The length of a rectangle exceeds its breadth by 9 cm.

Let the breadth of the rectangle be $x \mathrm{~cm}$.

Therefore, the length of the rectangle will be $(\mathrm{x}+9) \mathrm{cm}$.

$\therefore$ Area of the rectangle $=\mathrm{x}(\mathrm{x}+9) \mathrm{cm}^{2}$.

If the length and breadth are increased by $3 \mathrm{~cm}$ each,

a rea $=(\mathrm{x}+3)(\mathrm{x}+9+3) \mathrm{cm}^{2}$.

Now,

$(\mathrm{x}+3)(\mathrm{x}+12)-\mathrm{x}(\mathrm{x}+9)=84$

or $\mathrm{x}^{2}+15 \mathrm{x}+36-\mathrm{x}^{2}-9 \mathrm{x}=84$

or $6 \mathrm{x}=84-36$

or $\mathrm{x}=\frac{48}{6}=8$

Thus, brea $d$ th of the rectangle $=8 \mathrm{~cm}$.

Length of the rectangle $=(8+9)=17 \mathrm{~cm}$.