Derivation for r=∆/s Derivation of r = delta/s

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Before going to derivation, let us know some definitions related to triangles.

Incircle: The circle that touches the three sides of a triangle internally is called ‘Incircle’ of that triangle. It is also called as ‘Inscribed circle’. The centre and radius of this incircle are called incentre and inradius respectively.

Incentre and inradius are denoted by ‘I’ and ‘r’ respectively. 

Inradius formula in terms of area and semi perimeter is r=∆/s.

Where, ∆ is the area of triangle and

              s is the semi perimeter of triangle


Derivation:

Let us consider ∆ABC


As shown in the above figure, let ‘I’ be the incentre and ‘r’ be the inradius.

IF,ID, and IE be the perpendicular line segments drawn from I to AB,BC and AC respectively.

Let the lengths of sides AB, BC, and AC be c, a and b respectively.

Then semi perimeter = s = (a+b+c)/2  .

Here we can write,

Area of ∆ABC = Area of ∆AIB + Area of ∆BIC + Area of ∆AIC be equation no.1

Here, Area of ∆AIB =  1/2 × base × height

For ∆AIB, let the base = length of side AB = c

Then height = length of line segment IF = inradius = r

Therefore, base = c and height = r

Therefore area of ∆AIB =  1/2 × c × r

Similarly,

Area of ∆BIC =  1/2 × a × r and

Area of ∆AIC =  1/2 × b × r

Let ∆ = area of triangle ABC

 From equation no.1 and above information,

∆ =  1/2 × c × r +  1/2 × a × r +  1/2 × b × r

Therefore, ∆ = r × (a+b+c)/2

and ∆ = r × s [since s = (a+b+c)/2 ]

Therefore, r = ∆/s  .

 

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