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# Pyramids - How to Find Their Height and Volume

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## Let my People go

We once again take a look at the kinds of numbers we can obtain from the study of shapes. This time we draw our attention to a consideration of the dimensions of Pyramids. By these I mean three dimensional constructions with square bases, and triangular sides, four of which slope up at an angle to a central peak at the very top ( vertex ) of the Pyramid, as found commonly in the deserts of Northern Egypt.

I came across a way of working out the heights and other dimensions of Pyramids based upon other information, when I was made aware of the fact that the formula for a Pyramid’s volume is ⅓ base area × height. That is to say, a three dimensional box with the same base area and height of the Pyramid that would fit inside it, has three times the volume of the said Pyramid.

But then I thought, only a third ? Doesn’t really sound like enough, so on more than one occasion I went and constructed Pyramids out of cardboard, and compared their volume with that of a three dimensional rectangular type shape with exactly the same base and height, and considering that, as well as working it out graphically and algebraically, the Pyramid was indeed found to be exactly one third the volume of such a box that would fit it. Proof of this is given later.

So, as much as being a lesson in finding out the truth of statements for oneself, it also led me to consider comparisons between the dimensions of the Pyramid, to find out if there was a pattern. There is.

## Getting to the Point of the Matter

Let us imagine a Pyramid with a base having sides each of one foot, and the four identical triangles that spring from this base to make the slanted faces of the Pyramid are equilateral in this case, so that both of their other two sides are also one foot in length. Now sure, a Pyramid that small is not much good for putting Pharaoh in, but let us use these lengths just for the sake of simplicity :

## Explanation of the Diagram

We see the centre line of the triangle, its height, is not to be confused with the height of the Pyramid, since when this triangle forms a face of the Pyramid, it does so at a slanted angle, which we shall see.

The length of this height for the triangle is measured here at ½ of the square root of three, or .866 feet. This is worked out simply by applying the theorem of Pythagoras, and realising that a right angled triangle is formed by half of the base, the height of the equilateral triangle, and the hypotenuse is its one foot in length side.

From this we see that - .5² = height² = 1 - .25 = .75, and this means that the height is equal to the square root of .75, which is .866. Having worked all this out, the next thing to consider is how this relates when the Pyramid is upright. Let’s therefore look at a side view of the Pyramid, and ponder it for a while :

## What is really going on

Try to visualise what is happening. Think three dimensionally. The triangular shape on the left hand side of the illustration is the Pyramid as if from side on, where the height of each of the four triangles whose bases form the square base of the Pyramid, is now the slant height of the Pyramid – lines that meet diagonally now at the vertex. The three dimensional shape at right is not to scale, but is a good enough representation of the whole Pyramid as it would be viewed from a greater height.

The Pyramid’s height is worked out through Pythagoras again, so that .866² ( = .75 ) minus .5² ( the square of half the base = .25 ) equals height², which is .5, so our height is the square root of .5, and that is .7071. Thus the height of a Pyramid with base and triangle sides all equalling one foot, is .7071 feet, or just under 8½”.

## If the dimensions differ somewhat

Now, to cut a long story short, if we let the length of the sides of the triangle all equal two, then the height of the triangle that forms one of the faces will equal √3, and when this becomes the Slant Height for the Pyramid, it forms the Hypotenuse of a right angled triangle that makes up one half of the regular triangle it becomes when viewed form side on. This will have a base which is half this Pyramid base of two, so that the Height of the Pyramid ends up being the square root of ( √3² - ) = √2 – as in the illustration below.

Remember that this Slant Height is the height of the triangle that forms the face of the Pyramid. It is not the length of the edge of that face where it joins with the face next to it. That length will in fact be that of the Hypotenuse of the Triangle face. Remember for this Pyramid now with a two foot base, where all sides are also two feet, so this edge is also that length.

But also be aware that these face triangles do not need to be equilateral, and for the formulas to follow, whether or not the triangle sides are the same as those of the base, we use different letters to symbolize them, allowing these formulas to be used for all occasions.

## On to the Formulas

There is a pattern here. The height of a given Pyramid is the square root of : the square of the height of the triangle that forms each of its four visible faces, minus the square of half its base.

So our first formula is :

## Working out the Height of the Pyramid using Height of Triangle Face

To find the height of any Pyramid, using the height of its triangles that make up the faces, follow these instructions :

Say we have a Pyramid with a base 4’ × 4’, and a triangle face, the height of which equals the square root of sixty feet :

1. Find the value of the base ( = 4 )

2. Square this value ( = 16 )

3. Quarter it ( = 4 )

4. Subtract this ( 4 ) from the square of the height of the triangle : 60 - 4 = 56

5. Take the square root

height = √56

This actually works whether or not the triangle is equilateral or isosceles. Certainly, in order to be one of the four visible faces of the Pyramid, the triangle can only be one of those two. If not, it may still form a three dimensional shape, but such a shape will not be a genuine Pyramid with respect to the definition I have given. We’ll be getting back to this.

## An alternative Algorithm for a slightly different Formula

But then, what if we try to work out a Pyramid’s height from the length of this triangle’s side, as opposed to the triangle's height ? To do such, we first look at how we work out the height of the triangle forming one of the faces, as this involves the use of the length of one of the sides, which is the value we need for our alternative formula. Be very aware that this formula is not for the height of the Pyramid, but just for each of the triangles that attached to the square base. This would be :

## Plugging Value for Slant Height of Pyramid ( Height of Triangle Face ) into formula for Pyramid Height

The value for the height of the triangle – the slant height of the Pyramid – could then be plugged into the first formula for Pyramid Height, that uses the height of the triangle face, but let us put it in algebraically.

To find the height of any Pyramid, using the side length of its triangle, follow these instructions :

Say we have the Pyramid as before, with a base 4’ × 4’, and this time, a triangle whose side equals eight feet. For the formula involving the side length :

1. Find the value of the base ( = 4 )

2. Square it ( = 16 )

3. Halve that ( = 8 )

4. Subtract this ( 8 ) from the square of the side of the triangle : - 8 = 56

5. Take the square root

height = √56

This is the same value as before, since one such as above whose side equals eight feet will have the height that equals the square root of 60 from the previous example, so either formula and therefore algorithm used for applying them, works.

## Summary of what we have so far

So the formulas for finding the height of a Pyramid using either the height of the triangle forming one of its four faces, or the side of such a triangle, whether or not it is the same as that of the base of the Pyramid, are :

## For example

To make things interesting, say the height of the face triangle is now one foot, rather than one foot being the length of its side. This would obviously change not only the length of the side of the triangle to longer than one foot, but would also make the height of our Pyramid taller.

Letting the length of the base be also one foot, and using Pythagoras, this would make the side of the triangle √1¼’ in length, since if the base is one foot, then we use half this length as the shortest side of the right angled triangle. This triangle, as was shown for triangles of different dimensions, is formed by this half base, as well as the length that becomes the height of the triangle, and finally by the length of its side, this last part being the triangle's hypotenuse.

Now since this triangle is one foot high, when we look at it side on, as we did with the first diagram, this one foot length now becomes the hypotenuse of our right angled triangle for determining the height of the Pyramid. So, - .5² = height² = .75, therefore the height this time is higher, at √.75 = .866.

Checking this with both formula :

Pyramid Height = √h² - .25b² or √ s² - .5 b²

h = 1 b = 1 √h² - .25b² = √1² - .25x= √.75

s = √1¼ b = 1 √ s² - .5 b² = √1¼ - .5x= √.75

So it works. Excellent.

## A Caution

Now if You go and put rubbish values into these formulas, you will get rubbish answers. The term for this is GIGO ( Garbage In - Garbage Out ). Say You for instance decided your base was going to be eight feet, but your triangle sides are one foot only. How the thunder freaking hail could that come about, seeing that if the edge of the base was indeed these eight feet, then half the base has to be four, and in order for one to form a triangle out of that, you have to have two sides that would be long enough to reach. Look at the diagram.

In the first case shown in the illustration above for the Pyramid with a base length of 8 feet, one would plug these numbers into the formula used for the sides and get

( - ½b² ) = ( 1² - ½ x 8² ) = ( - 31) , but this is an imaginary number, and unacceptable for the real geometry of a three dimensional Pyramid. So make sure Your dimensions make sense.

For the one below, with base lengths of four feet, and a triangle height of 1.5 :

( - ¼b² ) = ( 1.5² - ¼ x 4² ) = ( - 1.75) , also complex, and therefore, at least for real, three dimensional, physical world measurements, unacceptable. In this case there ends up a one foot gap between the vertices of triangles stemming from opposite base lengths, and a gap of ( 2) feet between vertices for adjacent base lengths.

## Pyramids with non isosceles triangle faces

Previously I stated that in order to be a genuine Egyptian style Pyramid of the kind I stipulated to begin with, having a square base out of which emerge four identical triangles, to form the four visible faces of the Pyramid, the triangles have to be equilateral or isosceles in order to count as such. If not, and say at least one triangle if not more are scalene, then the situation will be different.

The illustration below shows how a Pyramid where the four triangle faces are not all the same would look. Either the two side ones as we look at them face on are identical, or not, but in any case, one would have to work out every detail of this one before coming up with a formula.

## PROOF OF PYRAMID VOLUME AS ONE THIRD

Now at the beginning of this Hub, I expressed amazement at the fact that the Pyramids dealt with here have one third the volume of the three dimensional box that holds them, such that the box has the same base and height as the Pyramid. But I also added that I then went about trying to see whether or not it was true with the use of models.

If one does make cardboard mock ups of Pyramids and then tries to compare them with shapes making up either the volume of the box, or of the shapes that could be left of such a volume when that of the Pyramid has been taken out, then one may be able to visualize the truth of this.

A better way, however, is to do so algebraically, taking into account in non specific terms what the relevant volumes can be given as. Now imagine a Pyramid housed inside such a box. The illustration below may help.

## Explanation

So what we could do, is try to prove that the volume of the half Pyramid type shape on the right hand side of the above illustration is half that of the Pyramid itself. But since it might be too difficult to do so directly, what we actually need to show is that if a box were put around such a half Pyramid shape, as it has been to the Pyramid itself, the box would have half the volume of the box that is around the Pyramid.

Now that in itself might not be enough, since we are dealing with two different three dimensional solid shapes. The idea, though, is that if the Pyramid is one third the volume of any box it is in, then so would the half Pyramid be one third the volume of the box it would fit into.

For a start, let us see if the box this half Pyramid would fit into could be half the volume of the box the Pyramid fits into.

## Just to be absolutely certain

So far so good, but as stated, we have only proven that two different solid shapes, one shape alleged to be half the volume of the other, will fit respectively into two different rectangular boxes, where one box is half the volume of the other one. What we really need is to try to find a shape that is more like the one we took from the remaining space within the box - the right hand side image in the illustrations above. If we can form from the Pyramid itself a similar three dimensional shape, and prove that the box this fits into relates to that our shape in question does, then we will be in business.

This is because sometimes it may be necessary to have identical shapes in the sense they are made up of the same two dimensional shapes in the same relationship to each other, even if they differ in the lengths of their dimensions. After all, we could fit a sphere into a cube, and using the formulas for spheres and cubes, we can determine what fraction of the cube volume the sphere makes up, but this may not work if we try to prove it by then taking an ellipsoid and relating its volume to a cube. Just to be safe, let us as they say, compare apples to apples and oranges to oranges. Below is a variation on the illustrations above, but this time we will cut the Pyramid in two, and deal with a half Pyramid.

## In summing up

This proves my point. We were trying to prove that the Volume of the Pyramid is one third that of the Volume of the box it would fit into, if both it and the box had the same height and base. If this is so, then the Volume that takes up the rest of the three dimensional space within the box, which is not occupied by the Pyramid, must make up the other two thirds of the volume of the box.

We see that the Pyramid has four faces, and that there are four corresponding, identical volumes around the Pyramid, which combined do supposedly ( if the statement is true ), make up this remaining two thirds volume, and each of these four volumes is bounded by one of the triangle faces, a quarter of the ceiling of the box, and two triangular sides, each bounded by an edge of the Pyramid made when two faces meet, a line going up from a corner of the base where two base sides meet, and a line perpendicular to this, which goes from the top of the height line to the vertex of the Pyramid.

We said that if the statement is true, then these four volumes make up two thirds the box Volume, which would also equal twice the Pyramid Volume if the Pyramid Volume is one third that of the box, so one of these four must make up half the Pyramid volume. That is the assumption.

We prove it by stating that if there is a box that fits the whole Pyramid in it, likewise there can be a box that fits this half Pyramid Volume also, and we went on to prove that the Volume of the box that could hold half the Pyramid was the same as the Volume of the Box that held one of the four solid Volumes making up the remaining two thirds of the Box able to hold the whole Pyramid, where this meant this particular Volume would also be equal to half the Volume of the Pyramid.

Good. The only concern is we have not proven it directly, but by the idea that if we have identical three dimensional solids, in the sense one is an exact half Pyramid solid shape, and the other has the same features - a rectangular base, a perpendicular triangular face, two slanted right angled triangular faces, and one slanted isosceles triangular face, twice the area of each of the right angled triangular faces - then the idea that the boxes that could contain them have the same Volume seems reasonable proof.

## The Parthian Shot

From each of these formula, we can now work out the Pyramid’s volume by calculating its height from the length of the side of the triangle, then multiplying it by the area of the base, and dividing that answer by three. It is this length of the triangle face's side which also forms the edge of each of these four faces of the Pyramid. In addition, we could now work out the angles at which each side rests in relation to the others by applying the principles of Trigonometry to all lengths in respective triangles.

We can see from the illustration below of a Pyramid with a base of four, and a triangle side of eight, that the half base is two, so that the angle formed by these two sides is the arcus cosine of .25, which equals just over 75.5°. Now this is NOT the angle the slanting side or face of the Pyramid forms with the ground, but rather, if you go to a corner of the Pyramid, this is the angle that the edge of the face forms with the Pyramid’s base.

Try to visualise or even construct a Pyramid, if you cannot understand my meaning. Now the other angle if you are looking at the face of the Pyramid, is where this triangular face meets at the top, and this will be at an angle of just under 29°.

The next angles are those that the slant of the face forms with the ground if we look side on, and this will determine the gradient we would climb at if we tried to scale the Pyramid.

If we stick with what we have, the height of this Pyramid is √56 ’’, or just under 7’ 6". Another, different right angled triangle is formed by this height, half the base, and the length of the height of the triangle, which is √60, or just under 7’ 9". The arcus cosine of 2 ÷60, will be our angle. This is slightly over 75°, which gives us a slope of 3.74 to climb. Then the last angle, formed by the meeting of opposite faces at the Pyramid’s vertex, will be just under 30°. These angles are very similar to each other, but the differences are not to be ignored.

## Disclaimer

As much as some of this Hub contains certain Mathematical knowledge accessible in the public domain, and not subject to any Copyright, other information has been drawn from textbooks which themselves are Copyright, but only in the sense of how they deliver the information which itself is shared and sometimes ancient Mathematical knowledge. Other information has also been found on Wikipedia ( Copyright 2013, the Wikimedia Foundation ), which is a good source of information. Part of this is also my own discovery, but may also independently have been found by others.

Some of the illustrations in this particular Hub are my own, and have primarily been done using Microsoft™ Paintbox, edited from illustrations done using Microsoft™ Word. Others are, as noted, from Wikipedia. Any quote or part of this material which seems to belong to any other author should be treated as such, and I claim no ownership of anything I did not myself invent or discover, nor of any obvious copyright, trade mark, or registered trade mark.

The Adventure continues in the next Hub on Geometry, which is How to find the Area of Regular Polygons,

Also, feel free to check out my non Maths Hubs :

There will also be many More to come on a wide variety of Subjects.

Just take a good look at it, and note how interesting it all is, then see if you can come up with anything else along the same lines. As usual, I would appreciate any comments, feedback and suggestions which would be given due credit, or indeed have a go and publishing Your ideas Yourselves, but firstly, by all means, add Your comments - it's a free Country.