 Hi, I'm Zor. Welcome to Unisor Education. Today we will solve some problems about prisms. Obviously, this is part of the Advanced Mathematics course on Unisor.com. The problems are actually described on the corresponding webpage as notes to this lecture. And solutions I'm going to discuss right now. So it's not really a lecture, it's kind of a problem-solving exercise. Alright, so let's just go. The problem number one. You have a cube which is just a particular case of a prism. So this is our cube. What's known about this cube is the length of its main diagonal from A to C prime. Which is actually the same as from B to D prime. Doesn't really matter. So let me put it this way. This is my main diagonal. So this is what's known about this cube. It's equal to D. What I have to find out is its volume and the total surface area. Well, obviously, if I know the side of the cube, then I can always find the area of each face. Now there are six faces. So it's time six. So that's our total surface area. And the volume is equal to area of the base, let's say A, B, C, D, which is also known if I know the sides of this square, times the altitude, which is just another edge on this prism, which is again the same thing as every other edge. So everything goes into knowing the edge. Now we don't know the edge. We don't know this A, B or B, C or whatever. But we do know the main diagonal. And now let's resort to what is the main diagonal of a right rectangular prism in terms of its three sides, which are going from the same point. Let's say this is A, this is B and this is C. Well, for a cube, they're all the same. But let's just consider it's a general right rectangular prism. Now in terms of these, we can approach this problem. Actually, we did solve this problem before, but I'll just remind the solution. If I will connect B and D, I will consider first I would like to find out what's the BD. And BD is just diagonal in this rectangle. So BD square is equal to A square plus B square. Now if you consider a right triangle BDD prime, well, it's right because DD prime is perpendicular to DD, obviously. So in this right triangle, the square of the diagonal BD prime is equal to BD prime, BD square plus DD prime, which is C square. And substituting BD square with this, you will see this. So my main diagonal square is equal to sum of the squares of three sides, three edges, which are originated at the same point. Now back to the cube, D square is equal to, this is A, this is equal to A, and this is equal to A because all edges are the same in the cube. So it's 3A square from which A is equal to D divided by square root of 3, right? So now knowing the diagonal, the main diagonal, we have determined the edge. Now knowing the edge, we can very easily find the volume and the square and the surface of the prism. So the volume is equal to area of the base, which is A square, obviously, right? Times the height, which is D, sorry, which is A, which is A cube. So the volume of the cube, as we know, is equal to a cube of its edge. And now I know what A is in terms of D. So it's D cube, square root of 3 in cube, it's 3 square root of 3 and divided by 3 cube 27, right? Which is D cube square root of 3 divided by 9, am I right? D cube 3, yeah, it's about right. Now the total surface area of the cube is equal to 6 areas of each base. They're all the same and they're A square, which means it's 6 times D square times one-third, right? Which is 2D square. That's total area. Remember this, total area of the cube is equal to double square of the main diagonal. We will refer to this a little later in another problem. So that's it for this problem. Now, next problem. Let's consider a hexagonal prism, which is number one, right? Which means all the side edges are perpendicular to the base. And it's regular, which means that the hexagonal in the base is the regular six-sided polygon. So let me, well, now I'm challenged by this picture. So let's consider this is the plane where the upper base is and this is the plane where the bottom is. Now here we have, now, yes, my graphical abilities are not that good. How about this? Something like this. And let's connect it. No. All wrong. Sorry about that. Let me do it without these planes. So this will be my first and this will be my second. Something like this. And now let's connect it. Does it look like a hexagonal? Yes. Let's consider. Now, right? Because all the edges are perpendicular and regular hexagonal because this is a regular six-sided polygon. Now, what I do know about this is I know that all edges are equal to D. This is G, although it doesn't look like they're equal to each other in my picture, but they are. So question is again, we have to determine the volume and total surface. Okay, now let's start with the area of the base because we need it to calculate the volume, right? Now the base is the right hexagonal. Now, this is easier. And I know that all sides are equal to D. Now, since it's hexagonal, the whole 360 degrees divided in six parts, which means that every angle is 60 degrees and obviously we have the equilateral triangle, so these are all D as well. Now, the area is six times the area of a triangle. Six times S of triangle. Now, this is a base. Now, what is the area of triangle? Well, if you have a triangle, equilateral triangle with a side D, then D, well, this is obviously 30 degrees and this is 30 degrees, this is 60 and this is 60. So this is the right triangle. If this is 30 degrees, this is D, this is D. So this is D over 2 and this is D square root of 3 over 2, right? So the area is equal to D times D square root of 3 divided by 2 and times 1 half because it's a triangle, right? So it would be square root of 3 divided by 4 times D square. So that's the area of this triangle. So this one, therefore, is equal to 6 times greater, so it's 3 seconds square. So that's my base. Great. My altitude is D, which means that the volume is equal to 3 square root of 3 D cube divided by 2. I think it's right. Okay, so that's the volume. How about the total surface area? Well, total surface area is two bases, the bottom and the top, which is this times 2, which is 3 square root of 3 D square, plus six sides. Now, each side is actually a rectangle D by D, which is a square. So it's 6 G square. Or if you wish, you can factor out 3 G square root of 3 plus 2. So that's the formula for the total surface. That's it. Next problem. Okay, we have a right rectangular prism. Let's say A, B and C. Right rectangular prism with any lengths of the edges. Now, let's say this is main diagonal and its length is G. So my problem is I have to prove that D square is greater or equal than one half of the total surface of the prism. Now, first of all, in terms of A, B and C, we have just calculated in the previous problem that D square is equal to A square plus B square plus C square. So that's what we can substitute here. Now, how is area expressed in terms of A, B and C? Well, two bases, each one has A times B, right? So it's 2AB. That's the area of the two bases. Now, left and right are BC, right? So it's 2BC. Now front and back are AC. So what I have to do is I have to prove that double this or half of this is D square, which means this is greater than half of this or double of this is greater than this. Let's put it this way. 2 times A square plus B square plus C square is greater than equal than 2AB plus 2BC plus 2AC. Now, is that right? Well, let's rewrite this differently. A square plus B square plus B square plus C square plus A square plus C square. Now, that's the same thing, right? 2A square, 2B square and 2C square, that's the same thing. Now this and this. Obviously, we all know that A square plus B square is greater than equal to 2AB, why? Well, because if you will do it differently, this is an obvious inequality because on the left you have A minus B square. Square is always non-negative. So A square plus B square is greater than 2AB, B square plus C square is greater than 2BC So each member is greater than corresponding and that's why the whole inequality is proven. Now, when is the equality? When the equality comes to this being equal all the time, which means A should be equal to B, B should be equal to C and A should be equal to C, which means it should be a cube. And if you remember, just in one of the previous problems, we had an equality that the square of a diagonal is equal to double total surface. I told you to pay attention to this. That's exactly what happens. So this is a general inequality and the equality is reached when it's a cube. Okay, next we will have a couple of problems with prism which is not right prism. I mean we all kind of used to think about prism as right prism. But if you remember, the definition of a prism doesn't really imply it's being right. So the age is not necessarily perpendicular, I mean the side age is not necessarily perpendicular to the base. So these two problems which I'm going to talk about, they are related to this case when it's a slanted prism. So let's consider a slanted triangular prism. And what's interesting about this prism is that it's not slanted enough. This would be slanted enough. And this is my side ages. Alright. So it's slanted. But however what's important is the following. Now I actually draw a triangular prism. It doesn't really matter. Whatever I'm going to talk about is related to any prism. So let's assume that I have cut the prism by a perpendicular section to the side. So it would be something like this. So now I know that this is perpendicular. This is not. Although it looks like a perpendicular but it's not. I intended to do it in a slated fashion. You know what? Let me just do it a little bit more slanted. A little bit even better. Something like this. And here I will do the same. Maybe this looks like more like slanted. So what I'm going right now to do is I'm cutting the prism with a plane perpendicular to the side age. Well, to all side ages, obviously. Because they're all parallel to each other. Now, the point which I would like to make is that the side area is equal to, which means area of all side faces, bases we are not including. It's equal to the side age, which is A A prime, times P, which is a perimeter of whatever the polygon comes when I cut the whole prism with this plane. So this section is some kind of a polygon because we are in the plane, which is cutting perpendicularly to edges. So whatever the perimeter of that thing is would be this way. Now, let's just put some letters. In this particular case I will put M and P, let's say. Now, it's actually a very simple problem. I mean, I'm talking about the conditions for a long period of time, but the problem is actually very simple because we know that every face, every side face of this prism is parallelogram. Now, the side is, the face is a parallelogram and its area is equal to one side times the altitude to this side. Now, M and N, since the whole plane, which is cutting my prism, is perpendicular to all side edges. So M and N is perpendicular to A A prime, which means that the area A B B prime A is equal to A A prime times the altitude M N. Similarly, the next face B C C prime B prime is equal to what? It's also a parallelogram, right? Now, it's B B prime times altitude towards it, which is P N, since P N is perpendicular to B B prime. Finally, A C C prime A prime is equal to C C prime times M P. Now, if you will add these together, what will you have on the left? Sum of the areas of all side faces, which is basically as a side. What do you have on the right? Well, A A prime, B B prime and C C prime are all equal exactly the same. So that's why I can put A A prime. This is also A A prime and this is A A prime because all side edges are the same in the prism. So I can factor it out and what's inside of the parenthesis? M N plus P N plus M P. M N plus P N plus M P, which is perimeter of this, in this case triangle, but in general case it can be prism with any number of vertices on the base. That's it. This is our first problem with a slanted prism. And here is another problem, the last one for this lecture. So again you have a slanted, in this case I'm specific, this is a triangular prism and it's slanted. Now, what's important about this prism is the following. First of all, base is a socialist triangle. These are equal. Obviously both bases because they are congruent to each other. What else is important is that this particular H, A A prime, has the same angle with both this and this. So it's kind of equally slanted towards both sides of this isosceles triangle. So the whole prism, if you consider first the right prism and then you slanted, so you're slanted towards middle basically between these two sides of the isosceles triangle. So that's how these two angles are equal to each other. So angle A prime A B equals angle A prime A C. A prime A B is equal to A prime A C. Now, what you have to prove is the following. That A A prime is perpendicular to B C, which again seems to be intuitively obvious because again, if this is your prism and these are two equal sides of the isosceles triangle and you are slanting it this way. So basically this used to be vertical, now it's tilted forward but still it will be perpendicular to the opposite side. So that's what we have to prove. Okay, how can we do it? Well, and then there is another little side theorem about this but I will mention it later on. Here is what I'm suggesting to do. First of all, let's draw a perpendicular from A prime to A B within this front face. So from A perpendicular to A B within the A A prime B. That would look like this. Yeah, not very straight. It would be nice if it's also a straight line. Let me try again. Okay, now let's drop a perpendicular from A prime to the plane A B C. It's somewhere here. You're dropping the perpendicular. Okay, so let's say this is P and this is Q and connect them. Now what can we say about this? So Q is a projection from A prime down. It's a perpendicular to the plane. Now A prime P is perpendicular to A B. What I would like to say is the following. Since A prime P is perpendicular to A B. I'm stating the following property. PQ, the projection of A prime P, right? Since A prime is projected to Q then the whole A prime P is projected to A prime Q because P is projecting to itself on the plane. So this line A prime P is perpendicular to this line. Then I'm stating that this projection which is PQ also is perpendicular to A B. So A A prime P is perpendicular to A B by construction. Now I would like to prove that PQ is also perpendicular. Now how can it be proven? Well, actually it's very simple. Consider a plane A prime PQ and the line A B. Now line A B is perpendicular to one line which is A prime P because that's how we constructed it, right? But it's also perpendicular to A prime Q because A prime Q is perpendicular to the triangle at the base and therefore it's perpendicular to any line on the base, right? So it's perpendicular to A B. P is perpendicular to both A prime P and A prime Q which means it's perpendicular to an entire plane A prime PQ and therefore to every line on this plane including PQ. So as a consequence we see that PQ is perpendicular to A B. Now if I will do exactly the same towards AC which means first I will put a perpendicular to AC. Let's call it M and connect it with Q. Similarly again A prime M is perpendicular to AC by construction and as a consequence I see that MQ is perpendicular to AC. So again if the line is perpendicular to a line then its projection is also perpendicular to the same one. Now let's talk about triangles. Now A A prime P is a right triangle, right? Because this is perpendicular and this is an angle which we will consider. Now A A prime M on that side of the prism on the back side is also a right triangle with a Q-tangle A prime AM and since these angles are equal to each other A prime AP and A prime AM they are equal to each other. Therefore the triangles are congruent, right? We have a hypotenuse and an acute angle in the right triangle that's sufficient to congruence of the triangles which means the length of A prime P and A prime M is exactly the same. Now we go into another pair of triangles A prime MQ and A prime PQ. So we have two right triangles, hypotenuse A prime M and A prime P are the same and casualties A prime Q is common, they share it which means triangles are the same, which means these two are the same. So MQ is equal to PQ. Finally I connect A and Q and I will consider everything in the area, in the plane of ABC. So I have ABC, I have Q in the middle and I have two perpendicular from Q, M and P and they are equal to each other. Obviously the triangles AP, Q and AMQ are congruent because they share hypotenuse and one is equal to another. So triangles are congruent which means these two angles are the same which means Q is on the angle bisector of this triangle ABC, angle bisector. And we know that in a socialist triangle, angle bisector on the top is actually the same as median and is the same as altitude of this triangle which means that this line which is basically a projection of AA prime is perpendicular, AQ is perpendicular to opposite side to BC. This line is perpendicular, this is right angles, which is actually what we wanted to prove. And now since we have proven that AA prime is perpendicular to BC we can also say that AA prime is parallel to BB prime and to CC prime, right? Which means these two lines also perpendicular to BC. Which means that BC, C prime, B prime is not just a parallelogram but it's a rectangle since these angles are perpendicular. These lines are perpendicular, angles are right angles, 90 degrees. That's the end of it. So not only this line is perpendicular to the opposite side on the base but also one particular side face remains a rectangle as it was if it was not slanted. So whenever we are slanting right in the middle of this isosceles triangle this particular face remains rectangular. Well, that's it for today. Thank you very much. What I suggest you to do right now is, and that's a very, very important exercise, write down a proof in every case of these five problems especially the last one. It's a little bit more involved. Well, actually if you do it only for the last one that's also very useful because when you are writing your thoughts using whatever the symbology, letters, signs of mathematics, etc. it actually contributes to develop the clarity and the rigid logic whatever logic you're using in this case. So I do suggest you to just write it down in a piece of paper and if you want you can send it to me and if I consider it really to be the really good explanation I'll just put it as a solution offered by somebody and somebody I'll put it on the website. Basically that's it. Thank you very much and good luck.