SOLUTION: A rectangular parking lot is 50 ft longer than it is wide. Determine the dimensions of the parking lot if it measures 250 ft diagonally.

Question 331016: A rectangular parking lot is 50 ft longer than it is wide. Determine the dimensions of the parking lot if it measures 250 ft diagonally.
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
A^2+B^2=C^2
50^2+(X+50)^2=250^2
2,500+X^2+100X+2,500=63,500
X^2+100X+5,000-62,500=0
X^2+100X-57,500=0
AMP Parsing Error of [x = (-b +- sqrt( b^2-4*a*c ])/(2*a) ]: Invalid function: opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 88. .
X=(-100+-SQRT[100^2-4*1*-57,500)/2*1
X=(-100+-SQRT10,00+230,000])/2
X=(-100+-SQRT240,000)/2
X=(-100+-489.9)/2
X=(-100+489.9)/2
X=389.9/2
X=194.95+50=244.96 ANS. FOR THE DIAGONAL.
PROOF:
50^2+244.95^2=250^2
2,500+60,000=62,500
62,500~62,500

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