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4 May, 20:03

The rectangle below has an area of 4 (x+3) [4 is width, x+3 is length] square units. When the dimensions of the rectangle are doubled, the area of the new rectangle in terms of x is 8 (2x+6). [ 8 being the width and 2x+6 being the length.] Will the ratio of the area of the original rectangle to the larger rectangle be the same for any positive value of x?

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  1. 4 May, 20:09
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    Yes it will be. The ratio will always be

    4 (x+3) / 8 (2x+6)

    4x+12/16x+48

    ratio = 1/4

    You can test this out by substituting some values of x.

    For example,

    x=1

    Orig=16 New=64

    16/64=1/4

    x=2

    Orig=20 New=80

    20/80=1/4

    and so on ...
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