# The Wonder and Amusement of Triangles - Part One

## A Lesson In Origano

Previously I mentioned a discovery due to folding a paper triangle. I did this on the suggestion of one of my maths teachers many years ago, and after some pondering, made yet another interesting discovery.

The problem was to find out which kind of triangle could be folded into the shape of a square without cutting off any paper, or overlapping in any way, where you folded it onto the square shape. Let’s draw, Lucky Luke, to get a better idea from the illustration over to the right.

I realise that the apex of the triangle had to fold over and just reach the base of the square, which itself at first forms part of the area of the original triangle, and that we are folding over a triangular top part. If you consider the situation from what you can see, you also realise that the height of the original, larger triangle is exactly * twice* that of the height of the square, so that the height of the part of the triangle you are folding over, itself triangular in shape, is equal also to the height of the square, and also therefore, the length of its side. In addition, we have two right angled triangle shapes on each side to fold over horizontally.

## Getting right into comprehending the Situation

The best way to see this is to try it yourself. We realise that the original triangle has to be** Isoceles**, but that it cannot be an** Equilateral**. But for such a thing to be possible, that a triangle can fold into a square in this way, it can have only certain possible internal angles, and no other combinations of such angles will be able to allow the triangle to fold over exactly.

What we need to understand also, is that the area of all three triangular shapes that fold over the square will equal exactly the area of the square, and if you think about it, that this square will equal half the area of the largest triangle - the original one we fold parts of over to form the square.

What we do is draw the square first, then draw the triangle around it, to fit the square. Only one triangle with unique internal angles will do that. Let us then say that our square has a base of **six** inches, and therefore an __area__ of ** 36**. So then do all three triangular parts to be folded, two of which are of the same __area__, which are folded horizontally, while the third is the largest, and is folded over the top. The __area__ of the overall triangle, that is to be folded into a square is thus equal to **72 ** __square__ inches.

So let us have ourselves another diagram so that we can clearly and more easily understand exactly what is going on, since if the explanation doesn’t suffice, well, as the proverb says,

**" A picture is worth a thousand words. "**

## Figuring out the Paper Folding Triangle

The angle **θ **is found by realising that the height of square, divided by the base of either of the smaller right angled triangles beside the square, is equal to its **tangent**.

**Arcus tangent ** ( **2 **) = ** 63.43° ** = ** θ ** and also, ** δ ** = **53.13°**

The angle delta ( **δ** ) is found by doubling **θ**, and subtracting this from **180°**, since all three internal angles of a triangle add up to this.

## All things being Equilateral . . .

So, having found exactly which triangle folds over into a square, and knowing it is not equilateral, I then wondered how you would fit a square into an equilateral, and knowing the angles already to be **60° ** each, the question in this case is rather, how do the dimensions compare with each other ? What I wanted to find out was, if I knew the lengths of the sides of my equilateral, is there a way to work out the dimensions of the square that fits snugly into it ? We of course would realise, that with this arrangement we cannot fold this particular triangle over nicely to fit the square.

Once again, and fun as always, we are left here to draw a diagram in order to clarify the situation, so that we can see exactly what is going on.

## Paper Folding Triangle for Equilateral

## Explanation

It is clear from this that the side of our square, which we shall call **x**, also forms the base of a smaller equilateral triangle, so that, if as we said the sides of the greater triangle equal **10"**, then the part of that side from the top right hand corner of our square down to the angle on the bottom right of our greater triangle will equal **10** -** x**. What we are trying to do here is find out the value of **x**, being the side of our square, as well as the height of a right angled triangle comprising the bottom right hand side area of our greater triangle, with sides **10 **-** x**, **x**, and its base ½(**10 **-** x**).

This is because we can see that the base of our larger triangle equals **ten **inches, while the base of our right angled triangle is one of two identical lengths left over when the length **x **( the base of our square ) is subtracted from this value of **ten**. On the two diagonal sides to the large equilateral, one **x **is subtracted from the length of the diagonal, which equals **ten**, what remains is **10 **- **x**, and since the base is the same as either of these two lengths, once **x **is taken from this base, that which is left also has to be **10 **- **x**, just this time in **two **equal parts. Since the two of these parts together equal **10 **- **x**, obviously one of them is equal only to half of that.

Now our square this time will actually be longer than half the height of our larger triangle, and to work out how high that is, we look at the right angled triangle at bottom right end of the greater triangle, and see that the angle at bottom right, as it is in equilateral triangles, is **60°**, and also that the two sides, **x**, ** **and **10 **-** x**, form both the opposite side and the hypotenuse, respectively.

Let us consider an illustration with its own explanation of how to work out **x**, followed by a text on what the values are.

## How to figure it

As we can see, **x **is shown to equal **10√3 **÷ (** 2 **+**√3 **), which in fact equals **4.641016151377546**. This means that **y **is equal to **10 ** minus** 4.641016151377546** =** 5.358983848622454**, which should be confirmed by it also being **20 **÷ (** 2 **+**√3 **), which works since **x **= **.5 **x **√3** **y**, where the **10 ** in the denominator for the **x **expression is **half **of the **twenty** in the one for **y**, then the **10 ** is multiplied by **√3**, giving us the proper value.

The identical bases for the smaller right angled triangles on either side of the square will be equal to half of **y**, which is **2.679491924311227**, and to work out the height of such a triangle we can use two ways - one backing up the other. First, if we say that **height ² ** =

**y**+

**²****base**, then

**²****height**=

**4.6410161513775465**, which is absolutely correct, as the

**height**of such a

**Triangle**has to also be the side of the square, which value is

**x**. The other way is

**Trigonometry**, by knowing that since, as noted in the illustration,

**x**=

**y sin**(

**60**),

**°****then we multiply our**

**y**by this to get this

**height**.

As for the **height** of the whole equilateral, we can simply find this by realizing that the length of this **height**, divided by half the base, equals the **tangent **of **60°**, which is also equal to **√3**. Half the base is **five **inches, so **height **= **5 tan** (** 60°**) =

**8.660254037844384**( or

**10**÷

**√3****).**

**2**This allows me to calculate the triangle's area, at half base times height, being **5 **x **10 √3 **÷

**2**=

**43.30127018922192**. Incidentally, this should work out to be the sum of all the parts - the smaller equilateral, the square, and both right angled triangles. The smaller equilateral, having

**x**for all three sides, will have an area equal, according to my regular polygon formula, to (

**½ base**)

**²**x

**3 tan**(

**30**) =

**°****9.32667397366058**. The square will be

**x**, which is

^{2}**21.539030917347247**, and both of the right angled triangles will be twice half their base times height, so added together, they would combine to base times

**height**=

**12.435565298214105**. This plus

**9.32667397366058**and

**21.539030917347247**gives

**43.30127018922193**, which, considering rounding error and it being accurate to a good number of decimal places, is right on.

## Let's pick a Side

That is very well and good, but what if we wanted to stipulate our **x **? We could do that, but then we would not know the side of our triangle. We did do it for the proper paper folder, the first example shown, where we wanted the square's sides to be **six **inches, but this was not **Equilateral**. If we wanted that, we can also do so, keeping the angle at **60** degrees, and this time, it would be the square's side that determined the sides of our triangle. Let us look at an example for a square with sides of **4.5 **inches, inside** an Equilateral Triangle**.

## Equilateral for Known side of Square

## Last Words

So we see that the side of the larger** Equilateral** for this one is no longer as long as **ten** **inches**, and can be worked out by realizing for a start that the identical bases to the two smaller **right angled triangles** either side of the square are such that their length divided by the **4.5 **inches of the square's side ( forming here the opposite side to the right angled triangle ), will be the cosine of our sixty degrees, then we can work out all else from there.

Well that is it for **Part One**. Check out **Part Two of this Series on Triangles** - an in depth look into the **Cosine** and **Sine Rules** and proofs of them, as well as triangle heights.

## 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.

Lucky Luke is a Cowboy drawn by the Belgian Morris in the 1940's, and later an animated TV Show. Origano is a deliberate misspeak of the word Origami.

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 Hubs, which continue the study of Triangles, and these are : The Wonder and Amusement of Triangles - Part Two, the Law of Missing Lengths, The Wonder and Amusement of Triangles – Part Three : the Sine Rule, and The Wonder and Amusement of Triangles - Part Four : the Cosine Rule.

If You are curious, then do not hesitate to take a good look at the other Hubs, The Maths They Never Taught Us - Part One, The Maths They Never Taught Us - Part Two , The Maths They Never Taught Us - Part Three, The Very Next Step - Squares and the Power of Two , And then there were Three - a Study on Cubes, Moving on to Higher Powers - a First look at Exponents, The Power of Many More - more on the Use of Exponents, Mathematics - the Science of Patterns , More on the Patterns of Maths , Mathematics of Cricket , The Shape of Things to Come , Trigonometry to begin with , Pythagorean Theorem and Triplets, Things to do with Shapes, Pyramids - How to find their Height and Volume, and How to find the Area of Regular Polygons.

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

Bartholomew Webb , They Came and The Great New Zealand Flag

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.