Is it possible to design a rectangular park of perimeter $80 \mathrm{~m}$ and area $400 \mathrm{~m}^{2}$ ?

Solution:

Perimeter of the rectangular park = 80 m

$\Rightarrow$ Length $+$ Breath of the park $=\frac{\mathbf{8 0}}{\mathbf{2}} \mathrm{m}=40 \mathrm{~m}$.

Let the breadth be x metres, then length = (40 – x) m

Here, $x<40$

$x \times(40-x)=400[$ Each $=$ area of the park $]$

i.e., $-x^{2}+40 x-400=0$

i.e., $x^{2}-40 x+400=0$

i.e., $(x-20)^{2}=0$

$\Rightarrow x=20$

Thus, we have length = breadth = 20 m

Therefore, the park is a square having 20 m side.