SOLUTION: Please show me how to solve this word problem: The width of a rectangular parking lot is 53ft less than its length. Determine the dimensions of the parking lot if it measures 2

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Question 726772: Please show me how to solve this word problem:
The width of a rectangular parking lot is 53ft less than its length. Determine the dimensions of the parking lot if it measures 220ft diagonally.

Found 2 solutions by Alan3354, DrBeeee:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The width of a rectangular parking lot is 53ft less than its length. Determine the dimensions of the parking lot if it measures 220ft diagonally.
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W%5E2+%2B+L%5E2+=+220%5E2
L = W + 53
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W%5E2+%2B+%28W%2B53%29%5E2+=+220%5E2
Solve for W
L = W + 53

Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
Let w = the width of the driveway.
Let l = the length of the driveway.
Let d = the diagonal of the driveway.
The diagonal of a rectangle is the hypotenuse of the right triangle with legs equal to length and width.
We are given that
(1) l = w + 53 and
(2) d = 220
Applying the Pythagorean theorem yields
(3) d^2 = l^2 + w^2 or
(4) 220^2 = (w+53)^2 + w^2 or
(5) w^2 + 106w + 53^2 + w^2 - 220^2 = 0 or
(6) 2w^2 + 106w - 45591 = 0 or
(7) w^2 + 53w - 45591/2 = 0
Using the positive square root in the quadratic equation gives us
(8) w = (-53 + sqrt(53^2 + 2*45591))/2 or
(9) w = (-53 + sqrt(2809 + 91182)/2 or
(10) w = (-53 + sqrt(93991)/2 or
(11) w = (-53 + 306.579...)/2 or
(12) w = (253.579...)/2 or
(13) w = 126.789...
Then using (1) we get
(14) l = 179.789...
Let's check w and l using (3).
Is (220^2 = (179.789...)^2 + (126.789...)^2)?
Is (220^2 = 32324.357... + 16075.643...)?
Is (48400 = 48400)? Yes
Answer: The parking lot is 179.789... ft by 126.789... ft.