1. ## 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?

2. ## question unclear.

Originally Posted by maxreality
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.

3. 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.

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# volume of frustum of pyramid

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