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# 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 = $${{400} \over {x}} = {{400} \over {20}} = 20 m$$