 Hi, I'm Zor. Welcome to InDesire Education. I would like to continue talking about Parallel Pippets, and today's lecture is about the volume of Parallel Pippet. Well, first of all, what is volume? Volume is obviously a quantitative characteristic of the space occupied by some body, solid body, like Parallel Pippet. What's very important about volume, this is an additive measure. Now, what it means is that if I have one particular solid body, which has certain volume, which characterizes how much space it occupies, and I have another one, solid body, with its volume, then editiveness of the volume means that both of these solid bodies together occupy the volume which is equal to a sum of the volumes of each individual one, which means that, for instance, if you have a large object and you divide it into two different pieces, then the volume of the whole object is equal to the sum of the volumes of its pieces. So that's what editiveness means. Okay, now, obviously, if we are measuring something, we have to have the unit of measurement, right? So the unit of measurement of the volume is, by definition, a little cube with edge equals to some linear unit, like one meter, for instance, in one system or one inch in another system or whatever. So if we have, as a unit of volume, a cube with the edge equal to one meter, it means we are measuring the volume in cubicle meters or in cubic meters. That's how many these cubes can fit into a particular solid body, like parallelepipus, for instance. Alright, that's easy to say, but it's not that easy to demonstrate, and here is why. Let's consider the first example. We have a rectangular parallelepipus with sides A, B, and C. You know what, it's probably right to say B on the bottom and C on the height. Let's say this is B and this is C. So this is a rectangular parallelepipus, which means that A, B, C, G is a rectangle, as well as A prime, B prime, C prime, and D prime. And every side face of this parallelepipus is also a rectangle. So all of them are rectangles. So these are right angles. All of those angles are right. Now, let's also consider that all linear dimensions A, B, and C, these three dimensions completely determine our rectangular parallelepipus, they are all integer. Why do they have to put it as an integer value? Well, because I would like this cube, which has an edge of 1, to fit completely inside this parallelepipus. Now, obviously, if my dimensions are integer A, B, and C, I can put A cubes along this edge and B cubes along that edge in one row, which will have a height of 1. That will be my first row of cubes. And again, considering A and B are integer, I will have exactly A times B, these little cubes, fit into the bottom layer of this parallelepipus. The second layer will also have exactly the same. And how many layers will I have? If this is C and C is integer, I guess I will have C layers. So my volume would be equal to A times B times C. Or if you wish, the area of the base, which is A times B, times height, which is C. So S is area of the base and H is height, or altitude of the parallelepipus. So in case this is a rectangular parallelepipus with integer sides A, B, and C, then the volume is measured as this or this. Okay, that was an easy part of it. It's very easy with integer. But what if it's not integer? What if A, B is equal to 5.5, let's say, of the same units as this one? So obviously I will not be able to put integer number of these cubes along that H. So what should I do? Okay, let's consider the next case. Let's say A, B, and C are rational numbers. Now, it's three rational numbers. I can obviously use one common denominator and say that A is equal to P divided by N where N is common denominator. B is equal to Q over N and C is equal to R over N. So these are three rational numbers. So any three rational numbers can be brought into this kind of form if as N I will use the common denominator. All right. Now, how should I approach measuring this particular volume in this case? Here is my suggestion. Let's divide this cube, the unit cube, into little cubes which have the side equals to 1 over N. The same N is here. Now, if the side of the cube is 1 and the side of the little cube is 1 over N, it means that I have exactly N little cubes fit along this edge and N little cubes along this edge and N little cubes along the vertical edge. So the total number of little cubes in this one cube would be N times N times N, which is N cube. Right? So that's how many little cubes. Now, little cubes are better than the big cube, the unit cube in this particular case because I can fit an integer number of little cubes with this edge into each side. Well, how many? Let's see. If my edge of the little cube is equal to 1 N and my side A is equal to P N, it means I can put exactly P little cubes along this edge. And obviously, similarly, I can put Q along this edge, the B, and R along this edge. So my volume would be equal to P times Q times R times where V is equal to volume of the little cube. By the way, this is not a volume, this is number of cubes. So the volume is equal to, I'm sorry, this is number of cubes. So if the volume is N to the cube, if number of cubes of little cubes is N to the cube, then the volume of each one of them, let's put it little x here. Dx will be equal to the volume of the little cube. If there are N to the power of three little cubes inside the cube which has a unit volume, it means that each one of them is this. So here, this is Vx. So here I will have P times Q times R little cubes, each one of them has volume Vx and this volume is N to the third degree. So it would be P times Q times R divided by N cube equals to P over N times Q over N times R over N equals to A times B times C. As you see, we have exactly the same formula as with integer ABC. Which is equal to area of the base times H. So even with rational A, B and C edges of the rectangular parallelly pipette, I still have the volume expressed as the product of these three lengths or the product as an area of the base times the height. Now, irrational. Well, that's difficult. I cannot do any manipulation of that side and I really have to go back to the definition of what is irrational number. And this is a very, I would say obscure part of the mathematics because it involves to approach it rigorously, you really have to approach it using the limit theory and stuff like this. So, because basically every irrational number can be approximated to any degree of precision with the set of rational numbers. And this process of approximation to a better and better precision is the root of the next statement which I'm going to make. So, the next statement is that even with irrational A, B and C, because each irrational number can be as close as possible to some kind of a rational number. And with rational number, this formula is correct. It means that with any degree of precision, I can actually say that A times B times C, even if these numbers are irrational, they do represent the volume. Because again, I can always replace A with certain rational number and B with certain rational number and C with rational number which are very, very close. I mean as close as I want to this irrational. And that's why we basically can feel and even if it's not 100% rigorously proven, but the approach is, as I was just saying, approach is to approximate any irrational number with a rational and this is true for rational so which means this will be true for irrational numbers. So, that covers the rectangular parallelepipeds. So, for any rectangular parallelepiped, the volume is equal to a product of its three sides with common vertex or area of the base times the height or altitude. Now, let's go to the next complexity level. From the rectangular parallelepipeds, we go to the right parallelepipeds. Now, the right parallelepipeds are different in one particular aspect. At the base, I don't have a rectangle. I have just any parallelogram. However, these edges are really perpendicular to the base because this is the right parallelepiped. So, any right parallelepiped has the side edges perpendicular to the bases. Bases are parallel, obviously, and congruent parallelograms. So, that's the case which I'm going to consider next. Let me wipe out this and I will leave my beautiful picture. So, now we assume that A, B, C, G is not a rectangle but a parallelogram. Now, if you view this particular parallelepiped from the top, what do you see? Well, you will see the parallelogram, right? A, B, C, G. Now, since these are perpendicular to the plane, A prime, B prime, C prime, and D prime will be projected to exactly A, B, C and G. These are all perpendicular, right? So, that's what you will see if you will look from the top. Now, here's what I'm going to do. I will do a perpendicular from the B to A, C, called point M. And then, on the continuation, I will put perpendicular from D to M. Now, it's very easy to prove that the triangles C, N, D and A, M, B are congruent. Number one, they are the right triangles, right? Because these are perpendicular to this line. Now, this is equal to this because it's A, B, C, D's parallelogram. And the angles are the same, obviously. And these lines are parallel, so this is also the same as distance between parallel lines. I mean, we have lots of elements here. Actually, it was enough, even the first two. It was hypotenuse and an acute angle are congruent to each other. So, triangles are exactly congruent, which means if I will take A, B, C, G parallelogram and I will wipe out this piece, but add this piece, I will subtract something and I will add something. These are congruent triangles, which means their areas are supposed to be the same, which means that the area of this rectangle is exactly the same as the area of the parallelogram. And we are writing the wrong, if you wish. So, from the parallelogram, which used to be before, I convert this parallelogram into a rectangle with the same area. Now, I would like to do it in three-dimensional pictures. So, from B, I drop a perpendicular to A, D, something like this. And on the continuation of this, I drop a perpendicular from A, B, C, G. Well, I have different letters here. It doesn't really matter. And so, from this continuation, you know what? I will probably redraw my bottom. It's too many lines here. So, this is A, B, C, and G. So, this is perpendicular to B, N is perpendicular to A, G, and C, N is perpendicular to the same line. So, everything, all these letters, A, B, C, G, M, and N, are in the plane, which is a base plane of this picture. So, what I'm basically doing right now, I would like to convert my right parallelepiped into rectangular parallelepiped. So, I did this operation on the base. Now, I will cut with the plane. So, from here, I will cut here. So, that would be my M prime. And here, I would add. That would be N prime. And I will connect this to this. And, okay. So, what I will cut from my parallelepiped is a right triangular prism. One second. So, I will cut from my parallelepiped a triangular prism A, B, A prime, B prime, M prime, A, B, M. So, I will cut it off. And I will add exactly the same in all the dimensions, and obviously the volume, prism which has base C, D, N, and then the top base C prime, D prime, N prime. It's exactly the same triangular prism as this one. Well, the same it means congruent in all respects, because bases are the same, ages are the same, they're all perpendicular. I mean, everything is the same. Which means I will convert my right parallelepiped A, B, C, D, A prime, B prime, C prime, D prime into a rectangular parallelepiped BCNM, B prime, C prime, N prime, M prime, which has exactly the same volume. And it has exactly the same volume, because I'm cutting a piece from the old one and adding that piece on another side, thereby converting my right parallelepiped into rectangular parallelepiped. Now, about rectangular, I do know that its volume equals to the product of area of the base times height. But the area of the base which is M and C, B is exactly the same as area of A, B, C, D. Because of this, we are adding triangle and subtracting triangle. So that's why my formula actually remains exactly the same. Because this is the volume of the rectangular pyramid, the new one, and the new base. That is exactly the same, the old and the new one. Now, the old volume is exactly the same as the volume of the rectangular parallelepiped, because I subtracted one and added one. Area of the base is also exactly the same, old and new, because, again, I added a triangle and subtracted a triangle. Which means the formula is still exactly the same. The formula for the right parallelepiped is volume is equal to area of the base times its altitude or height. And the base is just any parallelogram. So this is one step forward from rectangular parallelepiped to any. So this step was from rectangle to the right. So I still preserved the perpendicularity of the edges. Now, the last step is, forget about perpendicularity. Now, I did use the perpendicularity in this particular case when I switched from the rectangle to the right. Now we are going from the right to any, which means that the whole construction can be tilted. Let's draw another. And I will basically do exactly the same, maybe a little bit less rigorous. So we have, again, some kind of a parallelogram on the top. And then these are slanted, something like this. Okay. Can I write this particular wrong, if I can say so? Well, basically I should use exactly the same approach. What I should do is, first I should write the edges, which are side edges. So I will drop perpendicular from A prime, B prime, C prime and D prime onto the base, which is here. Well, this is parallelogram, so it will project into some kind of other parallelogram. Say, if I will drop a perpendicular here, a perpendicular here, perpendicular here, it's not a pretty picture. But anyway, perpendicular here. So this will be C second, B second, A second and D second. So now, within this plane, I have two parallelograms, original one, A, B, C, D, and the projection of the top one onto the bottom, which is A second, B second, C second and D second. Now, what I would like to say is that the volume of the original parallelogram, which has the base A, B, C, D and A prime, B prime, C prime, D prime is exactly the same as the volume of the new parallelogram, which has the base A second, B second, C second and D second, and the top base is exactly the same as the first one, A prime, B prime, C prime and D prime. How can it be visible? Well, unfortunately, it's not easy to imagine in three-dimensional space. But basically, you really have to consider this as, again, transformation when I'm cutting a piece and adding a piece. So this type of projection converts this slanted parallelogram into the straight one. And the difference between them is something which, like you, for instance, you add this particular, it's kind of a pyramid, I guess, for instance. So you add some pyramids here and you subtract some pyramids there. And by manipulation of these parts, you are, again, you are adding something on one side and you're subtracting something from another side. And whatever you add and subtract is exactly the same. So whatever you cut from here, you add to there. And whatever you add here, you subtract from that side. So I would probably recommend you to try to draw a nice picture, much better than I do. And then it will be kind of obvious. So whenever you're shifting the bottom part from whatever it was before, A, B, C, D, into the new location, A second, B second, C second and D second, your whole object is transformed and this transformation can be expressed as adding a piece on one side and subtracting from another side. Whatever that piece is, it may be a pyramid or it may be some kind of a more complex figure. But whatever it is, you're always adding and subtracting the same thing, which means the volume is exactly the same. And since the volume of the new, and in this case, the new one is the right parallel paper, right, because I dropped the perpendicular. And since the volumes are equal, the volume is still the same of the right pyramid, which is area of the base times height. Now area of the base, A second, B second, C second and D second is exactly the same as A prime, B prime, C prime, D prime, is exactly the same as parallelogram ABCD. So we still have exactly the same formula, which is area of the base times height. Now height, in case of any parallel pipette, is basically at distance between two parallel bases, because that's the length of this perpendicular, A prime, A second. Alright, so the point is that in any parallel pipette, this is the formula. It's a very general formula, area of the base, which can be a parallelogram, obviously not necessarily a rectangle, times the height of this parallel pipette. And the height of the parallel pipette is, in case of a right parallel pipette, it's obviously the edge lengths. But if it's a slanted one, it's just a distance between two parallel planes, which are making the two bases. Well, that's it. I just wanted to demonstrate this formula for any kind of a volume of any kind of a parallel pipette. And I do recommend you maybe to try to draw this a little bit better than I do. But at least you should understand that you always can find equal elements, like A prime is equal to C, C prime, and A prime, A second is equal to C prime, C second. And everything, every element and every angle, every length of a segment and every angle between segments and even angles between planes are exactly the same on that side and on that side. That's why I was talking about aiding piece and subtracting piece. Alright, that's it for today. Thanks very much and good luck.