WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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For the next lesson, we are going to go over surface area of pyramids and cones.
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First, let's talk about pyramids: this is a picture of a pyramid; we have a vertex right here;
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all of these lateral edges meet this vertex; all of the faces touch at the vertex.
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The base is the only side, or only face, that is not going to intersect at the vertex.
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It is going to connect all of the lateral faces together; so all of these sides here are all of the lateral faces, except the base,
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just like all of the other solids that we went over; we have lateral faces, and then we have bases.
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Pyramids only have one base; lateral faces always form triangles.
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Because we only have one base, and all of the lateral faces meet at the vertex, they are all going to form triangles.
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Lateral edges have the vertex as an endpoint; so again, one of the endpoints of each lateral edge is going to be the vertex.
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The altitude is the segment from the vertex perpendicular to the base.
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So, the segment from the vertex to the base so that it is perpendicular is the altitude.
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A **regular pyramid** is when the base is a regular polygon; so this is going to be a square,
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because a rectangle that is equilateral and equiangular is going to be a square.
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So, the base is regular; the endpoints of the altitude are the vertex and the center of the base; that is the altitude--only in a regular polygon.
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All of the lateral faces are congruent isosceles triangles, because it is regular; from the vertex,
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it is going to be the altitude, and then we have each of these sides of the polygon, of the base, being congruent.
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And that is going to make all of these triangles (because, remember, all of the lateral faces are triangles) isosceles triangles.
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So, this and this are going to be the same; from this triangle right here, each one of them is going to be isosceles.
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The **slant height** is the height of each lateral face; the slant height is not the same as the height of the solid.
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Here, this height, from the vertex down to the center--we call that altitude.
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The slant height is the height of the lateral face, the polygon; so if the lateral face is like this, this is the slant height right here.
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Remember: it is not going to the center of the pyramid; the slant height is to there--this would be the slant height.
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See how, when you just look at the triangle itself, the polygon, the face, it is the height, like this; it is the height of that polygon.
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But when you look at it as a solid, the whole thing as a pyramid, it is slanted;
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so be careful there to distinguish between the slant height and the height of the pyramid.
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To find the lateral area of a pyramid, it is going to be 1/2 times the perimeter of the base, times the slant height--
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meaning not the altitude, not this height; it will be the slant height, the height of the triangle.
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Now, to try to make sense of this formula, we know that lateral area would be the area of all of the faces except the base.
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So, for this one right here, I have four lateral faces: I have this front; I have this right here; I have the back one; and I have this side one.
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So, I have four of them; that means that I have four triangles: 1, 2, 3, 4.
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This is what my lateral area is going to be: the area of this, plus this, this, and that--all four together--is the lateral area (everything minus the base).
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Well, I know that, if this right here is s for side, and then my slant height I am going to label as l,
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then look at this triangle right here: that is the same as this triangle.
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The slant height is this right here, and then this would be the side.
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To find the area of this triangle, it is going to be 1/2 the base, which is s, times the l,
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plus the same thing here: 1/2 the base of s and the height of l, plus 1/2 sl, plus 1/2sl.
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And again, how did I get this? This is just 1/2 base times height; it is the area of a triangle.
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But the base is a side, s; and the height is l, for the slant height.
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So, all I did was to replace the base with s, and then the height with l; and that is the area of this triangle here, and the area of this triangle...
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And so, I have four of them together; and this is all lateral area.
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Now, if I factor out the 1/2 and the l from each of these, then from here, I am going to be left with s, plus, from here, s, plus s, plus s.
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Well, all 4 s's together make up s + s + s; it makes up the perimeter of the base, so it will just be 1/2 times the slant height times the perimeter of the base.
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And that is how we get this formula: 1/2 times the slant height times the perimeter of the base.
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And to shorten it, you can just do 1/2, capital P for perimeter of the base, times l, which is the slant height (l is the label we are using for slant height).
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This is lateral area; again, if you want to think of each of these triangles, we have four of them;
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to find the area of each one, we know that, to find lateral area, we have to add the area of each triangle;
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so I just wrote them out; I factored out 1/2 and the l; I am left with 4s, which is the perimeter of the base; and that gives you 1/2Pl,
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1/2 times the perimeter of the base times the slant height; and that is the formula for the lateral area of a regular pyramid.
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And then, to find the surface area of a pyramid, we know that it is the lateral area, plus the base.
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There is only one base, and that would be that right there; so just find the area of that base.
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And this is a regular pyramid, so it is going to be a square; so you find the area of a square.
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We add it to the lateral area, and we know that lateral area, again, is 1/2 the perimeter of the base, times the slant height;
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plus capital B for the area of the base is surface area.
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Next is a cone: now, we know what that looks like--we have had ice-cream cones before.
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It is a circular base, and then they meet at a vertex.
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Now, the axis of a cone would be the segment from the vertex down to the center of the circle; that is the axis.
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In a right cone, the axis and the altitude are the same; but when it comes to oblique cones,
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because it is a little to the side, slanted to the side, leaning over, the altitude has to be perpendicular;
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the axis, though, is still going to stay from the vertex of the cone down to the center of the circle.
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So, the axis for the right cone is going to be the same as the altitude, but not for the oblique cone.
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And we know that altitude is the perpendicular height.
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Now, for the lateral area of a right cone, we are going to be measuring everything around (not including) the base (not the circle).
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So, to find the lateral area, it is π times the radius, times the slant height.
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So, slant height would be the measure of this right here; that is the slant height, l.
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This right here is the altitude; that is the height of the cone; but going this way, that is the slant height.
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Think of the height being slanted; that is l; we know that this is r; so the lateral area is π times r times l.
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And then, of course, the surface would be just all of that, the lateral area, so π times r times the l,
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plus the area of the base (capital B for the area of a base); and since it is a circle, we know it is πr².
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So, it is πrl + πr².
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Now, as long as you remember just this, you know that surface area is just the lateral area plus the area of the base.
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So, this is what you actually really have to remember; and then, for surface area,
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just remember that it is lateral area, plus the area of the base, the area of the circle.
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Let's go over some examples: Determine whether the condition given is characteristic of a pyramid, prism, both, or neither.
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Remember: a pyramid is when we have a polyhedron (because all of the sides are flat surfaces), and a vertex;
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so if you have a base like this, it will go to each of these sides like that; that is a pyramid.
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A prism is if I have something like that; that would be a prism, where we have two bases opposite and parallel and congruent.
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So then, the first one: The lateral faces are triangles (let me highlight this and shade in that base).
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The lateral faces of a prism all have to be rectangular; they are all rectangles--all of these lateral faces here.
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For the pyramid, all of them have to be triangles; so this one here, we know, is "pyramid."
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And the next one: There is exactly one base.
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Well, prism, we know, has two bases; pyramid, we know, has only one; the opposite side of it would be the vertex; so this one is also "pyramid."
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The next one: Find the lateral area of the regular polygon.
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Lateral area is 1/2 times the perimeter times the slant height.
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If you ever forget the formula, just think of it as finding the area of all four triangles.
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If you just find the area of this triangle here, because it is a regular pyramid, each of the lateral faces is a triangle, and they are all congruent.
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So, if you just find the area of one of the triangles, then you can just multiply that by 4, because I have 4 of them.
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They are not always going to be 4; it depends on the base.
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Find the lateral area: 1/2 the perimeter--if this is 12, then this is 12, and this is 12, and this is 12.
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The perimeter is 12 times 4, which is 48; and then, the slant height...
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now, be careful here: this is not the slant height; the slant height would be the height of the triangle, of the face, which is the triangle.
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So, if that is one of our lateral faces, this is 12, and this is 10; then, slant height has to be that right there.
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Be careful: this is not the slant height, so you have to look for it.
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If the whole thing is 12, I know that this part right here is 6; so then, to solve for my (I'll just label that l) slant height,
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it is going to be...using the Pythagorean theorem...l² + 6² = 10².
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This is l² + 36 = 100; then l² is going to be 64, which makes l 8;
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so we know that the slant height, that right there, is 8.
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And that is the measure that I need, 8; so the lateral area is going to be 48 times 8, divided by 2.
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Or you can just do 24 (because I cut this in half), times 8; so on your calculator, 24 times 8 equals 192.
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And I don't see any units, so it will just be units squared.
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The next example: Find the surface area of the pyramid.
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Surface area is the lateral area, plus the area of the base; the lateral area is 1/2 times the perimeter times the slant height,
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and then, plus the area of the base...which is going to be side squared, because if this is a side, and this is a side...
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we don't even need this; this is a side, and it is a square; so length times width is just side squared.
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1/2...the perimeter would be...oh, do we know the side?
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Well, we are not given this, but since we know that this is the center, and we are given half of that--
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we are given from the center to this side, then if this half is 6, then the whole thing has to be 12, so this is 12.
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So, the perimeter would be 12 times 4, which is 48.
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The slant height...also be careful here: they give us the altitude--they give us the height--of this pyramid; but we don't know the slant height.
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So again, we have to solve for it; now, slant height, we know, is from the vertex;
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and just look at one of the triangles, one of the lateral faces, and then find the height of that.
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If you are looking at a triangle, the lateral face, then it has to be that right there; that is the slant height.
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OK, so then, here, how would we find the slant height?
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Here we have a triangle; this is a right triangle; so this is 8, 6, and then the slant height would be the hypotenuse.
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Using the Pythagorean theorem, 8² + 6² = the slant height, squared; as you can see, that is the right triangle.
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This is 64 + 36 = l²; this gives you 100, which is the slant height squared;
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and then, if you take the square root of that, then you get 10; so I know that this slant height is 10.
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It is 10, plus...the area of the base is 12 times 12, which is 144; then, you just solve it out; so 24...
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I just divided this by 2...times 10 is 240, plus 144 is going to be 384 units squared, and that is my surface area.
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And the fourth example: Find the lateral area and the surface area of the cone.
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My lateral area: the formula is π times the radius times the slant height--that is the lateral area.
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I have the radius, and I have the slant height; surface area is the lateral area, plus the area of the base, which is the area of the circle.
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First, let's look for the lateral area: LA is π times radius (is 5), and the slant height would be this right here.
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So, on your calculator, you are going to multiply out π times 5 times 13; and I get 204 and 20 hundredths inches squared.
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It is area, so make sure that it is units squared.
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Then, to find surface area, let's first find the area of this base right here.
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The area of the circle is B for area of the base; it is πr²; that is π, and then the radius is 5, squared, which gives me 78.6.
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And we'll just do π times 5 squared; make sure that you square this first, and then multiply it.
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Because of Order of Operations, you have to do the exponent before you multiply: 25 times π is 78.54, or 78.6;
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so, my surface area is going to be my lateral area, that number right there, added to my area of the circle.
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And then, when you add it together, we should get 282.74; and then here, it is inches squared.
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Now, if you remember from the prism, remember: when you find surface area, you are not just adding the base.
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If you have two bases, then you have to multiply this number, the area of the base, times 2, because you have 2 of them.
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You have to take that into consideration.
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I only have one base, so whatever my lateral area is, I can just add this number to that.
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But if I have two bases, then I have to multiply that base times 2 and then add it.
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Just remember: when you find surface area, it has to cover every single part of your solid--every single side, lateral faces and your bases.
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That is it for this lesson; thank you for watching Educator.com.