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Q5. Is it possible to design a rectangular park of perimeter 80 metres and area 400 \(m^2\). If so, find its length and breadth.
Answer :

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:
\( 2 ( x + {{400} \over {x}}) = 80 \)
\( 2 ( {{x^2 + 400} \over {x}})\)
\(2x^2 + 800 = 80x\)
\(2x^2 - 80x + 800 =0\)
\( x^2 - 40x + 400 = 0\)
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 \(m^2\).
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 = \({{400} \over {x}} = {{400} \over {20}} = 20 m \)