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anonymous

  • 5 years ago

How do the areas compare when the dimensions of one are 3 times the dimension of the other?

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    if you have a one by one square the area is 1. if you have a 3 by 3 square the area is 9. so it varies with the square of the side

  3. anonymous
    • 5 years ago
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    if your square is x by x then area is \[x^2\] if your square is 3x by 3x then area is \[(3x)^2=9x^2\]

  4. anonymous
    • 5 years ago
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    so second is 9 times the first.

  5. anonymous
    • 5 years ago
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    but one square is 15 cm and 1 is 135 cm

  6. anonymous
    • 5 years ago
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    if one square is 15 cm i assume you mean the side is 15cm yes?

  7. anonymous
    • 5 years ago
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    no the area

  8. anonymous
    • 5 years ago
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    oh the area is 15 square cm so the side is \[\sqrt{15}\]

  9. anonymous
    • 5 years ago
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    but the question is How do the areas compare when the dimensions of one are 3 times the dimension of the other?

  10. anonymous
    • 5 years ago
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    as i said, if one has side x and the other 3x the area of the second is 9 times the area of the first . just like 135 is 9 times 15

  11. anonymous
    • 5 years ago
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    so the answer is, "if the dimensions of one square are 3 times the dimensions of the other, then the area of the bigger square is 9 times the area of the smaller."

  12. anonymous
    • 5 years ago
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    thankss <3

  13. anonymous
    • 5 years ago
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    Wait sorry. The question was: How do the areas of two parallelograms compare when the dimensions of one are 3 times the dimension of the other?

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