In this video we are going to look at how to calculate the area of a regular polygon. Now remember a regular polygon is one in which all of the sides are the same length and also all of the interior angles are equal to each other.
Now there is a formula for calculating the area of a regular polygon, and we’re going to look at where it comes from first of all. So we’re going to derive this area formula in terms of a pentagon first of all, and then we’ll look at generalizing it to polygons with any number of sides later on.
So I have here a diagram of a regular pentagon and the side length of this pentagon is 𝑥. Now the first step in working out this area formula is to connect all of the vertices to the centre of the pentagon, so drawing in line like these ones I’ve done here.
Now what this does is it divides the pentagon up into five triangles, and in fact they’re congruent triangles meaning they’re all exactly the same size as each other. So if we want to work out the area of this total pentagon, we can in fact just focus on one of these triangles first of all, think about how to find the area of that, and then we’ll be able to multiply the answer by five to find the total area of the Pentagon.
So let’s just focus on one of these triangles first of all, and I’ll draw it out again at the side so we can visualise it more easily. So here is one of those triangles; it’s an isosceles triangle because those lines connecting each Vertex to the centre of the pentagon are all the same length, so we have five congruent isosceles triangles to think about.
Now I know the length of the base of this triangle is 𝑥, which is the side length of the pentagon. And at the moment I haven’t got any other information, so I want to think is there anything else that I can work out about this triangle. And in fact there is, I can work out the size or the measure of this angle here.
Now back in the context of the original pentagon, that angle would be this angle here. And what you see is that you have five of these angles cluster together around a point. Now as all of these triangles are congruent to each other, those angles at the centre must be the same. So if I want to work out the measure of just one of them I need to divide three hundred and sixty, which is the total of angles around a point, by five.
So this angle at the top of my triangle, which I’ve chosen to label 𝜃, is equal to three hundred and sixty divided by five. Now that’s obviously is equal to seventy-two degrees, but I’m going to keep it as three hundred and sixty over five during my calculations so that it becomes easier to generalise to a polygon with any number of sides at a later stage.
Now if I want to find the area of this triangle, well the area of a triangle remember is the base multiplied by the perpendicular height divided by two. So in the case of this triangle, that would be 𝑥 multiplied by whatever this distance here is and then divided by two.
Now the perpendicular height of this triangle, which I’ll also mark onto my diagram of the pentagon, well it has a very specific name in the context of polygons. It’s called the apothem. And what that refers to exactly is a line connecting the centre of the polygon to the midpoint of each of the sides.
So because its name is the apothem, I’m gonna use a lowercase letter 𝑎 to refer to it in this triangle here. So we’d like to know what the length of that apothem is because if I know that, then I can just do a base times height over two or in the context of the letters we’ve got here, 𝑥 multiplied by 𝑎 over two.
So it’s helpful to think about an even smaller subsection of the diagram and what I’m going to do is draw out half of this triangle. So I have this right-angled triangle here. Now some information we can put on, this height is 𝑎 for apothem; this base what it was previously 𝑥, but I’ve now only got half of that so this base here is going to be 𝑥 over two; and the angle well the full angle was labelled as 𝜃, which remember was three hundred and sixty over five; again I’ve only got half of that, so this angle here is going to be 𝜃 over two; or specifically, well if it’s three hundred and sixty over five and then I half it, I could think of this as one hundred and eighty over five.
Now again that does of course have a specific value, thirty-six degrees, but I’m going to keep it like this so that I can generalise more easily later on. Now I’d like to know the length of the apothem, assuming I know what the side length of this polygon is. So this is why I need to use a little bit of trigonometry, so let’s label up the sides of this right-angled triangle according to this angle, 𝜃 over two. So this side over here, 𝑥 over two, that’s the opposite of a triangle and 𝑎, the apothem, that is the adjacent because it’s between the right angle and the known angle.
So if I think about trigonometry and if I think about SOH CAH TOA, the trigonometric ratio that involves the opposite and the adjacent is tangent. So I’m gonna be using tan here, and I can write down what the tan relationship tells me for this triangle. So tan of this angle is the opposite divided by the adjacent, which is 𝑥 over two divided by 𝑎. Another way of writing that for the right-hand side would be 𝑥 over two 𝑎.
So I can write it like this. And then if I’m looking to rearrange this to work out what the apothem is, well 𝑎 is currently in the denominator of this fraction on the right-hand side. So if I multiply both sides by 𝑎, then I’ll have 𝑎 lots of tan one eighty over five is equal to 𝑥 over two. And the final step to get 𝑎 on its own is to divide both sides to this equation by that tangent ratio, which gives me this expression for the apothem: 𝑎 is equal to 𝑥 over two tan one eighty over five.
Now that tan is in the denominator, so at this stage you may recall that there’s another trigonometric ratio called cot, which is just equivalent to one over tan. So I could replace that one over tan with cot, and instead what I’ll now have is this expression for the apothem here: 𝑎 is equal to 𝑥 over two multiplied by cot one eighty over five.
So that gives me a method for working out the apothem of this polygon if I know the side length. Now back to working out the area of this pentagon, we said at this stage here that the area was equivalent to the base times the height divided by two for each of these triangles. So now I just need to substitute in what the base and the height are. So the base of this triangle is 𝑥 and the height, which is the apothem, is 𝑥 over two times cot one eighty over five.
So I can substitute that part in, and then remember I need to divide by two again so I’ll include another division by two. Now this does simplify a little bit, I’ve got 𝑥 times 𝑥 in the numerator so that can become 𝑥 squared. And in the denominator, there are two factors of two so that can become four, which gives me the area of 𝑥 squared over four times cot one eighty over five for each of these triangles.
But remember there are five of them. So if I want the total area of this pentagon, then I need to multiply by five. So that gives me this formula here: the total area of the pentagon is five 𝑥 squared over four multiplied by cot of one hundred and eighty over five, where 𝑥 remember represents the side length of this pentagon.
So if I knew that 𝑥 for example is four centimetres, then it’s just a question of substituting that value into this formula in order to find the area. Now that formula that we’ve just worked out; that was specifically for a pentagon, and we’d like to have a formula that works for any polygon. Now just a couple of very small tweaks that we need to make to this formula, if we look at the formula, there are two places with a number five appears. And that number five is because it was a pentagon, which had five sides.
If you want to make this work for any polygon, all we need to do is replace those fives with the letter 𝑛, where 𝑛 represents the number of sides that the polygon has. So here we have our general formula: the area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛, number of sides, multiplied by 𝑥 squared over four times cot of one hundred and eighty over 𝑛.
Now it’s worth just mentioning that this is assuming we’re working in degrees rather than any other type of measure. If you were working in radians for the angle, then this one hundred and eighty would be replaced by 𝜋 because a hundred and eighty degrees is equivalent to 𝜋 radians.
So you may also see versions of this formula written, where the angle is in terms of 𝜋 instead of in terms of one hundred and eighty degrees. So let’s look at applying this to a question. The question asked us to calculate the area of a regular hexagon of side length seven metres.
So as a reminder, here’s the formula that we need: the area is equal to 𝑛𝑥 squared over four multiplied by cot one eighty over 𝑛. So 𝑛 is the number of sides. Well for a regular hexagon, that’s going to be six. And 𝑥, remember, represents the side length, so in this question that is seven.
So all we need to do is substitute the values of 𝑛 and 𝑥 into this formula. So we have that the area of this hexagon is equal to six multiplied by seven squared over four multiplied by cot of a hundred and eighty over six, which gives me two hundred and ninety-four over four multiplied by cot of thirty. At this point I can use my calculator to evaluate the answer; remember I’m working in degrees here. so my calculator needs to be in degree mode.
So evaluating that gives me one hundred and twenty-seven point three one metres squared, and that’s rounded to two decimal places.
Now suppose you couldn’t remember that formula or perhaps you like to work a little bit more from first principles. Now you could just work this out as we did when we derived the formula. So here is the hexagon with sides of length seven. Each of those angles at the centre is sixty degrees, because they’re three hundred and sixty divided by six, and you could draw out one of these six congruent triangles.
So here it is and I’ve divided it in half. Now this length here would be three and a half metres because it’s half of that seven. This angle would be half of that sixty degrees, so it’s thirty degrees. And you could use trigonometry as we did in order to work out the length of the apothem there. So you would have tan of thirty is equal to three point five over 𝑎. Rearranging that formula for 𝑎 would give you 𝑎 is equal to three point five over tan thirty or three point five cot thirty.
And you would then have everything you need to work out the area of this hexagon. So the area of each triangle would be base times height over two, so seven times three point five cot thirty over two. But then remember for the total hexagon, we would need to multiply this by six as we have six of these triangles.
And of course that gives us the same value of a hundred and twenty-seven point three one metres squared. So you either learn this formula off by heart, committed to memory, or you remember the process and the understanding behind it so you could work it out yourself in a similar way.
Right the final question that we’ll look at here then the area of a regular decagon is a hundred and fifty-five point eight centimetres squared. What is the perimeter of the decagon? Now first of all a decagon remember has ten sides, so we’re interested in a polygon where 𝑛 is equal to ten.
Now looking at the question we’re not given the side length anywhere, but we want to work out the perimeter. So this is a question about working backwards from knowing the area, deducing the side length, and then using that to calculate the perimeter of the decagon. So we’re going to need our area formula, so let’s recall that from the previous work.
So there it is: area is 𝑛𝑥 squared over four multiplied by cot one eighty over 𝑛. Now we know that 𝑛 is equal to ten because we’re working with a decagon, and we also know that this area is equal to a hundred and fifty-five point eight centimetres squared. So we can set ourselves up an equation. And here it is: ten lots of 𝑥 squared over four times cot of a hundred and eighty over ten is equal to a hundred and fifty-five point eight.
So now we have an equation that we can solve in order to work out the value of 𝑥; 𝑥 remember presents the side length. And once we have that, we can then easily work out the perimeter. So let’s just simplify this equation a bit first of all. Ten over four will simplify to five over two. And also cot of a hundred and eighty over ten, well that’s just cot of eighteen.
So we actually have this equation here. Now we want to get 𝑥 squared on its own, so we got that factor of two in the denominator. We’ve also got a factor of tan eighteen in the denominator because remember cot is one over tan. So I’m gonna multiply both sides of the equation by two tan eighteen.
And that would give me that five 𝑥 squared is equal to two tan eighteen multiplied by a hundred and fifty-five point eight. Then I can divide both sides of the equation by five, which will give me this line here. And I’ll evaluate that on my calculator; remember I’m working in degrees here.
So that gives me 𝑥 squared is equal to twenty point two four eight nine nine, and so on. Next we need to take the square root in order to work out what 𝑥 is, so I use my calculator to evaluate the square root of twenty point two four eight. And that gives me 𝑥 is equal to four point four nine nine eight eight eight, and this value to any such a reasonable degree of accuracy is 𝑥 is equal to four point five. So I’ve worked out the side length of one side of this decagon, but I haven’t finished the question because the question asked me to calculate the perimeter. Now for the perimeter I just need to add up all the sides. Well there are ten of them, and they’re all the same length. Therefore the perimeter is gonna be ten 𝑥.
So ten lots of this value we’ve just worked out of four point five. So the perimeter of this decagon is forty-five centimetres.
So we’ve seen the formula for calculating the area of a regular polygon. Remember you either need to learn it off by heart or just be confident in the logic and the method behind it. We’ve seen how to apply it to the question. And then we’ve also seen the method of working backwards from knowing the area to calculating the side length or the perimeter of the polygon.