 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about similarity and using the similarity for calculating the volumes. But before talking about the volume of the pyramid, which is my next topic, I would like to just exemplify how the theory of limits can be used in two-dimensional case because in a three-dimensional case it will be basically the same from the idea standpoint. So in this lecture I would like to derive the formula for the area of a triangle using this limit theory approach, which is going to be used for pyramids as well. Well, obviously we all know that the area of triangle can be derived differently. For instance, if you have this triangle, you can always make the same triangle of the same area and make a parallelogram and the area of the parallelogram can be actually converted into the area of this rectangle. So the area of this triangle would be half of the area of this rectangle which equals to the product of its base, which is the same as the base of a triangle by height, which is the same as the height altitude of the triangle. So this formula, base times altitude to this base divided by 2 can be derived purely geometrically, which I did before. Now I'm going to derive it differently and again the purpose is not to derive the formula. The purpose is to explain the idea how it can be done using the theory of limits because this would be used in three-dimensional case. Alright, so let's assume you have a triangle, A, B, C, this is A, this is H. Now this altitude has a base, A, H. So B, C equals A, A, H, altitude, base and altitude. And again my purpose is to derive this formula. Now the way how I'm going to do it is the following. I choose some number N, integer positive number and I will divide A, H into N equal parts and draw horizontal lines. So this would be my B1, B2, B3 and B4 which is equal to B. This would be C1, C2, C3 and C4 which is C. So in this case N is equal to 4. Now what I will do next is I will convert each trapezoid which I have here and the triangle on the top into a rectangle. Now the way I will do it is the following. I will just make it this way. So now I will assume and this is a non-rigorous part of this lecture. I will assume that the area of this new object, I don't even know how to call it, it's kind of a step, ladder, whatever it is. So the area of this particular new object which is a combined area of all these rectangles as my number of divisions tends to infinity, increases to infinity would actually be closer and closer to the area of this particular triangle. So if my N is increasing, the number of these is increasing and therefore each one of them is decreasing. So the height of this little rectangle would be smaller and smaller which means that these steps would be smaller and smaller that would be closer to the triangle. And that's why I'm assuming and this is again an assumption I do not pretend that I rigorously proved it. So I assume that if I will go to a limit then as N goes to infinity the area, the sum of the area of these rectangles would approach closer and closer to the area of the triangle. Well, let's try to calculate exactly the area of this step-like object and see if we will have really this particular limit existing and it will show us whether we are approaching this particular formula that we know from the geometric considerations. Alright, so let's consider a particular number N which is from 1 to N. Let's say in this case it's B2 and equals 2 in this case. And let's consider a triangle A, B, N, C, N which is this triangle and triangle ABC. Now this line from BN to CN is parallel to BC. Obviously the consequence of this is that these two triangles are similar to each other because these lines are common, this is parallel so all angles are the same. Now, how about the lengths of segment BN, CN? Well, everything is proportional, right? So BN, CN relates to BC as, let's put letters H1, H2, H3 and H4 which is equal to H. It's related as altitudes for obvious reasons since triangles are similar then the corresponding altitudes are also of the same type. So it's AHN to AH. Now, but this we know because AHN is N pieces of these small things and AH, I mean lowercase N, and AH is uppercase so this is equal to N over N. So that's simple. Which means that BN, CN is equal to BC which is A, right? BC is A times N divided by capital N. And HN, AHN is equal to AH which is altitude also times N over N. Now, what follows, this is trivial, what follows is the following. Now, let's summarize all these rectangles, areas of these rectangles. Now, the area of each rectangle is the product of its base which is BN, CN times its height which is 1 eighth. The height of each one is 1 eighth, 1 nth of the total height. So, the area which is SN there, SN is equal to BN CN times HN minus 1 HN, right? This is from HN minus 1 to HN and this is from BN to CN. Which is equal to, now this we know, AN over N and this is H divided by capital N, right? AH over N square times N. N is a variable, that's why I put it separately. A and H and capital N are constants. Now, the area which I'm interested is, is sum of these. Sum of SN when N is changing from 1 to capital N. Which is equal to what? AH over N square, 1 plus 2 plus etcetera plus N. Right? Because this is changing from 1 to capital N, right? So, if I'm summarizing, for S1 it's 1, for S2 it's 2, etcetera. When I'm summarizing, I can just factor out this and I will have this in my, in parenthesis. Now, this is easy. This is just the arithmetic progression which we know how to deal with. And the, I will, I'll just write the formula which I remember, which is kind of a rare case, but I do remember it in this case. For those who don't remember the formula, go to the corresponding lecture on arithmetic progression and it's derived. Basically, it's very easy. Well, actually it's very easy, I can do it right now. If you write this formula in a different order and then summarize both, you will have N plus 1 plus N plus 1 plus etcetera plus N plus 1 N times. Right? That's how you do it. That's how all arithmetic progressions are summarized. Just write it in reverse and add vertically. So, since I added the same one, so I have to do it divided by 2, so it's N times N plus 1 divided by 2. That's the sum of one series of arithmetic progression. Alright. So, that's the formula. This is sum of all areas of rectangles. And again, let me just repeat, I assume that as N approaches infinity, this particular formula should have a limit, something like this. So, let's just do it and let's check if that's true. Let me just make it a little bit more palatable. I can factor, well, I can actually reduce by N, right? And I will write it in this way, A H over 2 N plus 1 divided by N. How about this, right? 2 is here, N is there, one of the N's is reduced, so that's the result. Or, if you wish, A H over 2, if I will divide, I will get 1 plus 1 over N. Or again, I can just simplify it easier. That's when I multiply by 1 and that's when I multiply by 1 N. Now, it's obvious that as N goes to infinity, this thing obviously tends to zero. And what's remaining is this. So, in the limit when capital N goes to infinity, some of these rectangles, some of the areas of these rectangles, has a limit which is this. So, that's basically an approach how I can calculate the area of this particular figure. Now, what's not really rigorous? You have to really understand my assumptions which definitely need some proof, etc. Well, assumption was that this step-like object has the area which is approaching the area of a triangle as N goes to infinity. Now, what if I will divide it differently? What if I will use not equal divisions, but divisions of some kind of division, which is not really making these equal, but still I can make a process when the height goes to zero. Because in this case, my height is 1 over N, right? 1 over capital N. But they are all equal. Maybe they are not supposed to be equal. I mean, there are many different ways to approach this purely philosophical dilemma. How should I divide this particular triangle and convert it into a sum of rectangles to make this particular limit? Equal to some really area of this triangle, because maybe if I will divide it differently, I will get a different limit. I mean, in this case, I've got exactly this particular formula, this particular formula. But maybe in other cases I will not. So, this is something which definitely needs to be addressed on a much more rigorous level. However, my purpose was, again, to just demonstrate this particular approach of approximating the area of a triangle with sum of areas with ultimately infinite sum of the area of small rectangles. And that's exactly what I'm going to use for pyramids as one of the ways to derive the formula for the volume of the pyramid. And there are others. There are other approaches based on the Kabalyeri principle, which is also kind of defined in this particular topic. Because it's related to the fact that in the similar figures, in two-dimensional cases, the linear dimensions are similar. Because that's exactly how I have derived the lengths of the BNCN. In the three-dimensional cases, I will have the areas proportional to each other. Okay, that's it for today. I would suggest you to read the notes for this lecture. It basically explains the same thing again. And it's on theunisor.com, of course. That's where I actually suggest you to watch any of my lectures because the notes are very important. So, basically, that's it for today. And again, prepare for making this particular approach a little bit more complicated when I will move to three-dimensional world for the pyramid. In this lecture, I would consider it to be as an introduction to three-dimensional case of the volume of the pyramid. Thanks very much and good luck.