Results 1 to 3 of 3

Thread: Pyramid Frustum Question

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    15

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    202

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    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 $\displaystyle b$,
    square top of side $\displaystyle a$, and height $\displaystyle h.$
    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 - - - - :

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


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


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


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


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


    The volume of the frustum is:

    . . $\displaystyle V \;=\;V_1-V_2 \;=\;\frac{1}{3}(h+y)b^2 - \frac{1}{3}a^2y
    $

    . . . . $\displaystyle =\;\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]:

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


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




    What happens if $\displaystyle a = b$ ? . If $\displaystyle a = 0$ ?

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

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




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

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


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

    **

    This formula can be generalized.


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

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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 31st 2011, 03:40 AM
  2. Lateral Area of the Frustum and Pyramid
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 17th 2011, 11:19 PM
  3. Frustum of Regular Pyramid..FlowerPot
    Posted in the Geometry Forum
    Replies: 6
    Last Post: May 9th 2011, 05:14 AM
  4. Frustum of Cone & Pyramid
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Jan 27th 2010, 11:20 PM
  5. [SOLVED] [SOLVED] Creating a frustum cone from a frustum cone
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Sep 17th 2008, 04:14 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum