 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue solving problems about pyramids. Well, today's problem would be a slightly different character. It's basically lots of calculations. And well, you know, sometimes it's boring. It's definitely boring for me to tell you the truth. But you know what, it's necessary. Sometimes you have to do the boring task just to achieve some results. Sometimes you have certain like spark of creativity where you can significantly reduce your calculations based on some little trick. And that was basically a subject of the previous lecture. One little trick allows us to reduce the calculations quite significantly. In this case, I just don't know that trick or whatever. So I'll just do it straightforward way. And well, my only concern is not to make mistakes. All right, so let's do it. So what's the problem? The problem is as follows. You have a triangular pyramid with equal side edges and equal angles of the faces at the top, all right angles. S, A, B, C. So S, A equals S, B equals S, C. Angle A, S, B equals angle B, S, C equals angle C. S, A equals 90 degrees. What else is known? Known altitude. Altitude A, H, sorry, S, H equals H. That's the only thing which is known as far as lengths are concerned. So again, all side edges are equal. All angles at the top for all faces, side faces are equal to 90 degrees. And we have to define everything. Basically, we have to determine everything about this pyramid, which means every edge, every area, and volume. So all edges, all areas of all faces and the volume are everything in terms of H. Well, that's why I'm saying it's kind of lots of calculations. All right, so what can we do about it? Well, let's just think about it. First of all, all these triangles, A, S, B, B, S, C, and C, S, A, R, all these right triangles with side edges being catative of these right triangles. And they are the same since all catatives are the same, all side edges are the same. So at the base, A, B, C, we have an equilateral triangle. Equilateral triangle, okay. So I think what would be easier, if we can start with just assigning some kind of an A for the side edge, and we will try to express H, the altitude, in terms of A, and that would be kind of an equation from which we will define A in terms of H. I think it's a little easier. At least it was easier for me. It may be done differently anyway, but here is my approach. Now, if this is A, and this is A, and this is A, this is obviously A square root of 2, according to the Pythagorean Pyramid, right? A, S, B is the right triangle, A and A are catative, angle at the top is right angle. So this is A square plus A square equals to this square, which means it's A square root of 2. That's fine. Now what I will do is, if I have an equilateral triangle with a known side, well, I assume it's a known side, equals to A square root of 2. Obviously, I can find out the A H. And then from the S-A-H right triangle, knowing k-partners and the characters A H, I will determine what my altitude is, and I will equate it to H, and that would be my equation. So all I have to do right now is, let's consider the equilateral triangle, ABC. Now, obviously, the point of intersection of all the medians and altitudes and bisectors, which is the center of this triangle, is exactly the point where altitude going into. Why? Well, first of all, you can do it very, very easy. For instance, it's very easy to prove that A H is equal to H C and equal to H B, because all these triangles are right triangles with equal k-partners and shared characters, right? So other characters are supposed to be equal. So S-A-H, S-B-H, and S-C-H are all right triangles. They share one character, characters, and their k-partners are the same. So they're all the same, which means H is the center of the circumference around ABC. And since it's an equilateral triangle, then this point is everything, the center of everything, including inscribed, circumscribed circle. Medians are intersecting in this point, altitude, etc., etc. So let's just determine this is H. So let's determine A H, that would be our key segment. Well, again, that's very easy. Now, the whole thing is, as we know, A square root of 2, so half of it, let's call it C prime, so it would be A square root of 2 over 2, and this is A square root of 2 over 2, right? These are halves. Now, what else I know? Well, I know that if this is X, this is X divided by 2, right? Because this is a 30-degree angle, remember? We're talking about equilateral triangle. So in the equilateral triangle with an angle of 30 degrees, the k-partners is twice as big as opposite cartridges. And they all supposed to satisfy the Pythagorean theory, which means X square is equal to X over 2 square plus A square divided by 2, right? So that's an equation for X. So let me just define what the X is. X square divided by 4 plus A square divided by 2. X goes here, so it's 3 quarters of X square is equal to A square over 2. Now, this is 2. This is 1. I've reduced by 1 by 2, sorry. So 3 halves X square is equal to A square. X is equal to A square root of 2 thirds, right? That's my X. Now, this X is actually A H, which is this. So I know this calculus in terms of A and this calculus in terms of A. And now I can use the Pythagorean theorem to assign it to H. So A square is equal to H square plus this square, 2A square divided by 3. Am I right? So 3A square equals 3H square plus 2A square. So I reduced by 2A square on both sides. I have 1. So H is equal to A divided by square root of 3. Am I right? Well, actually, I'm sorry. I have to define A. So A is equal to H square root of 3, right? I don't need to define H in terms of A. I have to define A in terms of H. Okay, so we got this. So all the side edges, side edges equals to H square root of 3. Done that. Next. Next is the base edges. This triangle. Now we know that this triangle has the side equal to A square root of 2. So the base H equals to, this is A and times square root of 2. So it would be H square root of 6. All right? Now the areas. Okay. Area of the side face is easy because it's A times A divided by 2. It's a right triangle, right? So it's, okay, area side face equals. So it's this square divided by 2, which is H square, 3H square divided by 2, right? So it's H square root of 3 divided by 2. Square is equal to 3, 2H square. Am I right? Yes. Now this area, area of the base. Area of the base. It's just slightly more complicated, but anyway it's not very difficult at all. So what do we know about this? We know all the sides, which is A square root of 2, which is base H. So let me just redraw it, and I put something which we already know. ABC equal at root triangle, and every one of them is square root of 6. Okay? Now what's its area? All right, do the same thing. What's the altitude? Let's call it Y. We know that this is equal, this piece, HAC prime is half of the AAC, right? Because this is such a degree angle. So we have this one equal to H square root of 6 divided by 2. And now Y square is equal to, this is a Goren's theorem, 6H square minus square this, which is 6H square divided by 4, which is equal to, what is this, 18? Well, let's just do it three seconds, that would be easier. So 2Y square is equal to 12H square minus 3H square. Y square is equal to 9H square over 2. So Y is equal to 3H, sorry, 3H divided by square root of 2. Now I have to multiply the base, which is H square root of 6 times 3H divided by square root of 2 and divided by 2, right? So what do we have here? 6 is 2 times 3, square root of 2 would be reduced, so it's 3 square root of 3H square over 2, am I right? Let me check. 3H square, yeah, that's exactly what I have received. So this is my area of the base. 3 square root of 3H square over 2. Okay, so we have three, no, four. We have four faces, area we have determined. All there is now left is volume. Well, volume is easy because right now we know the area of the base and we know the altitude. So we multiply area of the base, which is 3 square root of 3H square over 2, multiplied by altitude and divided by 3. So that would be H cube square root of 3 divided by 2. That's my volume. So that's it. Lots of calculations and I admit it's a little, maybe tedious, boring, whatever other words you can come up with. But let me repeat it again. Sometimes you've got to do it and if you've got to do it, you've got to do it, that's it. So I do suggest you to repeat all these calculations yourself. No matter how you might dislike it. Let me tell you personally, I don't like these calculations. However, again, you must do it. So suggest you to do it again just by yourself and that would be it for today. Thank you very much and good luck.