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What are the dimensions of a rectangle with an area of 204 and a perimiter of 58?
Wiki User
∙ 12y ago
Suppose the length is L and the breadth is B.
Then since perimeter = 58,
2*(L+B) = 58 so that L+B = 29 that is L = 29-B
Now Area = L*B = 204
That is (29-B)*B = 204
Or B2 - 29B + 204 = 0
which is equivalent to (B-12)*(B-17) = 0
So, using L+B = 29, L and B are 12 and 17.
Wiki User
∙ 12y ago
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