The length of a rectangle exceeds its breadth by 7 cm.

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

Then, its length will be $(\mathrm{x}+7) \mathrm{cm}$.

The area of the rectangle will be $(\mathrm{x})(\mathrm{x}+7) \mathrm{cm}^{2}$.

$\therefore(x+3)(x+7-4)=(x)(x+7)$

$\Rightarrow(x+3)(x+3)=x^{2}+7 x$

$\Rightarrow x^{2}+3 x+3 x+9=x^{2}+7 x$

$\Rightarrow x^{2}+6 x+9=x^{2}+7 x$

$\Rightarrow 9=x^{2}-x^{2}+7 x-6 x$

$\Rightarrow 9=x$

$\Rightarrow x=9 \quad$ (by transposition)

Breadth of the original rectangle $=9 \mathrm{~cm}$

Length of the original rectangle $=(\mathrm{x}+7)=(9+7)=16 \mathrm{~cm}$