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Pyramid Theorem

  1. Apr 13, 2007 #1
    My math teacher handed this out today without explaining, and told us to try and solve it. He gave us the formula to solve the volume of a pyramid (1/3 x area of base x height) and told us to explain WHY it was like that. I already figured out how to explain the surface area of the tetrahedron (he told us to figure out the surface area, without the base, and volume for a square based pyramid first, and then just multiply by two), but I'm a bit stuck on the volume part. He figured this one would stump us since, although we understand how to find a triangle's area, (b x h)/2, we wouldn't understand a pyramid. He also said we should follow the way of Eudoxus who claimed: a pyramid can be approximated to a collection of slabs of reducing size. For example, a square-based pyramid is comparable to pile of square based cuboids. By increasing the number of slabs and reducing their heights, the approximation would improve.

    The theorem is v=(Abase x h)/3, where "h" is the perpendicular height.

    The proof, or what I think it is, is:
    Consider any pyramid, perpendicular height "h" and with area of base "A"

    I believe, you then split the pyramid into n layers. Let n be the unknown number of layers and k be the layer you are measuring. So, by similarity the kth layer will have a base with dimension k/n as a fraction of the original base, right? Then the area of the base would be (k/n)squared x A.

    The area of the base of each layer will be (1/n)squared x A, (2/n)squared x A,..., (n/n)squared x A.

    That's about all I have so far.

    I'm not quite sure how that works out; my friend taught me it, so I don't totally understand it. Like what does it mean "by similarity the kth layer will have a base with dimension k/n as a fraction of the original base"?
     
  2. jcsd
  3. Apr 13, 2007 #2

    Hurkyl

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    Consider a triangle for simplicity; the idea is the same.

    Draw a triangle. Choose one side to be the base. Halfway between the base and the opposite vertex, draw a line parallel to the base. How long is the line segment on that line that lies inside the triangle? What if I drew this line one third of the way between the base and the opposite vertex?




    On an unrelated note -- if you have exceptionally good spatial visualization skills, (or the right objects to play with), you can actually pack three congruent tetrahedra inside a cube.
     
  4. Apr 13, 2007 #3
    I do not understand what the point of drawing the triangle with the line is. I wouldn't know how long the line segment is though because the pyramid is to have unknown values.

    I can visualize what you mean by the three congruent tetrahedra in a cube though.
     
  5. Apr 13, 2007 #4
    Now, find the volume of each layer and sum up the volumes. What happens when n is very large (think infinity)?
     
  6. Apr 13, 2007 #5
    Umm, you then consider each layer as a prism? So, each will be h/n units tall and the volume of the kth slab would be...er, (h/n) x (k/n)squared x A?
     
  7. Apr 13, 2007 #6
    Actually, each layer is a parallelepiped. And yes, the volume would be h/n x (k/n)2 x A.
     
  8. Apr 13, 2007 #7
    A parallelepiped?! What's that?

    So the volume of the pyramid would be approximately:

    V= (h/n)(1/n)squared x A + (h/n)(2/n)squared x A +...+(h/n)(n/n)squared x A.
    = (Ah/n cubed)(1squared + 2squared +...+ n squared)?
     
  9. Apr 13, 2007 #8
    Correct. Now simplify and take the limit as n goes to infinity.
     
  10. Apr 13, 2007 #9
    How do I simplify it? :eek:
     
  11. Apr 13, 2007 #10
    If I remember correctly, 12 + 22 + ... + n2 = n (n + 1) (2n + 1) / 6. (Google around for "summation formulas".)
     
  12. Apr 13, 2007 #11
    Erm, my friend said it was:12 + 22 + ... + n2 = 1/6n (n + 1) (2n + 1)

    But the problem is I have the formula I just don't know how to explain why the formula is the way it is. I just don't get the 1/6 part.
     
  13. Apr 13, 2007 #12
    Hurkyl, This got me to thinking since the area of a triangle is 1/2b*h and can be demonstrated by constructing a similar triangle as the original from the left over pieces in the rectangle of dimensions b,h and in an analogous fashion packing three pyramids into a rectanglar solid of dims a,a,h whether by extension such a relation may obtain for higher dimensional spaces of N dims where the enclosed space would be 1/N*a^(n-1)*h.
     
  14. Apr 14, 2007 #13
    That is exactly the same thing I wrote down. There are ways of deriving this formula but it's a bit tricky if you ask me. Anywho, just use it, simplify and take the limit.
     
  15. Apr 14, 2007 #14
    The problem is, my homework requires me to explain the formulas. I understand the rest, just not why there is a 1/6 in the middle of the equation. My teacher requires that we explain why the formula is the way it is; the rest of my class is stumped too...
     
  16. Apr 16, 2007 #15
    That's from the derivation of the formula. I'm sure you can find a derivation somewhere after googling a bit.
     
  17. Apr 16, 2007 #16

    AKG

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    You can prove the formula for the sum of the first n squares by induction.
     
  18. Apr 17, 2007 #17
    Induction?
     
  19. Apr 17, 2007 #18
  20. Apr 17, 2007 #19
    As for deriving the formula for the sum of the squares
    1/6n (n + 1) (2n + 1), maybe
    my post #4 in this thread here helps.
     
  21. Apr 19, 2007 #20
    Mm, I asked my teacher what induction meant. I think I get the theorem now. Thanks a lot, y'all!
     
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