Question

A triangle has sides with lengths of 6, 4, and 3. What is the radius of the triangles inscribed circle?

Answer 1

`r=sqrt(((p-a)(p-b)(p-c))/p)` with `p = (a+b+c)/2`
`r = 0.820413`

Explanation:

From Heron's formula, the area of a triangle giving their sides `a,b,c` is

`A=sqrt(p(p-a)(p-b)(p-c))` with `p = (a+b+c)/2`

Let `o` be the triangle orthocenter, then the distance between each side and `o` is `r` which is the inscribed circle radius. So

`A = (a xx r)/2+(b xx r)/2 + (c xx r)/2` then
`r((a+b+c)/2)=A = r xx p`

Finally

`r = (sqrt(p(p-a)(p-b)(p-c)))/p=sqrt(((p-a)(p-b)(p-c))/p) = 0.820413`

Answer 2

`~=0.820`

Explanation:

I created this figure using MS Excel

Suppose
`AB=6`
`BC=4`
`CA=2`

`m+l=6` [1]
`l+n=4` [2]
`m+n=3` [3]

[3]-[2]
`m-l=-1` [4]

[4]+[1]
`2m=5` => `m=5/2`
`-> l=7/2`
`-> n=1/2`

Applying law of cosines:
`AB^2=BC^2+CA^2-2*BC*CA*cos(A hat C B)`
`36=16+9-2*4*3cos(AhatCB)`
`24cos(AhatCB)=-11` => `AhatCB~=117.28^@`

`tan((AhatCB)/2)=r/n`
`r=1/2*tan(117.28^@/2)=0.820`