SOLUTION: A piece of wire 42 centimeters long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle.

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Question 29069: A piece of wire 42 centimeters long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle.
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
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A piece of wire 42 centimeters long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle.
Let the length of the rectangle be L cms and
let the width of the rectangle be B cms
42 cms bent into the shape of the above rectangle means
The perimeter of the rectangle is 42
That is 2(L+B) = 42
That is (L+B) = 21 ----(1)
It should be length twice its width.
(by convention the length is larger than the width)
That is L = 2B ----(2)
Putting (2) in (1), that is substituting for L in (1), we get
L+B= 21 giving 2B+B = 21 implying 3B = 21 which gives B=21/3 = 7
B=7 and putting B in (2), we have L= 2B = 2X7 = 14
Answer: Length = 14 cms and width = 7 cms.
Verification: 14 is of course twice 7
and 2X(14+7) = 2X21= 42 which is the perimeter.
Therefore our values are correct