Question

What is the radius of the incircle of a triangle whose sides are 5, 12 and 13 units?

Answer 1

`r=2` units

Explanation:

As `13^2=5^2+12^2`, the triangle is a right triangle.
Formula for the inradius (`r`) of a right triangle :
`r=(a*b)/(a+b+c)`, or `r= (a+b-c)/2`
where `a and b` are the legs of the right traingle and `c` is the hypotenuse.
`=> r=(a*b)/(a+b+c)=(5*12)/(5+12+13)=60/30=2` units

or `r=(a+b-c)/2=(5+12-13)/2=4/2=2` units

Proof 1 : `r=(a*b)/(a+b+c)`

Area `DeltaABC=1/2*(BC)*(AC)=1/2*r*(BC+AC+AB)`
`=> (r*(a+b+c))/2=(a*b)/2`
`=> r=(a*b)/(a+b+c)`

Proof 2 : `r=(a+b-c)/2`

Recall that the tangents to a cicle from an external point are equal,
`=> CD=CF=r, BE=BF=a-r`,
and `AD=AE=b-r`
`=> c=a-r+b-r`
`=> c=a+b-2r`
`=> 2r=a+b-c`
`=> r=(a+b-c)/2`