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Getting in a muddle

  1. Mar 7, 2007 #1
    hi, soz, this is way below my level but i havnt done these in age and are tired and getting in a muddle. thnx for clarifying things...

    1. The problem statement, all variables and given/known data

    Two mathematically similar frustums have heights of 20cm and 30cm

    The surface area of the smaller frustum is 450cm^2

    qu) calculate the surface area of the larger frustum

    2. Relevant equations

    surface area of a frustum = 2pi x dia. x (square root)[h^2 + d^2]

    3. The attempt at a solution

    right, i am being a moron i know, but i forget where the dimensional adjustment to

    20/450 = 30/ans

    is amplied. I was thinking of something like:

    20/450^2 = 30/ans^2

    but then i tried using my own example of stuff like cubes and swaures and stuff and it didnt work.

    then i tried

    20^2/450 = 30^2/ans

    but then using my own dimensional examples it didnt work.

    i also tried using the surface area formula for frustums, but that didnt work either.

    i know i am being a dumbhead, but hey, i guess its got i havn't answered these qus in years lol. and it called a frustum for a reason ;)

  2. jcsd
  3. Mar 7, 2007 #2


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    Science Advisor

    Irrelevant! You are told that these figures are "mathematically similar" which means that they have the same "shap": all lengths in the large one are 30/20= 3/2 the corresponding small one. Since the "area" of anything depends upon multiplying two lengths together, area always is proportion to the square of a length. If in going from the smaller to the larger, all lengths are multiplied by 3/2, what is the area multiplied by?

  4. Mar 7, 2007 #3
    would it be 9/4?
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