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 $b$,
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
$

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

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.