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Integral of a triangular pyramid

  1. Jan 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the volume of a pyramid with height 24 and with base an equilateral triangle with side 11.


    2. Relevant equations


    3. The attempt at a solution

    So I know the relationship

    h/x=b/l

    where h is my height 24
    x is simply x or height above base
    b is my base 11
    l is the length of the base at height x

    i also know the integral is

    024(A(x))dx

    I am having trouble finding A(x)

    I believe that it you would use the area of the equilateral triangle

    so i know area of the triangle is base*height/2

    I also know that base is l, and height is unknown, but i do know that

    l^2=(l/2)^2 + z^2 where i take z as the height of the equilateral triangle

    then z= √(l^2 -(l/2)^2 )

    so area of the triangle would finall be

    [(l^2)√(l^2 - (l/2)^2)]/2

    I would finally plug in l=bx/h from the identity h/x=b/l

    However, when i do this, i end up with an integral i am unable to solve. I feel like there is another way to do this, can anyone help?

    edit: I mean an easier way using integrals, not volume of a pyramid please

    thanks
     
  2. jcsd
  3. Jan 25, 2013 #2

    SteamKing

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    What kind of a triangle is an equilateral triangle?
     
  4. Jan 26, 2013 #3

    haruspex

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    You've one too many l factors in there - it's an area, not a volume. You should be able to simplify the term inside the square root (bringing the l outside). Once you've done that the integral should be straightforward.
     
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