In the given figure, a circle touches all the four side of quadrilateral ABCD with AB=6 cm, BC=7 cm and CD=4 cm. Find AD.

It is known that the length of the tangents drawn from an external point to the circle is equal.

$\mathrm{AP}=\mathrm{AS}$..............(i)

$\mathrm{BP}=\mathrm{BQ}$...............(ii)

$\mathrm{CQ}=\mathrm{CR}$..................(iii)

$\mathrm{DR}=\mathrm{DS}$...................(iv)

Now,

$\mathrm{AB}+\mathrm{CD}=\left(\mathrm{AP}+\mathrm{PB}\right)+\left(\mathrm{CR}+\mathrm{DR}\right)$

$\mathrm{AB}+\mathrm{CD}=\left(\mathrm{AS}+\mathrm{BQ}\right)+\left(\mathrm{CQ}+\mathrm{DS}\right)$

$\mathrm{AB}+\mathrm{CD}=\left(\mathrm{AS}+\mathrm{DS}\right)+\left(\mathrm{BQ}+\mathrm{CQ}\right)$

$\mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}$

$\mathrm{AD}=\mathrm{AB}+\mathrm{CD}-\mathrm{BC}$

$\mathrm{AD}=6+4-7$

$\mathrm{AD}=3\mathrm{cm}$