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#
