Consider the following triangle ABC, right angled at A. Let AB = 6 cm and BC = 8 cm. Let O be the centre of the incircle of triangle ABC. OP, OQ and OR are drawn perpendiculars to the sides of triangle ABC.

 

 

Using Pythagoras theorem in triangle ABC, it is found that BC = 10 cm.

 

Let radius of the incircle be r cm.

So, OP = OQ = OR = r cm

 

In quadrilateral OPAR, OP = OR = r cm each and all of its interior angle are right angle. So, OPAR is a square.

So, AR = AP = r cm

 

So, PB = (6 - r) cm and CR = (8 - r) cm

 

It is known that tangents drawn from an exterior point to a circle are of equal length.

 

So, CQ = CR = (8 - r) cm and BQ = BP = (6 - r) cm

 

Now, BC = 10 cm.

That is BQ + CQ = 10 cm

Or, (6 - r) + (8 - r) = 10

Or, 14 - 2r = 10

Or, r = 2

 

So, the radius of the incircle is 2 cm.