrt(s(s-a)(s-b)(s-c))/s=sqrt(((s-a)(s-b)(s-c))/s)`

Taking `a = 5.38516,b=4.47214,c=6.7082`

we obtain `r = 1.44879`