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

1 Answer
Sep 29, 2017

r=2r=2 units

Explanation:

As 13^2=5^2+12^2132=52+122, the triangle is a right triangle.
Formula for the inradius (rr) of a right triangle :
r=(a*b)/(a+b+c)r=aba+b+c, or r= (a+b-c)/2r=a+bc2
where a and baandb are the legs of the right traingle and cc is the hypotenuse.
=> r=(a*b)/(a+b+c)=(5*12)/(5+12+13)=60/30=2r=aba+b+c=5125+12+13=6030=2 units

or r=(a+b-c)/2=(5+12-13)/2=4/2=2r=a+bc2=5+12132=42=2 units

Proof 1 : r=(a*b)/(a+b+c)r=aba+b+c
enter image source here
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
enter image source here
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