Question

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?

Answer 1

`color(purple)("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:

`(W+50)^2+W^2=250^2`

`W^2+100W+2500+W^2=250^2`

`2W^2+100W -60000=0`

Divide both sides by 2 giving:

`W^2+50W-30000=0`

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

`ax^2+bx+c=0`

where
`x=(-b+-sqrt(b^2-4ac))/(2a)`

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

`W=(-50+-sqrt(50^2-4(1)(-30000)))/(2(1))`

`W=(-50+-sqrt(2500+120000))/2`

`W=(-50+-350)/2`

`W=-200 or +150`

`color(blue)("The "-200" is not logical so "W=150" feet")`

`color(blue)(=> L= 150+50=200" feet")`

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

`200^2+150^2->250^2`

`40000+22500->62500`

`62500->62500 color(purple)(" Which is correct")`

Answer 2

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 + (x + 50 )^2 = 250^2`

(distribute the bracket )

`x^2 + x^2 + 100x + 2500 = 62500`

(collect 'like terms' and equate to 0 )

`2x^2 +100x - 60000 = 0`

( divide equation by 2 ) : `x^2 + 50x - 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