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?

2 Answers
Jan 22, 2016

#color(purple)("The sides are 150 feet by 200 feet")#

Explanation:

Let length be #L#

Let width be #W#

Given #L=W+50#

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Tony B

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#

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Using the formula:

#ax^2+bx+c=0#

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

#a=1#
#b=50#
#c=-30000#
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#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")#

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Check

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

#40000+22500->62500#

#62500->62500 color(purple)(" 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 + (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