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

Ans. Let length of park = x metres

We are given area of rectangular park = 400 \(m^2\)

Therefore, breadth of park = \({{400} \over {x}}\) metres{Area of rectangle = length × breadth}

Perimeter of rectangular park = 2 (length + breath) = \( 2 ( x + {{400} \over {x}})\) metres

We are given perimeter of rectangle = 80 metres
According to condition:

\(\Rightarrow 2 ( x + {{400} \over {x}}) = 80 \)
\( \Rightarrow 2 ( {{x^2 + 400} \over {x}})\)
\(\Rightarrow 2x^2 + 800 = 80x\)
\(\Rightarrow 2x^2 - 80x + 800 =0\)
\( \Rightarrow x^2 - 40x + 400 = 0\)

Comparing equation, \( x^2 - 40x + 400 = 0\) with general quadratic equation \( ax^2 + bx + c = 0\),

we get a = 1, b = -40 and c = 400

Discriminant = \( b^2 - 4ac \)
\( = (-40)^2 - 4 (1) (400) \)
\( = 1600 – 1600 = 0\)

Discriminant is equal to 0.
Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter 80 metres and area 400 m².

Using quadratic formula \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to solve equation,

\(x = {{40 ± \sqrt{0}} \over {2}} = {{40} \over {2}} = 20\)
Here, both the roots are equal to 20.

Therefore, length of rectangular park = 20 metres
Breadth of rectangular park = \({{400} \over {x}} = {{400} \over {20}} = 20 m \)