A circle is inscribed in a triangle ABC. It touches sides AB, BC and AC at the points P, Q and respectively. If BP = 5 cm, CQ =7 cm and AR = 6 cm, then the perimeter (in cm) of the $$\triangle$$ABC is:

We know that,  Tangent segments from a common point external to a circle have the same length.

As above given Diagram, we can say

$$AP = AR , BP = BQ,  CQ = CR$$  

$$\therefore AP = AR = 6 cm,  BP = BQ = 5 cm,  CQ = CR = 7 cm$$ 

$$\Rightarrow AB = (AP + BP ) = 6 + 5 = 11cm$$

$$\Rightarrow BC = (BQ + CQ) = 5 + 7 = 12cm$$

$$\Rightarrow AC = (AR + CR) = 6 + 7 = 13cm$$

$$\therefore$$  Perimeter of the $$\triangle ABC = AB + BC + AC = 11 + 12 + 13 = 36cm $$