Surface area of an octagonal pyramid

0:00- [Instructor] We're told the base of the following right pyramid is a regular octagon, and we see that in the base right over here. They want us to figure out what is the lateral surface area of the pyramid and what is the total surface area of the pyramid. So, why don't you go ahead, pause this video, have a go at that and I'll warn you, it is a little bit hairy.

0:18All right, now let's do this together. So first, let's think about the lateral surface area. So, if we could figure out the surface area of each of these faces right over here, there's 8 of them. So, that we could just multiply by 8. But how do we do that? Well, we know the base of each of those faces.

0:34It is 6 meters. So, we need to figure out the height of each of those bases. So, this is the height over here. I'm trying to hand draw it, not so well. All right, so we need to figure out what that height is of each of the faces and I'll call that H. Not to be confused with the height of the pyramid, which they actually gave us.

0:54This is the height of each of those faces. So, how do we figure that out? Well, you might imagine there's a little bit of trigonometry involved. If I knew what this length was, then I could use the Pythagorean theorem to figure out our H there. Now, how could I figure out this length that I just drew?

1:15Maybe I'll do it in a different color. I'll do it in a deep blue right here. How can I figure out this blue length? Let me call that X. Well, I could imagine each of these triangles, each of these 8 triangles that make up that regular octagon. So, lemme just draw it right over here.

1:35So, this is a triangle that's making up that regular octagon. I want to figure out the height of this triangle that is going to be X. I know the base of this triangle is 6 meters. I also know some of the angles of this triangle. How do I figure that out? Well, if I were to go all the way around this octagon, it would be 360 degrees.

2:00But this total angle right over here is 360 divided by 8. So, if I take 8 into 360, 8 goes into 36 4 times, 4 times 8 is 32, subtract 36 minus 32 is 48 goes into 40 5 times, 5 times 8 is 40 and I'm done. So, this whole thing would be 45 degrees and this part right over...

2:28Lemme do this in a different color. This part right over here in red, that is going to be 45 over 2 degrees. This is a regular octagon. Each of these triangles are going to be isosceles triangles. If I were to drop this perpendicular like that, it directly splits the angle.

2:47It also splits this side into two. So now, I can use a little bit of trigonometry. If I think about this angle, I know the opposite side here, it's going to be 3. This is a right triangle. So, what involves opposite and adjacent? Well, tangent involves opposite and adjacent.

3:08So, if I said the tangent of 45 over 2, that's the same thing as 22.5 degrees, is going to be equal to the opposite side, 3 over X. Well, if I wanted to solve for X, I could multiply both sides by X and divide both sides by tangent of 22.5. And so I would get X is equal to 3 over the tangent of 22.5 degrees and I'll leave it like that for now.

3:41We have figured out X, which I said I would do it in blue, but I somehow switched into red right over there. Well, we know what X is, we know what the height of this pyramid is. So, we can use a Pythagorean theorem to figure out our H. We could write that H squared, H squared, is going to be equal to 9 squared.

4:02So, that is going to be equal to 81 plus this business over here squared, plus 9 over tangent of 22.5 squared. All I did is I squared this, I said this squared plus that squared is going to equal that squared. Or I could say that our height is going to be equal to the square root, this is gonna be a little hairy, 81 plus 9 over the tangent of 22.5 degrees squared.

4:35Now, why was that useful? Well, if I know the base and the height and I multiply that times one half, I get the area of one of these triangles, that's part of the lateral surface area and if I multiply that by 8 then I'll get all of them. So, let's do that. So, the lateral surface area, right over there, lateral surface area, is going to be equal to, I have 8 of these triangles, and then each of their areas is one half times the base.

5:02So, times 6 meters, times this height, that's the hairy part, times the square root of 81 plus 9 over the tangent of 22.5 squared. That is the lateral surface area. Now, if I wanted to find the total surface area, what would I do? Well, I'd have to add the area of this octagon right over there.

5:30And how do I figure that out? Well, I know the area or I can figure out the area of each of these triangles. I know its base and I know its height. Each of these triangles is going to have an area, so lemme just write, this is going to be, I'll just rewrite it over here, it's going to be 8 times one half times 6, so that's 4 times 6.

5:50So, I'll just write, it's 24 times the square root of 81 plus 9 over tangent of 22.5 degrees squared plus, plus, we're going to get 8 of these 8 times, the area of each of them is one half times the base, which is 6, times the height, which we already figured out was 3 over tangent of 22.5 degrees.

6:26So, 8 times one half times 6 is 24, and then you multiply that times, lemme make sure I'm doing that, 8 times one half is 4 times 6 is 24 times 3, I get 72. So, this part over here and I'm squinching it up, is going to be 72 over the tangent of 22.5 degrees. If you were to enter this expression for lateral surface area into your calculator and round to the nearest square meter, this is going to be approximately 277 square meters.

7:02And then if you were to take this total, that has both the lateral and the area of the base over here, and put that into your calculator, and I'm not doing it 'cause it's just gonna take so much time, but I think you know how to enter into a calculator. This is going to be approximately 451 square meters.