Respuesta :
Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet
Solution:
Given that
Area of rectangular parking lot = 15000 square feet
Length is 20 feet more than the width.
Need to find the dimensions of rectangular parking lot.
Let assume width of the rectangular parking lot in feet be represented by variable "x"
As Length is 20 feet more than the width,
so length of rectangular parking plot = 20 + width of the rectangular parking plot
=> length of rectangular parking plot = 20 + x = x + 20
The area of rectangle is given as:
[tex]\text {Area of rectangle }=length \times width[/tex]
Area of rectangular parking lot = length of rectangular parking plot [tex]\times[/tex] width of the rectangular parking
[tex]\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}[/tex]
But it is given that Area of rectangular parking lot = 15000 square feet
[tex]\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}[/tex]
Solving the above quadratic equation using quadratic formula
General form of quadratic equation is
[tex]{ax^{2}+\mathrm{b} x+\mathrm{c}=0[/tex]
And quadratic formula for getting roots of quadratic equation is
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
In our case b = 20, a = 1 and c = -15000
Calculating roots of the equation we get
[tex]\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}[/tex]
[tex]\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}[/tex]
As variable x represents width of the rectangular parking lot, it cannot be negative.
=> Width of the rectangular parking lot "x" = 112.882 feet
=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882
Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.