Question 282703: A certain rectangle's length is 5 feet longer than its width. If the area of the rectangle is 50 square feet, find its larger dimension
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! L = length
W = width
Area= L*W, for a rectangle
A = 50 ft^2 :: given
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L = W + 5 :: Length is 5 ft longer than the width, so you have to add 5 to W to make them equal
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Substitute L=W+5
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(W+5) * W = 50
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Collect and simplify
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W^2 + 5W = 50
W^2 + 5W - 50 = 0
(W + 10)(W - 5) = 0
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So our candidate solutions are W= -10 and W=5.
But a negative width is nonsensical, so our only practical option is W=5.
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Substituting W=5, we can find L.
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L = W+5
L = 5+5 = 10
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Checking the answer, we see
5*10 = 50, which matches the setup
10 = 5+5, which also matches the setup
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Answer:
The larger dimension is the length, which is 10.
The width is 5.
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Done.
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