# A rectangular parking lot is 50ft longer than it is wide. How do you determine the dimensions of the parking lot if it measures 250 ft diagonally?

Jan 22, 2016

$\textcolor{p u r p \le}{\text{The sides are 150 feet by 200 feet}}$

#### Explanation:

Let length be $L$

Let width be $W$

Given $L = W + 50$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Pythagoras

${L}^{2} + {W}^{2} = {250}^{2}$

But $L = W + 50$ giving:

${\left(W + 50\right)}^{2} + {W}^{2} = {250}^{2}$

${W}^{2} + 100 W + 2500 + {W}^{2} = {250}^{2}$

$2 {W}^{2} + 100 W - 60000 = 0$

Divide both sides by 2 giving:

${W}^{2} + 50 W - 30000 = 0$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the formula:

$a {x}^{2} + b x + c = 0$

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

$a = 1$
$b = 50$
$c = - 30000$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$W = \frac{- 50 \pm \sqrt{{50}^{2} - 4 \left(1\right) \left(- 30000\right)}}{2 \left(1\right)}$

$W = \frac{- 50 \pm \sqrt{2500 + 120000}}{2}$

$W = \frac{- 50 \pm 350}{2}$

$W = - 200 \mathmr{and} + 150$

$\textcolor{b l u e}{\text{The "-200" is not logical so "W=150" feet}}$

$\textcolor{b l u e}{\implies L = 150 + 50 = 200 \text{ feet}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check

${200}^{2} + {150}^{2} \to {250}^{2}$

$40000 + 22500 \to 62500$

$62500 \to 62500 \textcolor{p u r p \le}{\text{ Which is correct}}$

Jan 22, 2016

length = 200 ft and width = 150 ft

#### Explanation:

If we let the width = x , then length = x + 50

( I recommend you draw a sketch )

There is now a right-angled triangle with 2 sides of x and
( x + 50 ) and hypotenuse = 250.

Using Pythagoras' Theorem to obtain :

${x}^{2} + {\left(x + 50\right)}^{2} = {250}^{2}$

(distribute the bracket )

${x}^{2} + {x}^{2} + 100 x + 2500 = 62500$

(collect 'like terms' and equate to 0 )

$2 {x}^{2} + 100 x - 60000 = 0$

( divide equation by 2 ) : ${x}^{2} + 50 x - 30000 = 0$

We now require 2 factors of - 30000 which multiply to - 30000

and add to + 50.( These are 200 and - 150)

[You should use the quadratic formula if not sure .]

equation now becomes (x + 200 )(x - 150 ) = 0

solving gives : x = - 200 or x = 150

x ≠ - 200 hence x = 150

so width = x = 150 and length = x + 50 =150 + 50 = 200