A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres 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. Find the length and breadth of the rectangular park.

Let the length of the rectangular park be x m then, breadth be $(x-3)m$

$\therefore$ Area of rectangular park = $x(x-3)m^2$

Area of isosceles triangular park = $\frac{1}{2}(x-3)\times 12m^2$

= $6(x-3)m^2$

According to the given condition,

$x(x-3)-6(x-3)=4$

$x^2-3x-6x+18=4$

$x^2-9x+14=0$

$x^2-7x-2x+14=0$

$x(x-7)-2(x-7)=0$

$(x-2)(x-7)=0$

$x=2$ or 7

$\therefore x=7m$

(As breadth can't be negative)

and $x-3=(7-3)m=4m$

Hence, length and breadth of the rectangular park is 7m and 4m respectively.