$A$ rectangular park is to be designed whose breadth is $3\, m$ less than its length. Its area is to be $4\, m^2$ more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude $12\, m$ (see Figure). Find its length and breadth.

(N/A) Let the breadth of the rectangular park be $x\, m$.
Then,its length is $(x+3)\, m$.
Therefore,the area of the rectangular park is $x(x+3) = (x^2 + 3x)\, m^2$.
The area of the isosceles triangle is $\frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times x \times 12 = 6x\, m^2$.
According to the problem,the area of the rectangular park is $4\, m^2$ more than the area of the triangular park:
$x^2 + 3x = 6x + 4$
$x^2 - 3x - 4 = 0$
$(x - 4)(x + 1) = 0$
So,$x = 4$ or $x = -1$.
Since breadth cannot be negative,$x = 4$.
Thus,the breadth of the park is $4\, m$ and its length is $4 + 3 = 7\, m$.