SOLUTION: A rectangle is half as wide as it is long. If both the length and width are decreased by 2 cm, the area decreases by 68 cm(squared) . Find the length of the original rectangle.

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Question 202617: A rectangle is half as wide as it is long. If both the length and width are decreased by 2 cm, the area decreases by 68 cm(squared) . Find the length of the original rectangle.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let the original length = L and the original width = W, so that the original area:
A+=+L%2AW but the problem says that the "...rectangle is half as wide as it is long.", so this can be expressed as:
W+=+%281%2F2%29L Substitute this into the first equation and you get:
A+=+%281%2F2%29L%5E2
Now, if we subtract 2 cm from the length (L-2) and 2 cm from the width (W-2) the area is decreased by 68 sq.cm. (A-68). Lets find the new area:
A-68+=+%28L-2%29%28W-2%29 but substitute W+=+%281%2F2%29L and A+=+%281%2F2%29L%5E2
%281%2F2%29L%5E2-68+=+%28L-2%29%28%281%2F2%29L-2%29 Simplifying:
%281%2F2%29L%5E2-68+=+%281%2F2%29L%5E2-3L%2B4 Subtract 1%2F2L%5E2 from both sides.
-68+=+-3L%2B4 Subtract 4 from both sides.
-72+=+-3L Divide both sides by -3.
L+=+24 and
W+=+%281%2F2%29L
W+=+12
The original length of the rectangle is 24 cm.
Check:
Original area:
A%5Bo%5D+=+L%2AW Substitute L=24 and W=12
A%5Bo%5D+=+24%2A12
A%5Bo%5D+=+288sq.cm.
Now the new area is:
A%5Bn%5D+=+%28L-2%29%28W-2%29 Substitute L=24 and W=12
A%5Bn%5D+=+%2824-2%29%2812-2%29
A%5Bn%5D+=+%2822%29%2810%29
A%5Bn%5D+=+220sq.cm.
The difference is:
A%5Bn%5D-A%5Bo%5D+=+288-220=168sq.cm.