SOLUTION: A college is creating a new rectangular parking lot. The length is 0.12 mile longer than the width and the area of the parking lot is 0.054 square mile. Find the length and width

Algebra ->  Pythagorean-theorem -> SOLUTION: A college is creating a new rectangular parking lot. The length is 0.12 mile longer than the width and the area of the parking lot is 0.054 square mile. Find the length and width       Log On


   

Question 1085387: A college is creating a new rectangular parking lot. The length is 0.12 mile longer than the width and the area of the parking lot is 0.054 square mile. Find the length and width of the parking lot.
Found 2 solutions by math_helper, jorel1380:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
W%2A%28W%2B0.12%29+=+0.054+
+W%5E2+%2B+0.12W+-+0.054+=+0+
Using quadratic formula:
+W=+%281%2F2%29%2A%28-0.12+%2B-+sqrt%280.12%5E2+-+4%2A1%2A%28-0.054%29%29%29+
+W+=+%281%2F2%29+%2A+%28-0.12+%2B-+0.48%29+
+W+=+0.18+ and/or +W+=+-0.30+
Discard the W<0 answer as it is nonsensical.
+W+=+0.18mi+ —> +L+=+0.30mi
—
Check: 0.18mi * 0.30mi = 0.054mi^2

Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
Let w and w+0.12 be your width and length,respectively. Then:
w(w+0.12)=0.054
w^2+0.12w-0.054=0
500w^2+60w-27=0
(50w-9 )( 10w+3)=0
w=0.18, -0.3
w+0.12=30
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