The points $\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S}$  of the circle touch the sides $\mathrm{AB},\mathrm{BC},\mathrm{CD},\mathrm{DA}.$ If AP = 9 cm, BP = 7 cm, CQ = 5 cm, and DR = 6 cm, then find the perimeter of a quadrilateral ABCD. 

Assume a circle is inscribed in a quadrilateral ABCD that touches the circle at four points, and then we will use the property of tangents of a circle so that the lengths of tangents drawn from the external points are equal.  

The perimeter of the quadrilateral is adding all sides

$\mathrm{AB},\mathrm{BC},\mathrm{CD}$ and $\mathrm{AD}$ are the tangents to the circle at P, Q, R, and S

Given, AP = 9 cm, BP = 7 cm, CQ = 5 cm, and DR = 6 cm

We know that the lengths of the two tangents from a point to a circle are equal.

Thus, $\mathrm{AP}=\mathrm{AS}=9 \mathrm{cm}$

$\text{BP}=\text{BQ}=7 \text{cm}$

$\text{CQ}=\text{CR}=5 \text{cm}$

$\text{DR}=\text{DS}=6 \text{cm}$

Perimeter of quadrilateral $\mathrm{ABCD}=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{AD}$

$=\mathrm{AP}+\mathrm{PB}+\mathrm{BQ}+\mathrm{QC}+\mathrm{CR}+\mathrm{DR}+\mathrm{AS}+\mathrm{SD}$

$=9+7+7+5+5+6+9+6=54 \text{cm}$

Therefore, the perimeter of a quadrilateral ABCD = 54 cm