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Math Help - Pyramid Frustum Question

  1. #1
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    Pyramid Frustum Question

    Find the volume of the frustrum of a pyramid with square base of side b,
    square top of side a, and height h. What happens if a = b ? If a = 0 ?

    So far, this is what I've concluded.

    Am I on the right track?
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  2. #2
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    question unclear.

    Quote Originally Posted by maxreality View Post
    Find the volume of the frustrum of a pyramid with square base of side b,
    square top of side a, and height h. What happens if a = b ? If a = 0 ?

    So far, this is what I've concluded.

    Am I on the right track?
    the height h is the height of the pyramid or the height of the frustum?
    if it is the height of the frustum then you can directly apply formulae. otherwise you have another formulae involving the product of the height and the areas of the surfaces. in either case you simply need to apply formulaes.search wiki for the formulaes.
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  3. #3
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    Hello, maxreality!

    Your sketch is not quite right.
    The height is a vertical measurement, not along the slanted edge.


    Find the volume of the frustrum of a pyramid with square base of side b,
    square top of side a, and height h.
    I assume they want us to derive a formula "from scratch".


    We have this cross-section of the entire pyramid.


    Code:
        -           A
        :           *
        :          /|\
        :         / | \
        y        /  |y \
        :       /   |   \
        :      /   F|    \
        -   B *-----*-----* D
        :    /: a/2 |  a/2 \
        h   / :     |       \
        :  /  :     |        \
        - *---*-----*---------*
          C   H     G   b/2   E
          : - - - - b - - - - :

    BDEC is the frustum of the pyramid.
    . . The lower base is: CE = b\quad\Rightarrow\quad CG = GE = \tfrac{b}{2}
    . . The upper base is: BD = a\quad\Rightarrow\quad BF = FD = \tfrac{a}{2}
    . . Its height is h.


    ACE is the entire pyramid.
    . . Its base is: CE = b.
    . . Its height is:  h+y.
    Its volume is: . V_1 \;=\;\frac{1}{3}(h+y)b^2


    Draw BH \perp CE
    Note that: . CH \,=\,\frac{b-a}{2}


    Since \Delta AFB \sim \Delta AGC\!:\;\;\frac{y}{\frac{a}{2}} \;=\;\frac{h}{\frac{b-a}{2}} \quad\Rightarrow\quad y \:=\:\frac{ah}{b-a} .[1]


    The volume of the upper pyramid is: . V_2 \;=\;\frac{1}{3}a^2y


    The volume of the frustum is:

    . . V \;=\;V_1-V_2 \;=\;\frac{1}{3}(h+y)b^2 - \frac{1}{3}a^2y <br />

    . . . . =\;\frac{1}{3}\bigg[b^2h + (b^2-a^2)y\bigg] \;=\;\frac{1}{3}\bigg[b^2h + (b-a)(b+a)y\bigg]


    Substitute [1]:

    . . V \;=\;\frac{1}{3}\bigg[b^2h \;+\; (b-a)(b+a)\cdot\frac{ah}{b-a}\bigg] \;=\;\frac{1}{3}\bigg[b^2h \;+\; ah(b+a)\bigg] \;=\;\frac{1}{3}\bigg[b^2h \;+\; abh \;+\; a^2h\bigg]


    Therefore: . \boxed{V \;=\;\frac{h}{3}\left(a^2 + ab +b^2\right)} .**




    What happens if a = b ? . If a = 0 ?

    If a = b, we have a "box" with a square base (side a) and height h.

    Of course, the volume is: . V \;=\;\frac{h}{3}(a^2+a^2+a^2) \;=\;\frac{h}{3}(3a^2) \;=\;a^2h




    If a = 0, we have an entire pyramid with base area b^2 and height h.

    Of course, the volume is: . V \:=\:\frac{h}{3}(0 + 0 + b^2) \;=\;\frac{1}{3}\,b^2h


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    This formula can be generalized.


    . . V \;=\;\frac{h}{3}\left(B_1 + \sqrt{B_1B_2} + B_2\right)

    . . . . where B_1,\,B_2 are the areas of the two bases.

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