The post Taking our Polygon formula to the limit appeared first on Tackle Maths.

]]>Which is the familiar area of a circle! And if you think about polygons with many, many sides you will see they are very close to being circles.

This is why I love maths! Everything fits together!

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]]>The post Area of a polygon appeared first on Tackle Maths.

]]>Anyway, many of my posts have addressed specific exam questions but sometimes I just get a bug about something .. and then I think, why have I never thought about that before!!

I don’t know of any maths course this is on, but hey, I just wanted to know, as I wrote my last two posts…. Is there a formula for the area of a regular Polygon!

If it is 3 sided or 4 sided – a triangle and a square – then we know the formula for area, but I was thinking – what about a formula that works for any regular polygon – That is to say, one with all the sides the same.

Here is a polygon and lets say the length of all the sides is L. You can count the sides here and see there are 8 – this is an octagon – but let this represent ‘any polygon’ with a number of sides S.

What I will be looking for is a formula where A = something with L and S in, as they are the two ‘properties’ of the polygon that might change in our ‘general’ polygon

The art of finding the area of any unfamiliar shape is to divide it into shapes for which you know how to find the area. Any polygon can be divided into triangles by drawing lives from each corner into the middle.

How many triangles? S – One for each side of the polygon.

What is the area of each of these triangles? For this we need to know the base and the height; The base is L. To find the height we will need to use some trigonometry. Half of the triangle is a right angled triangle. I am going to use the angle at the top of the half triangle, which I have labelled x

The full angle at the top is 1/S of the 360 degrees at the centre =

360/S. x is 1/2 of this = 180/S

Tan(x) = L/2 / h (Opposite/Adjacent where the ‘opposite’ is half the base and h is what we will call the height, for now.

Add in some detail and rearrange

Tan(180/S) = L/2h

Rearrange again to give h = L/2Tan(180/S).

Area of the small triangle is (using 1/2 x b x h) = L^{2}/4Tan(180/S).

This is only 1 triangle out of S triangles, so the formula for the whole polygon is

Now, as I said at the start, I worked out this formula a week ago, bit I wanted to check my work because it looked a bit cumbersome….. But it does seem to stand up, and I’ll show you how.

There are two polygons for which we know another formula; The Triangle and the square (S = 3 and S = 4). Lets see what happens if we make S= 4

A = L^{2} x 4 / 4Tan(180/4)

Tan(180/4) = Tan(45) = 1 – so we cand leave this term out. Also the 4 and 4 can cancel, and we get

A = L^{2} – the simple and familiar formula for the area of a square!

Checking this for the triangle is more tricky because we to find the ‘normal’ area for a regular 3-polygon – or as we usually call it, an equilateral triangle. For this we need our old friend, Pythagoras’ Theorem.

Using the theorem in half the triangle, we get

(L/2)^{2} + H^{2} = L^{2}

H^{2} = L^{2} – L^{2}/4 = 3L^{2}/4

H = √3L/2

so A = √3L^{2}/4

Now lets see what we get from our polygon formula with S=3.

Tan(180/3) = Tan 60 = √3

A = (L^{2} x 3)/(4 x √3)

Remember that 3/√3 = √3

so A = √3L^{2}/4

The same as with the direct method!

That, I claim, justifies my formula. If anyone can think of an alternative way of finding the area of a pentagon or hexagon, then the formula can be checked for these shapes too.

Now, interesting things happen to our formula if S gets bigger and bigger, but that will have to wait for another post!

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]]>The post Just what is an external angle? appeared first on Tackle Maths.

]]>Actually its easier to understand what is going on if we look at the external angles.

It’s important to know what the external angle is. It is NOT the angle all the way round the outside

It is the angle between each line.. if it were drawn longer.. and the nest line round the shape in that direction.

It is useful to imagine you are walking around the shape. The external angle is the angle you turn your body round at each corner. Then once you have walked round the whole shape.. and ready to start again .. you have turned the full 360°

So the size of each external angle = 360°/Number of angles.

The number of angles is the same as the number of sides, of course.

That is an easier formula than the one we saw for the internal angles, but I always get curious in these situations. We have two formulas… do they work together?

For a given regular polygon lets say it has S sides (so also S corners, A internal angles and S external angles). Let A be the size* of each internal angle and X be the size* of each external angle.

*They will all be the same because this is a regular polygon. This sudden move into the language of algebra is because we don’t know how many sides our polygon has – we are looking at all polygons at the same time.

X = 360/S (Today’s formula)

A = (S-2) x 180/S (Yesterday’s formula for the size of each angle)

Also, X = 180 – A : The two angles make a straight line. Look at the diagram.

To show these formulas all say the same thing, we need to combine two of them and show we get the other one. This can be done a number of ways, but I’ll only show one here.

Take X = 180 – A and substitute in the formula for A

X = 180 – 180(S-2)/S

X = (180S – 180(S-2))/S – I’ve made the whole equation ‘over S’ by including the first 180 in the fraction)

X = (180S – 180S + 360)/S I’ve multiplied out the bracket on the top

X = 360/S – Because 180S – 180S = 0!

And so we get to the other formula for X

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]]>The post How many angles can you fit on one post? appeared first on Tackle Maths.

]]>the same length. Some of these shapes we might not immediately recognise as a ‘regular polygon’ because we have another name for the shape – Square and Triangle – specifically an ‘equilateral triangle’

In this post we are going to look at how big each angle is in each shape.

We know that angles in a triangle add up to 180° – That is something we learn – and that a square is made up of four right angles – 4 x 90° = 360°

The next polygons – the pentagon, hexagon, heptagon and octagon.

The hexagon is probably the most familiar of these words, the pentagon is perhaps most famous for the US department of defence. The other two words might be new to you (though where I was at university the Octagon was the main conference building – it was where I graduated!)

The sums of the angles in these shapes are 540°, 720°, 900° and 1080°.

You might see a pattern there – each number goes up by 180°.

We can turn this into a formula where S is the number of sides of the polygon

Total of Angles = (S – 2 ) x 180.

From this we can work out the size of each angle in a regular polygon. Because the sides are equal, so are all the angles.

Each angle is 180(S-2)/S.

In this post I have been looking at the ‘Internal Angles’. Actually, its easier to show that these formulas work if we consider the external angles. What are they, you might wonder?

Well that will be the subject of the next post!

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]]>The post How to construct a …. appeared first on Tackle Maths.

]]>This was the set of skills called ‘Constructions’ – which can be hard to recreate on a whiteboard, especially when was one my ‘Motor Skills’…

The premise of these skills is that the student needs to

- Construct a perpendicular bisector of a line
- Construct a perpendicular to a line from any given point
- Construct a line that bisects an angle

The extra catch is these have to be done using a pair of compasses and a straight edge only

There is something a little ‘old fashioned’ about these skills, but they remain on the GCSE Maths syllabus. I’ll also add that I haven’t seen them as much on recent exams papers – Maybe once at most across all three appears for any given season…

For all that they are rather fun to do and understand.

They are hard to do in front of a class; and on a blog too! So let me give you some links to show how to do them.

- Construct a perpendicular bisector of a line
- Construct a perpendicular to a line from any given point
- Construct a line that bisects an angle

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]]>The post Angling for some Fun – Part two appeared first on Tackle Maths.

]]>Then I will show how we can put all this together to answer a question.

For both of these situation, we need one line crossing two others which are parallel. If we don’t know the two lines crossed are parallel, we can’t use these rules, got it?

And you know how to tell if two lines are parallel? The two lines are never going to meet, however far we extend them.

So, the two angles marked in red – they are going to be equal. So if we are told the lower one is 50˚ then the higher one is also 50˚.

These is what are called ‘Corresponding angles’

This is like the rule in the last post where we know two angles must be the same if they are ‘opposite’. Angles are also the same if the are ‘corresponding’

Now look at this diagram.

The two red angles are ‘corresponding’ and the green angle is ‘opposite’ one of the red angles.

So all three angles will be the same.

We say the bottom Red angle and the green angle are ‘Alternate’. ‘Alternate’ is like a combination of the ‘Corresponding’ rule and the ‘Opposite’ rule.

Some people think of this as the Z rule – because the angles in a Z are the same.

I’ve been looking for an exam question that uses all of these but they are hard to find, and I only want one for this post, which would be too long otherwise… so… let’s have a look at this

A few points about the wording and notation here

1. I across two lines shows those lines are of equal length : AB = AC

2. > on two lines shows they are parallel. AP is parallel to CB

3. ‘Diagram Not Drawn to scale’ means take the information given as true. Don’t check them with your ruler or protractor.

ABC is an isosceles triangle, so <ACB = <CAB = 70˚. < ABC = 180 – 2 x 70 = 40˚ – because of angles in Triangle ABC add to 180˚

<BAP = <ABC because they are Alternate – (to see the Z shape turn it round a bit)

So x = 40.

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]]>The post Angling for some fun? Part One. appeared first on Tackle Maths.

]]>There are a lot of questions you will see that ask you to find the size of an angle, or to show that two angles are the same.

Answering these questions, you also have to give reasons, and in giving reasons you need the language for describing different patterns with angles, and different ways of how you can find the size of angles from other angles.

So over this weekend I’m going to post some of these rules, and the names you need to give while describing those rules…..

And I’ll be honest, I have to look some of them up! Not the rules themselves – I know all of them – but some of the names for them.

After these posts, I’ll give a couple of exam questions and show how to use the names of the rules in your explanations>

The first thing is to remember that there are 90˚ in a right angle and 180˚ in a straight line. How can we use that information to find missing angles?

In all these three examples, we are looking to find the number of degrees in angle x.

In this case x and 55˚ make up a right angle, or 90˚ – so x = 35˚ due the ‘angles in a right angle’. And that is what you should write, to show that the reason for the answer is understood.

In this example, x and 140˚ make up a straight line, which is 180˚.

So x = 40˚ due to ‘angles along a straight line’. Again, that is what you should write.

In this case, angles x and the one marked 47 are the same size. This is the law of opposite angles. You might be able to see that this is the angles on a straight line, applied twice, but remember the term ‘opposite angles’ and there no need to work out the other angles. We write

x = 47˚ , opposite angles.

In my next post I will add alternate angles and corresponding angles to the mix. Then we will be ready to tackle a lot of angle questions.

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]]>The post Factors and Multiples – and a clever disguise appeared first on Tackle Maths.

]]>

For today’s post I am going to look at the exam question here.

You can see from the question that we are in the area of common factors and multiples, but the question does come with a bit of a disguise!

Look at that last line – We have been asked to find the highest value of a/b. So how does a fraction have the highest value?

By making the top as high as possible, and the bottom as low as possible.

So a is the HIGHEST common factor of 72 and 96

and b is the LOWEST common Multiple of 6 and 9.

Its a question about our old friends LCM and HCM – we have see through the disguise!

Let’s find the answer now though. We start by giving our four naumbers in Prime Factor form.

9 = 3 x 3 = 3^{2}

6 = 2 x 3

72 = 2 x 2 x 2 x 3 x 3 = 2^{3} x 3^{2}

120 = 2 x 2 x 2 x 3 x 5 = 2^{3} x 3 x 5

For the LCM we take the largest power of each prime number included

LCM = 2 x 3^{2} = 18 (=b)

For the HCF we take the largest power of each prime number included

HCF = 2^{3} x 3 = 24 (=a)

So a/b = 24/18 = 4/3 or 1 1/3

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]]>The post To Infinity and Beyond appeared first on Tackle Maths.

]]>For today something different a look at something quite unusual… As the title suggests – INFINITY.

Today’s post is based on some of the work of German mathematician Georg Cantor, a man not often understand by his contemporaries – In fact one said he was ‘aha ed of his time by one hundred years.

What Cantor was trying to do was explore what infinity means, and his revolutionary idea was that there is more than one infinity! In fact… there are an infinite number of infinities!

I’m not planning to show where they all come from in this post – I’ll just show that there is more than one.

If you go out with six friends this week, how will you know there are six of them? Well, OK, that seems like a simple question – because you can count, right?. Yes, but understanding how we can count is the starting point of understanding where all the extra infinities come in. You know you have six friends with you because you have given each a number, starting with 1.. and you got up to six! OK so may not have done that explicitly, but that is what you did.

And that is what we are going to use in a moment to demonstrate infinities.

Two quantities are the same if you can match up one item from one ‘list’ with one ‘item’ of the other… and have none left over. So the numbers 1 to 6 can be matched with your 6 friends with no numbers left over.

Now, this is where we come to some surprising ‘facts’….. Ask, are there more whole numbers than ‘even’ whole numbers. You think, yes? Well look at this…..

Even numbers can be paired up with whole numbers with no numbers from each group left out. This is one of the surprising things we learn about infinity. The number of even numbers is the number all numbers.

Now imagine you are on a ‘chessboard’ floor that stretches for infinity in all directions.

How many squares are there. Infinity – or more than infinity. This, after all ‘infinity squared

Let’s try counting the squares in the same way that we counted your friends. Start with one square – it could be the one you are standing on. Thats 1. Now count the 8 squares around you – 2,3,4… 8

Then the squares around that.

You can keep going like that for ever – but that’s OK, we can keep counting for ever.

So even on this infinite board, the number of squares is the same as the number of numbers – Infinity.

Now, at this point, if I didn’t know what was coming next, I might be starting to doubt my earlier remark – that there is more than one sort of infinity.

Lets consider now, decimal numbers, by which I mean numbers of the form

1.5434677654433345656

I have 20 digits after the pint there. Lets say we mean numbers that DO end, but we will not specify how many digits they have before they do.

So lets start with the number 1.5

Where do we go to next to count; 1.6 or 1.55? Or 1.5000000001.

There are actually an infinite number of next steps, and thats before we get off our first ‘square’.

And THAT is where the next order of Infinity comes in. Its not possible to ‘count’ the decimals with the numbers 1, 2, 3 ……. even they go on for ‘infinity’.

There are more than infinity decimals!

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]]>The post Crazy Curves appeared first on Tackle Maths.

]]>I was going to include this in the last post, but that was already too long.

We have looked at curves which are continuous everywhere, and some which are not – but are continuous for most of the way.

Is it possible for a curve to be discontinuous everywhere? In theory yes, though we need to consider rational and irrational numbers.

A rational number is any that can be written down, accurately, with numbers. This includes numbers there are ‘recurring’ like 0.3333333 because this can be written as 1/3, and be accurate

Pi is an example of an irrational number

So if we say that y = f(x) where f(x) = 1 when x is rational and f(x) = 1/x where x is irrational… then that would define a curve, but one that is so chopped up is would be continuous in only very small sections between rational numbers

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