Given,

A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at points P, Q, R and S respectively.

$AB=6cm$,

$BC=7cm$,

$CD=4cm$

Let $AD=X$

As$AP$  and $AS$ are the tangents.
$AP=AS$

Similarly,

$BP=BQ$

$CQ=CR$

and $DR=DS$

So, In $ABCD$

$AB+CD=AD+BC$  (This is the property of a cyclic quadrilateral)

$(6+4=7+X)$

$\Rightarrow 10=7+X$

$\Rightarrow X=3$

Therefore, $AD=3cm$.