SOLUTION: I need help with this. This is the work I have done, but I want to see if it is correct. Thanks! A rectangular parking lot is 100 ft longer than it is wide. Determine the dimens

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Question 151993: I need help with this. This is the work I have done, but I want to see if it is correct. Thanks!
A rectangular parking lot is 100 ft longer than it is wide. Determine the dimensions of the parking lot if it measures 500 ft diagonally.
w + (w + 100) = 500
2w + 100 = 500
2w + 100 – 100 = 500 – 100
2w = 400
w = 400/2
w = 200
200 + (200 + 100) = 500

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Remember, in solving prolems such as this, where you are given the length of the diagonal of a recatangle and are asked to find the lengths of the sides, think "Pytagorean theorem". Because the diagonal is the hypotenuse of the right triangle formed by half of the rectangle.
So, c%5E2+=+a%5E2%2Bb%5E2
The diagonal, c, is given as 500 feet.
The length of the side, L, is given as W+100, and the width is, of course, just W. Putting this all together and usingc%5E2+=+L%5E2%2BW%5E2, we get:
500%5E2+=+%28W%2B100%29%5E2%2BW%5E2 Expanding this, we get:
250000+=+%28W%5E2%2B200W%2B10000%29+%2B+W%5E2 Simplifying.
250000+=+2W%5E2%2B200W%2B10000 Subtracting 250000 from both sides.
0+=+2W%5E2%2B200W-240000 Divide through by 2 to to simplify this quadratic equation:
W%5E2%2B100W-120000+=+0 Now factor this.
%28W-300%29%28W%2B400%29+=+0 so...
W+=+300 or W+=+-400 Discard the negative solution as widths are positive.
So the width of the rectangular parking lot is 300 feet and its length is 300+100 = 400 feet.
Now, along with the 500-foot diagonal, you should recognise these numbers as a Pythagorean triplet (like 3, 4, and 5) because:
500%5E2+=+300%5E2%2B400%5E2
250000+=+90000%2B160000