A circle with centre O is inscribed in a right angled triangle ABC in which B = 90^o. The circle touches AB, BC and AC at points P, Q and R respectively. If PB=2 cm, find the diameter of the circle.

Answer :

We form our diagram , As :

We know " A tangent to a circle is perpendicular to the radius at the point of tangency. "

So,
$\angle \mathrm{OPB} = \angle \mathrm{OQB} = 90° ---- ( 1 )$
And
$\angle \mathrm{PBQ}\hspace{0.17em} = 90° ( \mathrm{As} \mathrm{given} \angle \mathrm{B} = 90° )$
So,
In quadrilateral PBQO from angle sum property , we get

$$\begin{gathered} \angle \mathrm{OPB} + \angle \mathrm{OQB} + \angle \mathrm{PBQ}\hspace{0.17em} + \angle \mathrm{QOP} = 360° \\ \mathrm{Substitute} \mathrm{values} , \mathrm{we} \mathrm{get} \\ \Rightarrow 90° + 90° + 90°\hspace{0.17em} + \angle \mathrm{QOP} = 360° \\ \Rightarrow \angle \mathrm{QOP} = 90° \end{gathered}$$

So,
Quadrilateral PBQO is a parallelogram as its opposite angles are equal.
So, PB = OQ and OP = QB  [As opposite sides of parallelogram are equal]
Now, OP = OQ [Radii of same circle]
Hence, PB = OQ = OP = QB
So, PBQO is a square.
Now, PB = 2 cm
So, OQ = PB = 2 cm
Radius of circle  =  2cm
So,
Diameter of circle  = 2 radius =   4 cm                                  ( Ans )