In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S.

(a) 9 cm

T angents drawn from an external point to a circle are equal.

So, $A Q=A P=5 \mathrm{~cm}$

$\mathrm{CR}=\mathrm{CS}=3 \mathrm{~cm}$

and $\mathrm{BR}=(\mathrm{BC}-\mathrm{CR})$

$\Rightarrow \mathrm{BR}=(7-3) \mathrm{cm}$

$\Rightarrow \mathrm{BR}=4 \mathrm{~cm}$

$\mathrm{BQ}=\mathrm{BR}=4 \mathrm{~cm}$

$\therefore \mathrm{AB}=(\mathrm{AQ}+\mathrm{BQ})$

$\Rightarrow \mathrm{AB}=(5+4) \mathrm{cm}$

$\Rightarrow \mathrm{AB}=9 \mathrm{~cm}$