 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about pyramids and volume of pyramids. As part of the Advanced Mathematics course for high school students, it's presented on unizor.com website. That's where I suggest you to watch this lecture because it contains some comments, notes, whatever. You can read basically notes for all these lectures like a textbook. Alright, so last lecture was about Cavalieris principle applied to the volume of triangular pyramid. And we have derived a very nice formula about the volume of the pyramid, which is one-third product of the area of the base times altitude. Now, what about non-triangle pyramids? I mean, there are quadrilateral pyramids, pentagon-based, etc. Well, actually, the formula is exactly the same. And let me explain why. It's actually very simple. Let's consider you have a hexagonal pyramid. So, let me draw some kind of a hexagonal. And this is the side edges. Side edges. And obviously, there is one here. And these two are invisible. A, B, C, D, E, F. Okay, this is my hexagonal pyramid. Now, what I'm going to do right now, I will draw three diagonals from A to C, from A to D, and from A to E. Now, I can consider my hexagonal pyramid as a combination of four pyramids. S, A, B, C, D, E, F equals S, A, B, C. D, union, S, A, C, D, union, S, A, D, E, and union, S, A, D, F. The union of all these triangular pyramids is exactly our hexagonal pyramid, right? So, basically, I'm cutting my hexagonal pyramid by planes S, A, C vertically, S, A, D vertically, and S, A, E vertically. And this division gives me three different triangular pyramids. Now, all of them share the same altitude, because the altitude is basically the distance from the apex to the plane where the base is concerned. And the basis of all these pyramids are exactly in the same plane as the basis of the original hexagon. So, the H is the same. Now, what about the volume? Well, volume is an additive function. So, if I clearly divide my pyramid in four different pyramids, which have no common volume, basically, because they're just touching each other. So, the volume of the big pyramid, the hexagonal pyramid, is the sum of these four volumes. Now, the volume of S, A, B, C is equal to area of A, B, C times H divided by three, right? Where H is the altitude of the pyramid. Now, analogously, volume S, A, C, D is equal to area of A, C, D, which is a triangle, by the way. So, I can put here, with a triangle here, times the same H and divide by three. Analogously, S, A, D, E is equal to area of A, D, E times H divided by three. And the last S, A, E, F is area of A, E, F times H divided by three. Now, if I will add them together, what will I get? Well, obviously, H divided by three is outside of parenthesis. And what's in the parenthesis? Sum of the areas of these four triangles, which is actually an area of a hexagon, A, B, C, D, E, F. So, as before, we have received exactly the same formula. The volume is equal to one-third of area of the base times altitude. And that's it for today's theoretical part of the lecture. It's very short, so I have decided to complement it with some typical and relatively simple problems related to volume. So, let's go to the problems. I have six problems here, and they are just reinforced the knowledge about volumes and pyramids. The first problem is we have a tetrahedron with all edges equal to D. D, D, D, D, D, D, D, D. Alright. So, the question is what's its volume? Well, let's think about it. Again, volume is one-third of the product of altitude and the area of the base. Now, the base is a triangle which has all sides equal, equilateral triangle. Now, what's its area? Well, that's easy, right? You draw a perpendicular from one of the vertices onto the opposite side. Obviously, it divides it into D over 2 and D over 2. Now, this altitude of this triangle is equal to, from the Pythagorean theorem, D square minus D over 2 square root, right? From this, this is hypotenuse and these are two calculators. So, this is a, the hypotenuse is equal to D and one calculator is equal to D over 2. So, that's the, that's the formula, right? Now, this is the height and the area is base times height divided by 2. So, it's times D and divided by 2. That's my area. Okay. So, area is done. Yeah, obviously, we can make it a little bit simpler, which is what? This is D square, this is D square, this is one quarter. So, it's D out of the square root and this D is square and under the square root would be 1 minus 1 fourth, which is 3 fourths. And square root is square root of 3 divided by square root of 4, which is 2 and this 2, this is 4, right? So, this is my area. All right. Now, how about the altitude of the pyramid? Well, let's just drop the perpendicular from the top and this is a center of the base. Center is this, it's the point of intersection of all three altitudes or all three angle bisectors or medians or whatever. So, we would like to have this. So, the distance from this center to the side. Now, we know that all medians are intersecting in the ratio of 1 to 2. So, it's 1 third here and 2 thirds there. This is a very known property of medians in the triangle. So, this is the equilateral triangle. So, obviously, it has the same thing. Now, this is the altitude. So, we're talking about 2 thirds of the altitude. Now, we know what altitude is. It's this one. So, we need 2 thirds of this altitude, which is, again, d goes out of the square root and this is square root of 3 divided by 2. So, that's my segment between the center and the vertex, right? Which is what? d square root of 3 divided by 3. Now, so, this is this piece. Now, but I need an altitude. Now, this is actually the right triangle because this is the perpendicular to this and this is the hypotenuse. Hypotenuse is equal to d. So, my calculus is equal to d square minus d square 3 divided by 3 and square root. So, that's my altitude, which is equal to d goes out 1 minus 1 third to 3rd. So, it's d square root of 2 divided by square root of 3. Now, what's next? Next, I have to multiply my altitude to the area and divide by 3. So, it will be d to the power of 3. Square root of 3 would cancel each other. Square root of 2 will remain and 4 plus I have to divide by 3 would be 12. So, this looks like this is a volume of the tetrahedron with all equal edges. That's it. Next, next is consider we have an Egyptian pyramid which has a square as a base and some kind of height. This is my Egyptian pyramid. So, this is the regular pyramid in the sense that there is a regular square actually at the base and the apex is projecting exactly into the center of the pyramid. Alright, so what's known is d is the side of a square and s is the side edge of the pyramid. Now, obviously if you are in front of the Egyptian pyramid, you can very easily measure this and this. And that is sufficient to evaluate the volume. Now, obviously it would be even easier if you know the altitude but you can't really measure altitude directly because it's kind of inside. So, you can measure the side by just actually walking on it and you can measure the side of the square at the base which is also by walking but you can't measure the altitude of the pyramid. But you can calculate it obviously. Again, let's just drop the perpendicular down. It's supposed to fall right in the center of this square. Alright, so that would be something like this. Now, what would be the length of the altitude? Well, that's basically very easy because you have again right triangle. This is s. Well, this is uppercase s, this is lowercase s. Now, square has the side d which makes this which is half of a diagonal. Diagonal is d square plus d square square root, right? d square plus d square and this is the hypotenuse. Now, we have half of it. So, it's divided by 2 which is equal to d square root of 2 over 2. So, that's this particular piece, d square root of 2 over 2. Now, what we have to do is to find the altitude using the Pythagorean theorem which is s square minus, this is d square over 2, right? Square root of 2 is 2, this is 4 and square root of this. So, this is my altitude times d square which is area of the base and divided by 3. So, before any kind of simplification but I don't think we can simplify it even further because this is s and this is d so we can't really do anything about it. So, this is basically the final formula. That's it. All we need it is just to find out what's the altitude. That's the most difficult part which is actually not difficult at all. Okay, now let's do reverse. What if I know the volume of the pyramid? This Egyptian pyramid has the volume v and side of the square of the base is d and what I have to do is I have to find h. But let's just use the previous formula. Now, the previous formula was what? 1 third d square root of s square minus d square over 2, right? So, we just have to resolve it for s. This is pure algebra. So, what happens is 3b divided by d square is a square root, right? Now, we square it and we get s square minus d square over 2 and we have to plus d square over 2 to get s square and we have to do the square root of this and that would give me the s, right? Yeah, seems to be fine. I had the same answer before. All right, that's it. Very simple. Just algebraic resolving of the equation. Next, all right, let's go back to our hexagon at the base and let's consider this is the right hexagon. Now, what's known about this hexagon is the following. It's altitude, which is s o equals o a equals d. So, this is basically, you can consider it a radius of the circle which circumscribes our regular hexagon. So, all these are radiuses, obviously. So, the altitude and this radius, which doesn't look like the regular my picture, but supposedly they're all the same and equal to d. So, question is what's the value? Well, again, we know the altitude, so now all we have to do is to vary the area. What is the area? Area is six equilateral triangles, each of them having d as a side. Now, I have already derived this formula before. Now, obviously I forgot it. So, let me just derive it again. So, the altitude is d square minus d over 2 square root. That's altitude times base divided by 2. So, that's my area. And the altitude is d, so I have to multiply by d and divide by 3. So, that would be my answer for the volume. Right? So, it would be d from here would be d and d square would be d cube. So, minus one quarter is divided by 2 times 2 times 3, which is 8, right? Something like this. Looks like it. Hope I don't make a mistake. All right. Oh, sorry, sorry, sorry. I had to multiply by 6, right? So, I have 6 triangles like that. So, it should be multiplied by 6. So, d square root of 3 over 8. Was it 8? No, wait a minute. 2, 2 and 3. 2, 2 and 3, right? And then I have to multiply by 6. Okay, here it is. And it would be d square root of 3 over 2. That's what it is. That's the right answer. Okay. Yeah, I forgot that we have 6 triangles at the base. All right, next. Next we have a triangular pyramid. Next what we do is we have two midpoints. This midpoint between A and B and this midpoint between A and C. And what I do next, I draw a plane from S and Q. This plane. It divides our pyramid in two halves. One half, well, not two halves, two parts. Right, two pyramids. One has a triangular base, APQ. And another has a quadrilateral PQCB at the base. But the apex is the same and therefore the height is the same. The altitude is the same. Now, what's necessary to do is to find out the ratio between the volumes, this to this. But now let's think about it. If the volume is equal to area of the base times height divided by 3, height is the same and 3 is the same. So if we will divide one volume by another, let's say S1 would be area of this big quadrilateral. And then times H divided by 3. That's the volume of the bigger part. Divided by S2 which is area of APQ triangle times the same H divided by 3. So obviously this thing is cancelling out and the ratio is actually the ratio between two areas. Okay, now that's easy because now we have a plane geometry problem. So this is our triangle ABC and we draw from midpoint to midpoint. Now, what's the ratio between these? Obviously this triangle APQ is similar to ABC. Its base is half and its height is half which means its area is one quarter of the whole triangle. So the area of APQ is equal to one quarter of area of ABC. Well, if this is one quarter, now this is three quarters. So what's the ratio between them? Three quarters divided by one quarter which is three. The ratio is three. And I have the last problem which is actually not very difficult but it's challenging to draw a picture. Let me try. Again we have a triangular period. I'll try to do it as big as I can. So this is a tetrahedral. Now what I'm doing is I'm connecting all the midpoints of all edges. So from this midpoint I connect to this midpoint and to this midpoint. And to this midpoint. And to this midpoint. And this AC midpoint. Now an AC midpoint is somewhere here and I connected this midpoint to this. To this. Then it's invisible. This. And this. And this. And this. Now what happens drawn right now is, well if you can imagine it's two quadrilateral pyramids. One is having this as a apex and it goes that way. Another is that apex and it goes this way. And they are having the same, sharing the same base. So that's how it looks. Now my point is what's the ratio between the volume of this. It's actually octahedron because it has one, two, three, four on this side and one, two, three, four on that side faces. So it's octahedron. It's like this approximately. I should use four fingers rather than five, something like this. And these are, and these are epises. This is one, this is another. So the question is what kind of a volume is occupied by this inside octahedron relative to the whole tetrahedron. Well, let's just think about it. These are all lines which connect the midpoints, right? So basically I can say that pyramid A prime, B prime, C prime. S A prime, B prime, C prime takes what part of the volume of the total pyramid? Well, it's a scaling, right? Because if you will scale it with this center and the ratio two, you would get from this point to this because this is twice as long, twice as long. From B prime you go to B with a scaling factor of two and from C prime to C. So this is a true similarity, 3D similarity, three-dimensional similarity between S A prime, B prime, C prime and S A, B, C. And the ratio is two, which means that the volume ratio equals 1, 8 of B where B is the volume of the original period. Right? Two to the third degree. If you remember when we were talking about similarity, we were talking about the 3D similarity. If there is a scaling, then all linear dimensions are multiplied by the scaling factor, but the volume is multiplied by scaling factor to the power of three. I do suggest you to refer me to that lecture about 3D similarity. It's a previous topic actually. All right. Now, so let's just cut it off, this one-eighths of my volume. Now, in a similar fashion, we can consider a pyramid with the apex A and base A prime, B prime. Again, this particular pyramid A, A prime, B prime, D with A as an apex is similar in three-dimensional sense to our pyramid, but considered as A being an apex and S, C, B being the base. So we are cutting this piece and it's also one-eighths for the same reason because the scaling factor is two. So basically what I'm saying is that for each side, we have to actually cut the piece of the main pyramid which is equal to one-eighths. So we cut all these four corners from S, from A, from B, and from C. And what's left is actually our octahedron, right? You just have to properly imagine it to yourself. So the pyramid, actually, now the second one is not A, A prime, B prime, D, it's A, A prime. We are cutting parallel to this. It's this. Call it E. We are cutting with apex A and the base A prime, E and D. Yes, that one. A prime, E, D is the pyramid which we are cutting from the point A. And then from the point B, considered as an apex again, we are cutting this triangle which is G, B prime, F. And from the point C, we are cutting C prime, E, F. That's how it is. So by cutting these four corners, we actually have remained our octahedron. So it's four times, each one of them is one-eighths, so it's four times times one-eighths, so it's V divided by two. So we have to cut half of the volume basically to get to the octahedron. So octahedron occupies half of the tetrahedron in this case. And again, it's kind of a simple but you have to really properly position your viewing point whenever you are cutting the edges. Well, that was my last problem. I hope I didn't make a mistake here. Okay, so thanks very much. I think it's very useful exercises. And I do suggest actually to do the same maybe calculations a couple of times. The problems are listed on the website as notes for this particular lecture. So basically that's it for today. Thank you very much and good luck.