CTK Exchange
Front Page
Movie shortcuts
Personal info
Reciprocal links
Terms of use
Privacy Policy

Interactive Activities

Cut The Knot!
MSET99 Talk
Games & Puzzles
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Math as Language
Things Impossible
My Logo
Math Poll
Other Math sit's
Guest book
News sit's

Recommend this site

Manifesto: what CTK is about Search CTK Buying a book is a commitment to learning Table of content Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

CTK Exchange

Subject: "Demonstration for the volume of the pyramid?"     Previous Topic | Next Topic
Printer-friendly copy     Email this topic to a friend    
Conferences The CTK Exchange This and that Topic #697
Reading Topic #697
Member since Aug-7-04
Apr-03-06, 07:28 AM (EST)
Click to EMail ricardo Click to send private message to ricardo Click to view user profileClick to add this user to your buddy list  
"Demonstration for the volume of the pyramid?"
   For some time I ask myself why the volume of the pyramid is 1/3*B*h.

That the volume is directly proportional to the Area "B" and the height "h" seems reasonable enough but the 1/3 puzzled me. I know some different demonstrations for it but I wasn't completely satisfied with them.

Most probably this "new" demonstration is not new but since I liked it and can't find it on the web I think it may be of interest to someone. (If you already seen this demonstration somewhere else please tell me so)

First let's pick a rectangular prism with square base ABCD and truncate it with a plane not parallel to the base. Let's say that at each vertice ABCD the solid has heights h_a, h_b, h_c and h_d.

We can show using some geometry that (h_a + h_c)/2 = (h_b + h_d)/2 = "height at the center of the square"

then "average height" = (h_a + h_b + h_c + h_d)/4 = (h_a + h_c)/2 = (h_b + h_d)/2.

The volume of the truncated solid is most reasonably B*(average height).

(For more rigor this can be demonstrated dividing the solid in prisms or by some cutting and gluing to make it a rectangular prism)

And what happens if the base is an equilateral triangle ABC? The same idea should apply so its volume is B*(h_a + h_b + h_c)/3.
If the we have h_a = h_b = 0 and h_c = h we have a pyramid so it's volume should be

B*(0 + 0 + h)/3 = B*h/3

Unfortunately this demonstration is not rigorous.

Another idea for a more rigorous demonstration is

There are three different pyramids we can make namely when:

h_a = h_b = 0 and h_c = h
h_b = h_c = 0 and h_a = h
h_c = h_a = 0 and h_b = h

These pyramids are all equal so they have the same volume each pyramid is determined by a different plane. We can think the volume as the integral of the heights of the pyramid over ABC (under the linear function of R^2 in R) if we add the functions determined by each plane (and so the volumes) we arrive at a linear function where

h_a = h_b = h_c = h

That is a prism so the volume of the pyramid must be 1/3 of that of the prism.

Obs. I made the base of the pyramid an equilateral triangle to ease things up. We can use Cavalieriís Principle to get to the other cases.

  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

Conferences | Forums | Topics | Previous Topic | Next Topic

You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.

|Front page| |Contents|

Copyright © 1996-2018 Alexander Bogomolny