The length of a photograph is 1 cm less than twice the width. The area is 28cm^2. Find the dimensions?

I have a ton of these questions... I would like to know how to solve them.Thank you!!!

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  • 1 decade ago
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    From the question:

    let put "the width of the photo" = W

    area = length x width

    area = length x W

    length = 1 cm less than twice the width

    = 1 cm less than 2W

    = 2W - 1

    area = length x W

    28 = (2W-1) x W

    28 = 2W^2 - W ****subtract 28 from both side*****

    2W^2 - W - 28 = 0 *****possible factor for 28 is 4 & 7******

    (2W + 7)(W - 4) = 0

    now try both answer 2W + 7 = 0

    and W - 4 = 0

    the first width can't be true because it's a negative number (when you solve for W)

    2W = -7 then W = -3.5 which can't be the width

    then the second answer W - 4 =0

    W = 4

    we got the width of the photo is 4

    the length is 1 cm less than twice the width

    so we got the length is 2(4) -1 which is 7

    area is 4 x 7 = 28 just like given in the first place

    *****hope this helps**********good luck**********

  • 1 decade ago

    If the area a a photograph is 28cm^2 then yuou know that the length times the width must equal 28. Now you can go about this in one of two ways. So the easiest way to think about this would be that you need to find what lengths would give you an area of 28, or what are the factors of 28. You could just create a chart and list all the possible lengths and widths and then figure out which of those has a length the is one less then twice the width. In this case the two side would be 7cm and 4cm. (4*2)-1=7.

    You could also solve this using basic algebra, but I'm not sure what level yuou are at.

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