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How to find the Area of Regular Polygons

Updated on July 24, 2014

To begin with

We shall try here to work out a general formula for finding the area of any regular polygon having n sides. When we say regular, we mean those for which every one of the n sides is the same length, and all n internal angles are identical.

Let us look first at the most basic shape, the Triangle, and compare how to find its area with that of other shapes. We are told that its area is half base times height, and that is fair enough. But if we look at a square, we find that this won't work, and that in fact, the area for this in particular will be base squared, which twice half base times height, since for a square, its base is also equal to its height.

So to help us figure it out, let us employ the use of an illustration with diagrams of the triangle and the square to begin with, then see what we can find, and whether we may apply it also to shapes with many more sides.

The illustration below shows that both triangles and squares can be divided into n smaller triangles all of the same area. The area of these smaller triangles will also be half base times height, and looking at the diagram, we see that the height can be found by realizing that height divided by half the base equals the tangent of the angle at the triangle’s vertex. This vertex, which is the internal angle of the triangle, is halved here, so the angle we want is half the internal angle we normally see.


Explanation of how to get the Formula

Explanation

The above diagram is reasonably self explanatory. For triangles and all subsequent regular polygon shapes, with identical side lengths and internal angles, we see that they can be divided up into n triangles with all the same area, who combined area equals that of the polygon in question. Don't believe me ? Check out the following two pictures showing a square and then a Pentagon.

More explaining

There is actually a pattern, other than how we get them from the formula, as to how the internal half angles relate to each other.

The first one is multiplied by 3/2 to get the second, which is multiplied in turn by 6/5 to get the third. This is then multiplied by 10/9 in order to obtain the fourth angle. This angle one then times by 15/14 to get the one after it. If we take a list of all numbers, and embolden those we use :

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Then we see that we include two distinct whole numbers in each fraction, leaving out one more each time for those we do not use. This suggests the next two numbers will be 27 and 28, to be used in the fraction 28/27.

The higher number is always on top, ensuring the coefficient we multiply the previous angle by to get the next is greater than one, and we can see that as the numbers get higher, even though the numerator is always only one more than the denominator, the value of the fraction decreases, and would converge to one in the limit.

This is because, whereas for the first fraction, 3 is only one more than two, when forming a fraction, 2 is two thirds of three, but 20 is twenty twenty firsts of 21 - a lot smaller proportion.

Hang on a minute . . .

But what if, as was mentioned in one of the illustrations, we did have shapes with all sides the same, but some of the angles different ? Can we fit these into a formula designed for only one side length, and one value for the internal angle ?

Well, for a start, with a triangle, if all your sides are the same length, the nature of it having three sides forces the internal angles to all be the same at our favourite sixty degrees, so there would not be any change of angles unless we also changed the length of at least one of the sides. Now, as to putting something like that into the formula, I have been working on it, but so far cannot find how to do that yet. The thing with triangles, even if it were possible to keep all three sides the same but change an angle, do we change one from the other two, or all three, and then what ?

This problem also arises with Pentagons and the other odd numbered sided shapes, but with polygons having an even number of sides, I have found a way to have two different internal angles, at least, and yet still be able to use the formula. I am not sure as yet if I can work it out for more than two, or for an odd number of angles, but let us see what can be done.

At home I have so far fifteen maths quad type books into which, over the past nine and a bit years, I have written out ideas in Maths and Physics, some copied, but a lot my own, some of which have already become the basis for the Hub topics seen here. These books are among my most prized possessions - perhaps worth more to me than my car, and on a par with the 1500 to 2000 books I have in my flat, that over the past 25 years of active collecting have cost me thousands of dollars.

Now in the fourteenth of these I looked at an interesting problem, the diagram for which is below, to show You where the following idea for having more than one angle in the polygon has come from, such that the ( preferably so far, two ) angles can be manipulated trigonometrically to find a kind of derived angle I call gamma, that fits into the formula, and gives the correct area. Gamma is derived in a slightly complex, but not too hard way, and even taking into account the trouble obtaining it, it would still be easier to do so and use it in the formula above than work out by hand the area - especially of a multi sided hydra of a polygon.

About the Diagram above

So I drew this diagram just for the fun of it, to see what would happen, and then notice the eight sided figure in the middle, and set about trying to work out its area. I at first did not realise the sides of the octagon are all the same length, and was not sure about the angles. To work out the area I divided the diagram up into shapes whose are I could work out, and subtracting from the area of the square the area of the shapes not included in that of the octagon, so that square area minus other shapes area = octagon area. The following illustration shows this more clearly.

Explaining the Diagram above - Areas of the Green and Blue Triangles

Now I decided upon a base for the original square of six inches, making its total area thirty six square inches. Looking first at the two large green triangles, it can be determined that one of their sides is this base of six inches, and the other is half the base at three inches. Since a triangle's area is half base times height, each of these triangles has an area of nine square inches, and since there are two of them, their total is eighteen square inches - half the original square's area is already taken off.

The red triangle has a base of half that of the square, so three inches, and its height is also this three inches, since from the first diagram with all the intersecting lines, this is the height at which the red line forming the left hand diagonal of this red triangle goes from the midpoint up to where it intercepts the tan brown line forming its right diagonal, so the area of the red triangle is 4.5 square inches. Adding this to the eighteen already found gives us 22.5.


The Tan Brown Triangle

The tan brown triangle has a base along the right hand edge of the square, a length of 3 inches, and its height - the distance from its tip at its very highest point on this picture, to the right hand edge, is found by a slightly more complicated method, just to make sure, which we shall also use later to determine the octagon's side length, and this is to imagine the square is on a Cartesian Plane, with its bottom left corner at origin, and the top right corner at the coordinates ( 6, 6 ).

I understand the diagram below may now look a bit cluttered and confusing, but it is simply aimed at the three coloured shapes not as easily worked out by area. Begin at the tan brown triangle, look at it, and read what is said about it above and below.

Each line can then be stated algebraically as a linear equation, and the green line from the first octagon diagram, which forms the top sloping ceiling of this tan brown triangle, begins at the point ( 0, 6 ), and travels down at a slope of negative one half, to meet the right hand edge of the square at ( 6, 3 ). This function is y = 6 - .5x. The line it intersects with at the point we want to gauge the tan brown triangle's height from, was the line on the first octagon diagram that begins at the midpoint of the ceiling of the square, and travels down at a slope of -2 to the bottom right corner. But if this line were allowed to continue, it would have met the y-axis at ( 0, 12 ), so its equation is y = 12 - 2x. So what we want is to find algebraically where these two lines meet.

6 - .5x = 12 - 2x 2x - .5x = 6 ( subtract six from each side, add 2x to each side )

1.5x = 6 x = 4, and since y = 12 - 2x, y = 12 - 2 x 4, which also equals 4, so the lines intersect at ( 4, 4 ), meaning the height for the tan brown triangle equals two, and its area is equal to three square inches. Adding this to the 22.5 we already have gives us 25.5 .This is shown in the illustration below, further comments about which shall follow.

Having worked out this area, look next at what is drawn with regard to the purple chevron, and read what is said in the section below. Follow that with the information on the grey chevron.

Octagon inside Square in Cartesian Plane

The Purple Chevron

As for the purple chevron, we calculate its area by creating a triangle whose base is the yellow line of length two inches as can be seen. We can also see that this line is two inches from the bottom, so a line can be drawn from the midpoint of this yellow line to the vertex of this larger triangle, and this would be its height at two inches. Note the brown line drawn from that same yellow one is for the smaller triangle, blue in colour. What we do is find the area of the larger one, and subtract from that the area of the smaller. So the larger - half two by two equals two square inches. The area of the smaller is half two by .5, giving .5 square inches. Two minus .5 = 1.5 square inches for the smaller, purple chevron. So far our compound area outside the octagon is 27 square inches.

The Grey Chevron

The larger chevron is a bit more involved, but can be worked out with the same principle as the previous one.

We see that if we take a diagonal line from ( 1.5, 3 ) up to ( 3, 4.5), we create a base for a triangle that is √4.5 inches long. This triangle, with vertices at ( 1.5, 3 ) , ( 3, 4.5 ) and ( 0, 6), includes all of the grey chevron plus a smaller triangle, and it is the area of this smaller triangle whose base is also this √4.5 inch long line, that we want to subtract from that of the larger to get the area of the larger, grey chevron.

We form a height for the larger triangle by drawing a line from the midpoint of the √4.5 inch long base, up and left towards the point ( 0, 6), and again through Pythagoras, find its length to be √10.125 inches. Its area is half the base times the height, so with a height of √10.125 inches, and a base at √4.5 inches, we have .5 x √10.125 x √4.5 = 3.375. Now the height of the smaller triangle whose area we will subtract from this, is found by realizing that is goes from the midpoint of the base of the larger triangle, up to the point where the lines y = 2x and y = .5x + 3 meet, which after a simple bit of Algebra, turns out to be the point ( 2, 4).

The height of this smaller triangle will be a line from this point ( 2, 4), to the base midpoint at ( 2.25, 3.75), and using Pythagoras' Theorem, we can find that this line equals ( .25² + .25²). This gives us a value of √( .125), and multiplying this by half the base, we get √( .125) x .5 x √4.5 = .375, and if we subtract this from the 3.375 we got for the larger triangle, we get a grey chevron area equal to exactly three square inches, giving a total for area within the square that is not part of the octagon, of thirty square inches, making the area of the octagon the remaining six square inches. There it is. But is there more ? There certainly is, and so I draw Your attention to another feature of the diagram, and a way to work out its area using a variation on my regular polygon area formula.

The Secret Square

About this Square within a Square

If we were to take away some of the lines from the original diagram, we see that our octagon can be fitted into a square. Using this square and what we can work out about distances from continuing the Cartesian coordinate idea, we can find out the area of this square, as a matter of interest, and how it relates to that of the larger square, and also calculate the sides and angles of the octagon, to see if we can work it into my regular polygon area formula.

To do this, let us carry on in the way we did before - using simultaneous equations to find out where the lines meet, and this will tell us how long the square's base is. To see this proof, refer to the altered diagram below.

Looking at the red line, it can be found that this has the linear equation y = .5x + 3, and it first intersects the green line y = 6 - 2x at ( 1.2, 3.6 ), since .5x + 3 = 6 - 2x .5x = 3 - 2x → 3 = 2.5x, such that x = 6/5 = 1.2, and since y = 6 - 2x, then y = 6 - 2 x 1.2 = 6 - 2.4 = 3.6. This is the top left corner of the smaller square. If we continue along the red line, we see it then intersects the tan brown line y = 12 - 2x, and since 12 - 2x = .5x + 39 = 2.5x, then x here equals 18/5 ( 3.6 ), and from this, as y = 12 - 2x, then y = 12 - 2 x 3.6 = 12 - 7.2 = 4.8, so these two lines intercept at the coordinates ( 3.6, 4.8 ).

This gives us the ability to work out Δx and Δy, being our legs for the right angled triangle for which the length of the base will be the hypotenuse. Δx = 3.6 - 1.2 = 2.4, and Δy = 4.8 - 3.6 = 1.2. From these, the hypotenuse will be ( 2.4² + 1.2² ) = ( 7.2 ) ( = 2.683281573 ), and this of course is the base ( or side ) of our smaller square, bounded by these four different coloured lines, and the area of this square will simply be the square of this number, which is 7.2. This sounds right, as it is larger than the area of 6 we found for the pentagon.

Working out the sides of the Pentagon

Now the four blue areas not part of the pentagon appear to be all the same, and since their total area must be the difference between that of the square and that of the pentagon at 1.2 square inches, each of these four must have an area equal to .3 square inches. But let us be absolutely sure.

Refer to yet another version of the diagram below to visualize what is noted here.

Because the inner square is somewhat symmetrical, we can be confident that each of the four light blue triangles are equal in area. But we want to find their sides. Let us look at the one in the top right corner of the square to see what is going on.

This triangle is bounded by three lines, y = .5x + 3, y = 12 - 2x, and y = 6 - .5x. Now we have already seen how y = .5x + 3 and y = 12 - 2x intersect at ( 3.6, 4.8 ), and You can show Yourself how y = 6 - .5x intersects y = 3 +.5x at ( 3, 4.5 ) and it intersects 12 - 2x at ( 4, 4 ).

The Big Picture

We can better follow what goes on next by magnifying the section above bordered by the blue rectangle. The blown up image can be seen below, while an explanation of it follows.

The Little Picture

Checking the Inner Square's side length with the Formula

Remember that my regular polygon area formula is :

( ½ base )² × n tan(½ internal angle ),

and in the case of this octagon, it therefore becomes

( ½ × √1.25 )² × 8 × tan(67.5 ) = .3125 × 8 × 2.414213562373095 = 6.035533905932738

But we know the area is exactly six, so even though all sides of this octagon are the same, there must be something up with the internal angles.

Now the octagon in some way follows a pattern, and I suspect that it has two different internal angles, that alternate around the perimeter. Looking at the diagram above of the blown up section showing the crimson triangle, we can work out the internal angles by taking advantage of the trigonometric identity that states that two angles along the same line or line segment add up to 180°.

Looking at the internal angles of the crimson triangle in another diagram below, the one at the corner is of course 90°, so this is a right angled triangle. The sum of the other two angles is another 90°. Now since the sides of this crimson triangle are √.45 and √.80, respectively, the tangent of the angle φ will equal √.45 divided by √.80, making the angle φ to be 36.86989764584402°, so that the angle θ, along the same line, will be 180° minus this angle, and therefore is 143.13010235415598°, and thus one of the internal angles of this octagon, but this should normally be 135 °. The angle ψ, whose tangent equals √.80 divided by √.45, will then be 53.13010235415597, and on the same line, the angle α should be 180° minus this angle, which will be 126.86989764584402, which is the other internal angle for this octagon, but is different. So four of the angles will be this latter one, and the other four the former, alternating around the perimeter of the octagon. Now what do we do to these, to get them into the formula ?

Explanation

Actually, what if we rearranged the area formula, knowing the area to be equal to our six square inches, to see what angle would be required to make such an area ?

That is, where

( ½ base)² x n tan ( ½ angle ) = ( ½ x √1.25 ) ² x 8 tan ( ½ angle ) = 6

6 ÷ 8 x ( ½ x √1.25 )² = tan ( ½ angle ) → tan ( ½ angle ) = 2.4.

½ angle = 67.38013505195957°

Now when I tried to manipulate the two actual half angles of the octagon, that is,

126.86989764584402° ÷ 2 = 63.43494882292201°, and 143.13010235415598° ÷ 2 = 71.56505117707799°,

I could not immediately find a pattern between them, by taking from their values to get to this one derived from manipulating the formula.

It was only when I realized that while 71.56505117707799° = arcus tangent ( 3 ), and 63.43494882292201° = arcus tangent (2), this derived angle, 67.38013505195957°, between the other two in value, equals arcus tangent (2.4), meaning that it is the angle whose tangent equals 2.4. It is interesting to note that whereas 2.4 equals 3 minus .6, or that is to say, 3 minus , it also equals 2 plus , and I thought - yea, nah - it could not be that simple ! But I decided to see if this was true.

That is, if there was some value x for the angles, such that I could take a given value, minus this value divided by this x, to get an equal value if I took another value, and added that other value divided by this same x to it.

Algebraically, where I am dealing with the tangents of angles, such that

tan( α ) - ( tan( αx ) = tan( β ) + ( tan( β ) ÷ x ), where either equals tan( γ), and gamma ( γ ) is this new angle that the formula works for when the angles are different.

So that :

tan( α ) – [ tan( αx ] = tan( β ) + [ tan( β ) ÷ x ]

tan( α ) = tan( β ) + [ tan( β ) ÷ x ] + [ tan( αx ]

tan( α ) = tan( β ) + [ tan( α ) + tan( β ) ] ÷ x

tan( α ) - tan( β ) = [ tan( α ) + tan( β ) ] ÷ x

→ [ tan( α ) + tan( β ) ] ÷ [ tan( α ) - tan( β )] = x

Now this works whether we pick tan( α ) - ( tan( α ) ÷ x ) or tan( β ) + ( tan( β ) ÷ x ), and whatever we do, this value for x, which is given just above, is divided into either tan( α ) or tan( β ), or better, its reciprocal is multiplied by which ever of the two tangents we choose.

Let us use tan(α). ( Incidentally, α is designated such that it is always greater than β ).

We can also write tan ( γ ) = tan(α) [ 1 – ( 1 ÷ x ) ] ( also equals tan(β) [ 1 + ( 1 ÷ x ) ] ).

Since

x = [ tan( α ) + tan( β ) ] ÷ [ tan( α ) - tan( β )] ,

then 1 ÷ x = [ tan( α ) - tan( β )] ÷ [ tan( α ) + tan( β ) ],

so the whole thing ends up being

tan ( γ ) = tan(α) [ 1 – ( 1 ÷ x ) ] = tan(α) { 1 – { [ tan( α ) - tan( β ) ] ÷ [ tan( α ) + tan( β ) ] } }

( This formula will be shown in a simpler to read fraction form in an illustration also ).

If we check our octagon example, we would see that this works, as it does for other shapes where every side is identical, and there are two different, alternating angles, and an even number of sides. For this example, as noted, tan(α) = 3, tan( β ) = 2, so fitting these in the formula we get

tan ( γ ) = tan(α) [ 1 – ( 1 ÷ x ) ]

= 3 { 1 – { [ 3 - 2 ] ÷ [ 3 + 2 ] } } = 3 { 1 – { [ 1 ] ÷ [ 5 ] } } = 3 × = 2.4


Or if we wanted to use tan( β ), then

tan ( γ ) = tan(β) [ 1 + ( 1 ÷ x ) ] = tan(β) { 1 + { [ tan( α ) - tan( β ) ] ÷ [ tan( α ) + tan( β ) ] } }

= 2 { 1 + { [ 3 - 2 ] ÷ [ 3 + 2 ] } } = 2 × ( 6 ÷ 5 ) = 12 ÷ 5 = 2.4.

So it does work. There are even some for shapes having the same angle but 2 different lengths, where all we do work out one area for one length, and add it to the area for the other, halving n in the process. So say we have a shape with the one internal angle, and it has ten sides, five of which are three inches, the other five 3.5 inches, we make our n equal to 5, work out the area for three inches, then keeping it at five, add to this the area for 3.5 inches. We shall see this later. I have also had success with shapes having different length sides and different angles, but eventually this gets rather complicated, and it may serve simpler to work out the area a different way. Naturally, as complicated as this may somewhat seem, it may be very useful with shapes having a good number of sides, so it certainly is not impractical. So let us see a picture of the formula above.


Try it Out

Just to be sure, let us try this on a four sided figure, to make sure it does not just work for Octagons. Check out the illustration below.

Working out the four side area.

Now we can see the area for this can be relatively easily worked out at base times height, which here is 36cos( 3 ) = 31.176914536239792, but it is still of interest to see if the regular polygon formula could work it out.

We see we have two different internal angels, sixty degrees and 120 degrees, respectively, so the half angles are thirty and 60 degrees, where 60 = α, 30 = β, and from these we are wanting to find γ. This will be easier to follow on a diagram, considering the fractions involved, so let us see :

Conclusion

This is the same value for working out base times height, so it does work. Now of course, as stated before, it might be more practical in cases where the number of sides is small, to use easier means to work out the area, such as in this particular sample, base times height, but when n gets very large, the formula will be more useful. Even so, whether harder or not, just like the Cricket Formulae mentioned in a previous Hub, where run rates and such could be done more quickly the traditional way, it is still interesting to see that an alternative is there and works, and teaches us about the intracacies of maths, where such a thing might lead to other shortcuts elsewhere not yet obvious. Of course, if one could run it as code in a computer programme, or input such a formula into a calculator that accepts them, then one need only punch in the values, and then these kinds of formulae are more worth the effort.

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 Hubs, which are all about Triangles, the first of these being on the Paper Folding Triangle. This one, and the Rest, are : The Wonder and Amusement of Triangles - Part One, 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, and Pyramids - How to find their Height and Volume,

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 these, and note how interesting everything 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.

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