Question 144734
Let W=width, L=length


Since the "parking lot is 50 ft longer than it is wide", this means that {{{L=W+50}}}



{{{L^2+W^2=250^2}}} Start with Pythagoreans theorem



{{{(W+50)^2+W^2=250^2}}} Plug in {{{L=W+50}}}




{{{W^2+100W+2500+W^2=62500}}} Foil {{{(W+50)^2}}}.  Square 250 to get 62,500. 



{{{W^2+100W+2500+W^2-62500=0}}} Subtract 62,500 from both sides



{{{2W^2+100W-60000=0}}} Combine like terms




{{{2(W+200)(W-150)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{W+200=0}}} or  {{{W-150=0}}} 


{{{W=-200}}} or  {{{W=150}}}    Now solve for W in each case



So our answer is 

 {{{W=-200}}} or  {{{W=150}}} 



Discarding the negative width, the only answer is {{{W=150}}} . So the width is 150 feet.


{{{L=W+50}}} Go back to the first equation



{{{L=150+50}}} Plug in  {{{W=150}}}



{{{L=200}}} Add



So the width is 150 feet and the length is 200 feet