 Our last set of videos for chapter 12 cover two sections, the surface area of pyramids and the surface area of cones. We're going to start with pyramids. If we just look at these three different pyramids, just a reminder of how we would name these. The green one has a triangular base. We would call this a triangular pyramid. This has a rectangle base. We would call that a rectangular pyramid. And this has a pentagonal base, so we would call that a pentagonal pyramid. It's named by the base. We're going to start off by identifying some parts in our regular pyramid. Remember, regular means all sides of the base are equal, and that's going to come in handy as we derive the formula for a pyramid. So for number one, if we're identifying the base, the base in this figure is a square because we're told it's a regular pyramid, so all sides are equal. And we would name that square, that's the symbol for square, and then each of those vertices BCDF. It doesn't matter the order. Number two, the lateral face. This pyramid has four lateral faces because each of these triangles that aren't part of the base are going to be a lateral face, and we don't need to name all of them. Let's just look at this one right here that I've highlighted, and we would call one lateral face CAD, but there's actually going to be four of them for this pyramid, and in fact you'll have the number of lateral faces will coincide with the sides of the base. Lateral faces for a pyramid are always going to be triangles because they're always going to come up to that apex point. And then the base edges, we said the base was a square, so a square has four edges, and we would just name each of those, again doesn't matter what the order is, and we have four segments that represent the base edges. The next set, we start with lateral edges. Lateral edges, again, are the edges that create those lateral faces, and because there's four sides on my base, I'm going to have four lateral edges, AF, AD, AC, and AB. You'll notice each of them include A because each of those edges come up to a point there. And then we come to a familiar term, the apathome of the base. Whenever we're using pyramids, we're going to probably be talking about the apathome. And if we're looking at that base, which is a square, the apathome of that square, remember, goes from the center point out to the middle of one of the base edges, not to the vertices but to the middle. So the apathome of the base is GE. We are going to have to use that to calculate some surface area problems. And then we get to the height or the altitude of the base versus the slant height. These two are often switched around, so we want to make sure that you can identify them correctly. Whenever we're talking about height or altitude, those are interchangeable. That's going to come from the apex of my pyramid straight down, and it's going to make the perpendicular to my base right there. It's the vertical distance of the height of my base, and in this case, that's going to be AG. And the slant height is going to be coming from the apex A and then going down the middle of one of the lateral faces. It's easy to get those mixed up. It's not going to be on an edge. It's going to go down the middle of one of those lateral faces. So my slant height here is AE. Don't get that confused with AD or AF. Now we're asked to draw a net of the regular pyramid, and this is the last net that we'll draw for these videos, and then we're going to derive the formula. Like before, I'm going to show you the finished net and then walk you through it. So if you want to pause the video and try to draw that yourself, go ahead. Otherwise, I'm going to show you what that net looks like, and then we're going to walk through it together. Whenever you're drawing nets, I would recommend that you start with the base. We know that the base is a square because it's a regular pyramid. We're told some values here, CD equals 6 and AE equals 5. AE remember was the lateral or the slant height of the pyramid that equals 5, and that's all the information we need for us to draw this netting with the correct dimensions. So if I start with that base, which is a square of 6, right in the middle of your grid paper, you can draw a square that's 6 by 6 by 6 by 6. So I can label those edges 6, and that's a good place to start, and then we can add the lateral faces. Remember we have four lateral faces in the shape of triangles, and they're going to have their base of the triangle coincide with one of the edges of my square base. So I'm going to start with that value of 6, and then what we know on this pyramid, if the slant height is 5, that represents this distance right here. So you can go out 5 and then just connect the dots to the vertices of that base edge. And you can do that all the way around. Go out a perpendicular distance of 5, and then just connect around. We have four triangles that are lateral faces, and I can label those LF all the way around, label all of your lateral faces, and then of course the square is going to be my one base. And so I'm going to use this information to answer the questions. I'm just going to have to do a little bit of math as I go, but let's go ahead and answer the questions to get to the formula for a pyramid. We're first asked to find GE, and remember if this value is 6 for my base edge all the way around, GE is the center point of that base, and so half of 6 that represents, again, the apathome of the base is going to be 3. You're just counting on the grid paper there, that half of that base edge is 3. And then AG, we are going to have to do a little bit of math. If we look on our figure up here, AG represents the height or the altitude of that pyramid. Remember to look for right triangles when we're calculating pieces of our solids here. If I'm trying to find AG, I can make this right triangle here, because I just figured that this apathome was 3, and I'm told the slant height is 5, and so I have that little right triangle here. If I'm trying to find the altitude AG, I can just do some quick calculations. 3 squared plus x squared equals 5 squared. And when I solve that, I'm going to get x equals 4. That's a Pythagorean triple, 3, 4, 5. So AG, the height of my pyramid is 4. And then the last value we're trying to find is AF, and that represents one of the, um, one of the lateral edges of my pyramid. In this one, I'm going to create a right triangle again, but this one is a little trickier, so make sure you understand all of the right triangles available in these pyramids. So, trying to find AF here, I can create a right triangle on this lateral face. That's kind of slanty. And if I pull that out, that's actually AEF. And I know that AE represents the slant height of my pyramid, and so that's going to be 5. EF is half the length of an edge, and that's going to be 3. And now I'm using the same numbers, but they're just mixed around a little bit to find x. And this one I will calculate out to show you that I get x squared equals 34. And so when I take the square root of both of those, I have AF equals root 34. So you have to be careful of what right triangles you're forming when you're doing these calculations. Then the next couple questions are asking me first about the perimeter of the base. We know that each of these base edges are 6. There's 4 edges, so 6 times 4 is going to be 24. The perimeter is just going to be in units because it is not area. And now we're asked to find the lateral area of my pyramid. And remember the lateral area is adding up all the lateral faces, everything that is not a base. So I'm going to add up the area of each of those four triangles. I only have to find the area of one of them, and then I'm just going to multiply that by four. And so in order to find the area of one of the triangles, area of a triangle, one half base times height, the base and the height of a triangle when I'm finding the area are always going to be perpendicular to each other. We know that this is 6, and this distance we know is represented by the slant height of the pyramid. We were told is 5. So the area of one of those triangles is going to be one half times 6 times 5, which is going to be 15. That's the area of one triangle. And so the total lateral area is just going to be multiplied by the four lateral faces. And we have lateral area of sixty because it's area units squared. The next two now are going to get us into actually deriving the formula for the pyramid. And so when we talked about that lateral area we multiplied half base times height four times around here. And when we do the half base times height four times, the height remember was considered the slant height. And half of the base when we do that four times ends up being half of the perimeter no matter how many sides that polygon has on the base. And then we're going to multiply at times the slant height. This is represented by the lateral area for any pyramid. This is going to give you the total So regardless of how many lateral triangles, lateral face triangles you have, if it's a four-sided base or a fifty-five-sided base this will give you the total lateral area of any pyramid. And then part H this should actually read on your note sheet, calculate the total surface area for the pyramid. And we know that in any surface area for a figure it's going to be the lateral area plus the area of the bases. We know that the lateral area we just calculated was sixty. We just have to calculate the area of the base because this is a square we already figured out that the area of that is just going to be six times six, thirty-six and so the total area of this pyramid is ninety-six units squared. So I'm going to show you what our final formula will look like for any pyramid and this is where you should write this down on your formula sheet or in your formula flipbook. The surface area of a pyramid then is going to be one-half the perimeter times the slant height plus the area of the base since there's one base. And the lateral area is represented by just the one-half PL without the base. Make sure you know the difference between slant height and altitude because those are easy to get mixed up and we're going to use these in our next set of practice problems.