Question

A triangle has corners at `(1 , 3 )`, `(4 ,8 )`, and `(9 ,7 )`. What is the radius of the triangle's inscribed circle?

Answer 1

Radius of incircle is `r=(2A)/(a+b+c)`
`A=` area of Triangle
`a,b,c` are side length of Triangle

Points of Traingle
`A(1,3), B(4,8), C(9,7)`
Side length `AB=c=sqrt((4-1)^2+(8-3)^2)=sqrt(9+25)=sqrt34`
Side length `BC=a=sqrt((9-4)^2+(7-8)^2)=sqrt(25+1)=sqrt26`
Side length `CA=b=sqrt((9-1)^2+(7-3)^2)=sqrt(64+16)=sqrt80`

Semi-perimeter of triangle `s=(a+b+c)/2=(sqrt34+sqrt26+sqrt80)/2`
Area of Triangle `A=sqrt(s(s-a)(s-b)(s-c))`
`A=sqrt(((sqrt34+sqrt26+sqrt80)/2)((sqrt34+sqrt26+sqrt80)/2-sqrt26)((sqrt34+sqrt26+sqrt80)/2-sqrt80)((sqrt34+sqrt26+sqrt80)/2-sqrt34)`

`A=sqrt(((sqrt34+sqrt26+sqrt80)/2)((sqrt34-sqrt26+sqrt80)/2)((sqrt34+sqrt26-sqrt80)/2)((-sqrt34+sqrt26+sqrt80)/2)`

`A=sqrt((((sqrt34+sqrt80)^2-(sqrt26)^2)/4)((-sqrt34^2-sqrt34sqrt26+sqrt80sqrt34+sqrt26sqrt34+sqrt26^2-sqrt26sqrt80+sqrt80sqrt34+sqrt80sqrt26-sqrt80^2)/2))`

`A=sqrt((((sqrt34+sqrt80)^2-(sqrt26)^2)/4)((-88+2sqrt80sqrt34)/2))`
`A=sqrt(((88+2sqrt34sqrt80)/4)((-88+2sqrt80sqrt34)/2))`
`A=sqrt(((-88^2+4(34)(80))/8)`
`A=sqrt(((-7744+10880)/8)`
`A=sqrt(((3136)/8))=sqrt(392)=14sqrt2`

Radius of in circle is `r=(2A)/(a+b+c)=(28sqrt2)/(sqrt34+sqrt26+sqrt80)`
`=(28)/(sqrt17+sqrt13+sqrt40)approx1.98`