A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions

Mathematics

1 answer:

faust18 [17]2 years ago

3 0

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3>Solution:</h3>

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:

$\text {Area of rectangle }=length \times width$

Area of rectangular parking lot = length of rectangular parking plot $\times$ width of the rectangular parking

$\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}$

But it is given that Area of rectangular parking lot = 15000 square feet

$\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}$

Solving the above quadratic equation using quadratic formula

General form of quadratic equation is

${ax^{2}+\mathrm{b} x+\mathrm{c}=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}$

$\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}$