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Thread: Perimeter

  1. #1
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    Perimeter

    How would I solve this?

    "The perimeter of a rectangle is 40 cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm. Find the dimensions of the original rectangle."
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  2. #2
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    Quote Originally Posted by largebabies View Post
    How would I solve this?

    "The perimeter of a rectangle is 40 cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm. Find the dimensions of the original rectangle."
    Let l denote the length of the rectangle and w the width. The you know:

    $\displaystyle 2l + 2w = 40$
    $\displaystyle 2\cdot (2l)+2\cdot \left(\frac12w\right) = 40+16$

    That means you have to solve for l and w:

    $\displaystyle l+w=20$
    $\displaystyle 4l+w=56$

    I've got l = 12 and w = 8
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  3. #3
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    $\displaystyle P=2l+2w, $where l = length and w = width
    $\displaystyle 40cm=2l+2w$
    and
    $\displaystyle 56cm=2*2l+\frac{2w}{2}$

    $\displaystyle 56cm=4l+w$

    $\displaystyle 40cm-2w=2l$

    $\displaystyle 20cm-w=l$

    $\displaystyle 56cm=4(20cm-w)+w$

    $\displaystyle 56cm=80cm-4w+w$

    $\displaystyle -24cm=-3w$

    $\displaystyle w=8cm$

    now plug $\displaystyle w=8cm$ into the original equation

    $\displaystyle 40cm=2(8cm)+2w$

    $\displaystyle 24cm=2w$

    $\displaystyle w=12cm$
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