Is it possible to design a rectangular park of perimeter $80\, m$ and area $400\, m^2$? If so,find its length and breadth.

(A) Let the length and breadth of the park be $l$ and $b$ respectively.
The perimeter is given by $2(l + b) = 80\, m$,which simplifies to $l + b = 40$,or $b = 40 - l$.
The area is given by $l \times b = 400\, m^2$.
Substituting $b = 40 - l$ into the area equation: $l(40 - l) = 400$.
This gives $40l - l^2 = 400$,or $l^2 - 40l + 400 = 0$.
Comparing this with the standard quadratic equation $al^2 + bl + c = 0$,we have $a = 1, b = -40, c = 400$.
The discriminant $D = b^2 - 4ac = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0$.
Since $D = 0$,the equation has real and equal roots,meaning the situation is possible.
The roots are given by $l = -b / (2a) = -(-40) / (2 \times 1) = 40 / 2 = 20$.
Thus,the length $l = 20\, m$ and the breadth $b = 40 - 20 = 20\, m$.