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A piece of wire 42 cm long is bent into the shape of a rectangle whose width is twice...

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samerrima | Student, Grade 9 | eNoter

Posted June 22, 2010 at 7:36 AM via web

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A piece of wire 42 cm 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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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted June 22, 2010 at 7:40 AM (Answer #1)

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The length of the wire used to make the rectangle is 42 cm

That means that the circumference  of the rectangle is 42

But the circumference of a rectangle = 2L + 2W where L is the length and W is the width

Then 2L +2W = 42

But we know that  W = 2L

Now substitute w=2L

==> 2L + 2(2L )= 42

==> 6L = 42

==>  L = 42/6= 7 cm

Then W = 2(7) = 14 cm

To check the answer:

2L + 2w = 2(7) + 2(14) = 14 + 28 = 42 cm

 

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brianevery | High School Teacher | (Level 2) Adjunct Educator

Posted June 22, 2010 at 7:44 AM (Answer #2)

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With out getting into the semantics of whether the width can be longer than the length - if we write the equation 2l=w (two lengths equal one width) and the equation 2l+2w=42 and substitute 2l from the first equation for w in the second equation we get this equation

2l+2(2l)=42   simplify

2l+4l=42   now add like terms

6l=42 and then divide both sides of the equation by 6

l=7 so if the length is 7, then the width must be 14 (2*7) and we can answer the question

The width is 14 and the length is 7.

We check this by substituting 7 and 14 into the original second equation 2*7+2*14=42

do the arithmetic 14+28=42 => 42=42 it checks.

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