 Hi, I'm Zor. Welcome to a new Zor education. Today's lecture is about the area of a circle. It immediately follows the circumference of the circle, and I will use basically the same kind of approach to the area which I was using for circumference of a circle. Now, this lecture is part of the Advanced Mathematics course for high school students presented on unizor.com. I do suggest you to watch this lecture from this website, primarily because there are very nice notes written for this lecture and as well as for other lectures as well. So they can serve as a textbook if you wish, and it's always good to be able to read something. About whatever you have listened during the lecture. So, area of a circle. Now, first of all, how the area is measured. You know, we usually have something like a unit of measurement of the area, which is a square with the side of one. So that represents one unit of area. One can be anything, like one meter, one centimeter, one inch, whatever. Now, then if you have some area of a very irregular shape, how can you measure this area? Using this particular measurement unit. Well, you take this measurement unit and position it inside one by one, wherever you can. Then you see that there is some space left after that. So instead of this measurement unit, you're using a smaller measurement unit. Let's say the size of the side would be like one tenths, for instance, instead of one. So it will be a very small one. And you fill with the small ones here. Again, up to a certain point when you cannot put any more smaller squares. And then you're using smaller and smaller and smaller. So basically what I'm saying is that you cannot, in cases like this, you cannot exactly measure this area. You need a process. A process of measuring in bigger units than smaller and smaller and smaller units. Gradually filling up the whole area. And as a limit, basically, of this process, as you continue this process to, well, basically infinity, you will come up with more and more precise evaluation of the area. So circle is basically of the same type. You cannot fit certain number of squares, no matter how small the squares you will use to fill the circle. So I'm suggesting a different approach, which would obviously depend on the fact that this is a circle. And the approach is similar to the one which we were using, evaluating the circumference. Now, if you remember about circumference, I have inscribed a hexagon first as my first approximation. Then I have divided each side in half by dropping a perpendicular and replacing my six-sided polygon with 12-sided. So I will have this. So instead of one bigger edge, I will have two smaller edge. So then I evaluate the perimeter on each step. So first for a hexagon, then 12-sided polygon, then 24-sided polygon, etc. And we saw that we were approaching a certain limit. And then I referred to a theorem, which basically states that no matter which polygon you use in the beginning, as long as you are increasing the number of sides and the maximum size of the polygon on each step is decreasing to zero, then the limit will exist in this process. And no matter how you approach, no matter how you organize this process, as long as the longest edge of the polygon, which you are inscribing tends to zero, the limit will be exactly the same. So there is an existence of this limit and there is a uniqueness of this limit. These two theorems, which I referred to without the rigorous proof, but using these two theorems, we can state that the limit of this process is, by definition, a length, the circumference of a circle. Absolutely in the same key, we will approach the area. So I will start with, let's say, a hexagon, for instance, as I did it for circumference, and I will calculate the area of the hexagon. Now, what is the area of the hexagon? Well, look at it this way. Let's say this is a, this is a, this is a, a, a, and a. So let's consider, in this case, we have six different triangles, right? Which are obviously congruent to each other. Now, what's the area of a triangle? Well, let's consider this one. A1 to A2. Now, actually I can use this piece. I already have this altitude. This is A2. So I have an altitude. Let's call this point B. And this would be our B1 whenever we are doubling the number of vertices. So the area of one triangle is equal to A times H. So A1, A2 is A, that's the side of my polygon, and Po is my altitude. So the area of this particular triangle is A times H divided by 2, right? As well as the area of each other triangle. Now, let's not think about this as a six-sided polygon. Let's say this is some polygon on N's step. So it has N on some step during this process of doubling the number of vertices. So it, for instance, it has N sides. N sides. So the area of the polygon would be N times a times H, right? Now, what is N times A? A, A, A, etc. N times. This is a perimeter, right? So this is a perimeter. So this is the area of the polygon. So in our, in our case the process was start with a hexagon and then divide each each arc by 2 using the perpendicular and then we go to 12, 24, etc. So at certain point and well actually on every point, perimeter times the altitude, now obviously altitude will change, but perimeter times A times altitude divided by 2 is the area of the polygon. Now, as N goes to infinity, how this particular thing is changing? Well, we already know and that's exactly the previous lecture that the perimeter goes to a circumference of the circle, right? If we increase number of sides, then we are getting closer and closer to a circumference. So the perimeter will go to 2 pi r, as we know. How about this H? Well, actually the H will go to the radius and here is why, very, very easy. O, let's consider triangle O, P, A. Now in this triangle, well, you can use different ways. You can use, for instance, the fact that since it's a triangle, then H is less than r, right? Because H is perpendicular and r is a radius. But at the same time, it's greater than r minus 8. Because in every triangle, one side is greater than the difference between two other sides, right? It's a the inequality of triangle. And now as the maximum H of the polygon, the longest H, goes to zero with increasing number of sides, this A goes to zero. So that's why we have H to be between two things and both of them are actually, this is equal to r and this is becoming more and more like r as n goes to infinity. So H is going to r. In this case, it's just bounded by r and something which goes to r. So there is no other way for H but to go to r. On another hand, you can use the Pythagorean theorem that H square is equal to r square minus A square and this goes to zero, right? That's why H and r in the infinity will be the same, the limit of H is r. So what our formula actually is transformed into in a limit case? Well, that's very easy. It's 2 pi r times r divided by 2, right? Which is pi r square and that's exactly the area which you know. So it's a very easy and very fast actually way to derive the formula for the area if you know the formula for for the circumference. Basically, that's it. I just wanted to present this particular formula and approach to get this formula, to derive this formula as a simple limit, using limit theorem in geometry. Actually, it's used in many other cases as well in geometry, but in the in case of a circle it's kind of a very obvious that without this process in this case the process of inscribing polygons into a circle, you cannot really derive what's the values of circumference and the area. And ancient Greeks actually spent a lot of time thinking about this. Now this particular approach to area is ancient and it goes to the Greeks, to Euclidus and other mathematicians. So they were all thinking about this and they did not really had, they did not have the value of pi calculated in any precise way. They were actually considering it about three at the time. It's only contemporary mathematics can prove that well, first of all we can rather rigorously define what is a circumference of a circle and what is the area of a circle using the theorems of existing and uniqueness of the limit of this process of inscribing polygons. So that's very important thing. And another important thing is that pi is actually irrational number. Greeks did not have such a good understanding of what irrational numbers are. They did have the notion of periodic decimals, for instance, but not the irrational numbers. I don't think so. So in any case, that's it for today. Thank you very much. I do suggest you to read the very short notes to this lecture. They basically explained the same thing which I was just talking about. And don't forget that unisor allows you not only to basically listen to lectures, but together with certain supervision you can engage in a real educational process which includes above everything else exams which you and your supervisor can can take a look at the results of how you did it and you can repeat any exam any number of times until you will feel comfortable with any particular topic which you have learned. Well, that's it. Thanks very much and good luck.