SOLUTION: The length of a rectangle is 6cm more than its width. The area is 216 sq cm. What are the dimensions? I have w= width, l= w+ 6cm = length A= 216cm^2
I set it up like this: 216cm^
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-> SOLUTION: The length of a rectangle is 6cm more than its width. The area is 216 sq cm. What are the dimensions? I have w= width, l= w+ 6cm = length A= 216cm^2
I set it up like this: 216cm^
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Question 643879: The length of a rectangle is 6cm more than its width. The area is 216 sq cm. What are the dimensions? I have w= width, l= w+ 6cm = length A= 216cm^2
I set it up like this: 216cm^2= w+ 6cm(w)= 216cm^2= w^2+ 6cmw= 216cm^2- w^2-6cmw=0
and that's when i just got stuck I'm not sure if i was doing it right in the first place. Can I please have a bit of guidance?
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The length of a rectangle is 6cm more than its width. The area is 216 sq cm. What are the dimensions? I have w= width, l= w+ 6cm = length A= 216cm^2
I set it up like this: 216cm^2= w+ 6cm(w)= 216cm^2= w^2+ 6cmw= 216cm^2- w^2-6cmw=0
and that's when i just got stuck I'm not sure if i was doing it right in the first place. Can I please have a bit of guidance?
Let width be W
Then length = W + 6
A, or area = LW
Since area = 216, then: 216 = W(W + 6)
(W - 12) (W + 18) = 0
W - 12 = 0, or W = - 18 (ignore as measurement CANNOT be negative)
W, or width =
Length = 12 + 6, or cm
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