Question

A triangle has corners at `(2 , 9 )`, `(3 ,9 )`, and `(4 ,8 )`. What is the radius of the triangle's inscribed circle?

Answer 1

Radius `r` of an inscribed into a triangle circle can be calculated using the lengths of its sides `a`, `b` and `c` as
`r=sqrt(((p-a)(p-b)(p-c))/p)`
where `p=(a+b+c)/2`

Explanation:

Assuming a triangle has sides `a`, `b` and `c`, its area can be expressed in two ways.

1. Using Heron's formula for an area of a triangle:
   let `p=(a+b+c)/2`, then
   `S = sqrt(p(p-a)(p-b)(p-c))`

2. Area of a triangle equals to half of a product of its perimeter by a radius of an inscribed circle:
   let `p=(a+b+c)/2` and `r` is a radius of an inscribed circle, then
   `S=p*r`

The derivation of both formulas for area of a triangle can be found in the course of advanced mathematics at Unizor

These two expressions for an area must be equal, therefore
`sqrt(p(p-a)(p-b)(p-c)) = p*r`

From this we derive `r`:
`r=sqrt(((p-a)(p-b)(p-c))/p)`

For this particular problem we can get the lengths of all sides using the coordinates of the vertices:

if vertices are `A(2,9)`, `B(3,9)` and `C(4,8)`

`a=BC=sqrt((4-3)^2+(8-9)^2)=sqrt(2)`
`b=AC=sqrt((4-2)^2+(8-9)^2)=sqrt(5)`
`c=AB=sqrt((3-2)^2+(9-9)^2)=1`
`p=(a+b+c)/2=(sqrt(2)+sqrt(5)+1)/2`
`p-a=(-a+b+c)/2=(-sqrt(2)+sqrt(5)+1)/2`
`p-b=(a-b+c)/2=(sqrt(2)-sqrt(5)+1)/2`
`p-c=(a+b-c)/2=(sqrt(2)+sqrt(5)-1)/2`

Putting all these values into a formula above for a radius of an inscribed circle, we calculate the radius.
These calculations we leave to a student who submitted a problem.