Question

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

Answer 1

Hint: In this question we are given a quadrilateral named rectangle. Firstly we know the perimeter i.e. sum of the sides of the rectangle and area is the region covered by the sides of the rectangle. In the rectangle opposite two sides are equal and are named as length and breadth. The formula used for the perimeter of the rectangle is $2 \times \left( {L + B} \right)$ and the formula used for the area of the rectangle is $L \times B$. Here ‘$L$’ denotes the measurement of the length of the rectangle and ‘$B$’ denotes the breadth of the rectangle.

Complete step by step solution:
Let the length of rectangle be$ = x$$m$
Let the breadth of rectangle be $ = y$$m$
As we know that the perimeter of the rectangle given i.e. $80m$
So, formula for perimeter of rectangle is $2 \times \left( {L + B} \right)$
So, $2 \times \left( {L + B} \right) = 80$
$2\left( {x + y} \right) = 80$
$x + y = 40$
$x = 40 - y$ --- $1$
The area of rectangle given is $400{m^2}$
The area of rectangle is given by $L \times B$
So, $L \times B$$ = 400$
$ \Rightarrow $$x \times y = 400$
Now, put value of $x$ from $1$in this equation
$ \Rightarrow \left( {40 - y} \right) \times y = 400$
$40y - {y^2} = 400$
${y^2} - 40y + 400 = 0$
Now, by the solving the quadratic equation by Discriminant method
Formula used,$\dfrac{{ - b \pm \sqrt D }}{{2a}}$
$D = $${b^2} - 4ac = $Discriminant
On comparing the above quadratic equation i.e. ${y^2} - 40y + 400 = 0$
We get, $a = 1,b = - 40,c = 400$
On solving, value of $D$is
$
  D = {b^2} - 4ac \\
  D = {\left( { - 40} \right)^2} - 4\left( 1 \right)\left( {400} \right) \\
  D = 1600 - 1600 \\
  D = 0 \\
 $
Now, we will apply the formula of $y$as we written below
$
  y = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\
  y = \dfrac{{ - \left( { - 40} \right) + 0}}{{2 \times 1}} \\
  y = \dfrac{{40}}{2} \\
  y = 20m \\
 $
Now, value of $x$ is
$
   \Rightarrow x = 40 - y \\
  x = 40 - 20 \\
  x = 20m \\
 $
So, $L = 20m,B = 20m$
Which is the required solution.

Note: One can assume the length and breadth to be any variable, it is not necessary to assume length and breadth to be ‘$L$’ and ‘$B$’.On the solving the quadratic equation you may get one value of $y$ to be negative, but we have to neglect that value as dimension of any length cannot be negative.