Question

A triangle has corners at `(9 , 2 )`, `(4 ,7 )`, and `(5 ,8 )`. What is the radius of the triangle's inscribed circle?

Answer 1

If `a,b` and `c` are the lengths of the sides and `S` is the area of the triangle the radius of the inscribed circle is given by the formula:

`r=(2S)/(a+b+c)`

Explanation:

The area `S` can be computed with Eron's formula:

`S=sqrt(p(p-a)(p-b)*(p-c))`

where `p=(a+b+c)/2`.

Substituting in the formula:

`r=(sqrt(p(p-a)(p-b)*(p-c)))/(p)`

Start calculating the lengths of the sides:

`a=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`b=sqrt((x_3-x_2)^2+(y_3-y_2)^2)`
`c=sqrt((x_1-x_3)^2+(y_1-y_3)^2)`

`a = sqrt((9-4)^2+(2-7)^2) = sqrt(25+25) = 5sqrt(2)`
`b= sqrt((5-4)^2+(8-7)^2) =sqrt(1+1) =sqrt(2)`
`c= sqrt((9-5)^2+(8-2)^2) =sqrt(16+36) =2sqrt(13)`

`p=(5sqrt(2)+sqrt(2)+2sqrt(13))/2 = 3sqrt(2)+sqrt(13)`

`r=sqrt((3sqrt(2)+sqrt(13))(3sqrt(2)+sqrt(13)-5sqrt(2))(3sqrt(2)+sqrt(13)-sqrt(2))(3sqrt(2)+sqrt(13)-2sqrt(13)))/(3sqrt(2)+sqrt(13))=sqrt((3sqrt(2)+sqrt(13))(sqrt(13)-2sqrt(2))(2sqrt(2)+sqrt(13))(3sqrt(2)-sqrt(13)))/(3sqrt(2)+sqrt(13))`

and I leave the rest of the calculation :D