SOLUTION: The length of a rectangle is 10 feet longer than it is wide. If each side is increased 10 feet, then the area is multiplied by 2. What was the size of the original rectangle?

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 10 feet longer than it is wide. If each side is increased 10 feet, then the area is multiplied by 2. What was the size of the original rectangle?      Log On


   

Question 1060532: The length of a rectangle is 10 feet longer than it is wide. If each side is increased 10 feet, then the area is multiplied by 2. What was the size of the original rectangle?
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
l = w+10 —> Area = lw = (w+10)w = w%5E2+%2B+10w+ (substituted w+10 for l)
The other relation given is:
(l+10)(w+10) = 2*Area
Again, substitute w+10 for l:
+%28%28w%2B10%29+%2B+10%29+%28w%2B10%29+=+2%2AArea+
++%28w%2B20%29%28w%2B10%29+=+2%2AArea+
++w%5E2+%2B+30w+%2B+200+=+2%2AArea+
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2*Area is +2%2A%28w%5E2%2B10w%29+ or +2w%5E2+%2B+20w+
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So +w%5E2+%2B+30w+%2B+200+=+2w%5E2+%2B+20w+
+-w%5E2+%2B10w+%2B+200+=+0+
++w%5E2+-+10w+-+200+=+0+
++%28w-20%29%28w%2B10%29+=+0+
+w=20 or +w=-10
Discard w=-10 because it makes no sense.
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Ans: width is 20, length is 30
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Check: Area = 20*30 = 600
and for the expanded rectangle:
(20+10)(30+10) = (30)(40) = 1200 = 2*Area