Question

A triangle has corners at `(3 , 3 )`, `(1 ,2 )`, and `(8 ,9 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`.351`

Explanation:

For the points `A(3,3), B(1,2)` and `C(8,9)`
`AB=sqrt((1-3)^2+(2-3)^2)=sqrt(4+1)=sqrt(5)~=2.2`
`BC=sqrt((8-1)^2+(9-2)^2)=sqrt(7^2+7^2)=sqrt(98)=7sqrt(2)~=9.9`
`AC=sqrt((3-8)^2+(3-9)^2)=sqrt(25+36)=sqrt(61)~=7.8`

The case is schematically represented by the figure below

As the sides of the triangle are 5, 8 and 9:
`x+y=7sqrt(2)`
`x+z=sqrt(61)`
`y+z=sqrt(5)` => `z=sqrt(5)-y`
`-> x+sqrt(5)-y=sqrt(61)` => `x-y=sqrt(61)-sqrt(5)`

Adding the first and last equations
`2x=7sqrt(2)+sqrt(61)-sqrt(5)` => `x=(7sqrt(2))/2+sqrt(61)/2-sqrt(5)~=7.737`

Using the Law of Cosines:
`(sqrt(5))^2=(7sqrt(2))^2+(sqrt(61))^2-2*7sqrt(2)*sqrt(61)*cos alpha`

`cos alpha=(98+61-5)/(14*sqrt(122))=154/(14*sqrt(122))=11/sqrt(122)`

`alpha=5.194^@`

In the right triangle with `x` as cathetus, we can see that
`tan (alpha/2)=r/x`

`r=x*tan (alpha/2)=7.737*tan (5.194^@/2)` => `r=.351`