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.

Algebra ->  Quadratic Equations and Parabolas -> 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.       Log On


   

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) About Me  (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
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%5D%29%2F%282%2Aa%29+
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