### Surface Area of Pyramids and Cones

- Pyramids
- The vertex is the point where all the faces except the base intersect
- The base is the polygon that connects all the lateral faces
- Lateral faces always form triangles
- Lateral edges have the vertex as an endpoint
- The altitude is the segment from the vertex perpendicular to the base
- Regular pyramid
- The base is a regular polygon
- The endpoints of the altitude is the vertex and the center of the base
- All the lateral faces are congruent isosceles triangles
- The slant height is the height of each lateral face
- Lateral area of a regular pyramid = Â½(perimeter of base × slant height)
- Surface area of a regular pyramid = lateral area + area of the base
- Cone
- The base is a circle
- The axis is the segment whose endpoints are the vertex and the center of the base
- The altitude is the perpendicular height
- Lateral area of a right cone = Ï€ × radius × slant height
- Surface area of a right cone = lateral area + area of the circle

### Surface Area of Pyramids and Cones

All the faces of a pyramid intersect at the vertex.

All the faces of a pyramid are triangles.

All the lateral faces of a regular pyramid are congruent isosceles triangles.

- LA = [1/2]Perimeter*slant height
- LA = [1/2]*(4*4)*6

^{2}

- LA = [1/2]Perimeter*slant height
- LA = [1/2]*(7*4)*8
- LA = 112m
^{2} - SA = LA + (area of the base)
- SA = 112 + 7*7

^{2}

The axis of a cone is the same as its altitude.

The base of a cone is _____ a circle.

- LA = p*radius*slant height

- radius = [1/2]diameter = 6
- LA = p*radius*slant height
- LA = 3.14*6*9 = 169.56
- SA = LA + (area of the circle)
- SA = LA + p*radius
^{2} - SA = 169.56 + 3.14*6*6

The slant height is always larger than the altitude of a cone.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Surface Area of Pyramids and Cones

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Pyramids
- Regular Pyramids
- Lateral Area of a Pyramid
- Surface Area of a Pyramid
- Cone
- Lateral Area and Surface Area of a Right Cone
- Extra Example 1: Pyramid and Prism
- Extra Example 2: Find the Lateral Area of the Regular Pyramid
- Extra Example 3: Find the Surface Area of the Pyramid
- Extra Example 4: Find the Lateral Area and Surface Area of the Cone

- Intro 0:00
- Pyramids 0:07
- Pyramids
- Regular Pyramids 1:52
- Regular Pyramids
- Lateral Area of a Pyramid 4:33
- Lateral Area of a Pyramid
- Surface Area of a Pyramid 9:19
- Surface Area of a Pyramid
- Cone 10:09
- Right and Oblique Cone
- Lateral Area and Surface Area of a Right Cone 11:20
- Lateral Area and Surface Are of a Right Cone
- Extra Example 1: Pyramid and Prism 13:11
- Extra Example 2: Find the Lateral Area of the Regular Pyramid 15:00
- Extra Example 3: Find the Surface Area of the Pyramid 18:29
- Extra Example 4: Find the Lateral Area and Surface Area of the Cone 22:08

### Geometry Online Course

### Transcription: Surface Area of Pyramids and Cones

*Welcome back to Educator.com.*0000

*For the next lesson, we are going to go over surface area of pyramids and cones.*0002

*First, let's talk about pyramids: this is a picture of a pyramid; we have a vertex right here;*0009

*all of these lateral edges meet this vertex; all of the faces touch at the vertex.*0021

*The base is the only side, or only face, that is not going to intersect at the vertex.*0037

*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,*0046

*just like all of the other solids that we went over; we have lateral faces, and then we have bases.*0053

*Pyramids only have one base; lateral faces always form triangles.*0059

*Because we only have one base, and all of the lateral faces meet at the vertex, they are all going to form triangles.*0068

*Lateral edges have the vertex as an endpoint; so again, one of the endpoints of each lateral edge is going to be the vertex.*0080

*The altitude is the segment from the vertex perpendicular to the base.*0095

*So, the segment from the vertex to the base so that it is perpendicular is the altitude.*0099

*A regular pyramid is when the base is a regular polygon; so this is going to be a square,*0114

*because a rectangle that is equilateral and equiangular is going to be a square.*0127

*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.*0132

*All of the lateral faces are congruent isosceles triangles, because it is regular; from the vertex,*0150

*it is going to be the altitude, and then we have each of these sides of the polygon, of the base, being congruent.*0162

*And that is going to make all of these triangles (because, remember, all of the lateral faces are triangles) isosceles triangles.*0168

*So, this and this are going to be the same; from this triangle right here, each one of them is going to be isosceles.*0175

*The slant height is the height of each lateral face; the slant height is not the same as the height of the solid.*0186

*Here, this height, from the vertex down to the center--we call that altitude.*0199

*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.*0206

*Remember: it is not going to the center of the pyramid; the slant height is to there--this would be the slant height.*0225

*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.*0242

*But when you look at it as a solid, the whole thing as a pyramid, it is slanted;*0255

*so be careful there to distinguish between the slant height and the height of the pyramid.*0262

*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--*0275

*meaning not the altitude, not this height; it will be the slant height, the height of the triangle.*0286

*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.*0294

*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.*0309

*So, I have four of them; that means that I have four triangles: 1, 2, 3, 4.*0321

*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).*0335

*Well, I know that, if this right here is s for side, and then my slant height I am going to label as l,*0346

*then look at this triangle right here: that is the same as this triangle.*0364

*The slant height is this right here, and then this would be the side.*0372

*To find the area of this triangle, it is going to be 1/2 the base, which is s, times the l,*0385

*plus the same thing here: 1/2 the base of s and the height of l, plus 1/2 sl, plus 1/2sl.*0396

*And again, how did I get this? This is just 1/2 base times height; it is the area of a triangle.*0410

*But the base is a side, s; and the height is l, for the slant height.*0417

*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...*0425

*And so, I have four of them together; and this is all lateral area.*0434

*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.*0441

*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.*0468

*And that is how we get this formula: 1/2 times the slant height times the perimeter of the base.*0492

*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).*0501

*This is lateral area; again, if you want to think of each of these triangles, we have four of them;*0516

*to find the area of each one, we know that, to find lateral area, we have to add the area of each triangle;*0526

*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,*0533

*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.*0548

*And then, to find the surface area of a pyramid, we know that it is the lateral area, plus the base.*0561

*There is only one base, and that would be that right there; so just find the area of that base.*0570

*And this is a regular pyramid, so it is going to be a square; so you find the area of a square.*0577

*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;*0585

*plus capital B for the area of the base is surface area.*0598

*Next is a cone: now, we know what that looks like--we have had ice-cream cones before.*0611

*It is a circular base, and then they meet at a vertex.*0619

*Now, the axis of a cone would be the segment from the vertex down to the center of the circle; that is the axis.*0629

*In a right cone, the axis and the altitude are the same; but when it comes to oblique cones,*0641

*because it is a little to the side, slanted to the side, leaning over, the altitude has to be perpendicular;*0647

*the axis, though, is still going to stay from the vertex of the cone down to the center of the circle.*0657

*So, the axis for the right cone is going to be the same as the altitude, but not for the oblique cone.*0663

*And we know that altitude is the perpendicular height.*0676

*Now, for the lateral area of a right cone, we are going to be measuring everything around (not including) the base (not the circle).*0682

*So, to find the lateral area, it is π times the radius, times the slant height.*0695

*So, slant height would be the measure of this right here; that is the slant height, l.*0703

*This right here is the altitude; that is the height of the cone; but going this way, that is the slant height.*0710

*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.*0721

*And then, of course, the surface would be just all of that, the lateral area, so π times r times the l,*0744

*plus the area of the base (capital B for the area of a base); and since it is a circle, we know it is πr ^{2}.*0751

*So, it is πrl + πr ^{2}.*0763

*Now, as long as you remember just this, you know that surface area is just the lateral area plus the area of the base.*0768

*So, this is what you actually really have to remember; and then, for surface area,*0777

*just remember that it is lateral area, plus the area of the base, the area of the circle.*0784

*Let's go over some examples: Determine whether the condition given is characteristic of a pyramid, prism, both, or neither.*0792

*Remember: a pyramid is when we have a polyhedron (because all of the sides are flat surfaces), and a vertex;*0803

*so if you have a base like this, it will go to each of these sides like that; that is a pyramid.*0817

*A prism is if I have something like that; that would be a prism, where we have two bases opposite and parallel and congruent.*0824

*So then, the first one: The lateral faces are triangles (let me highlight this and shade in that base).*0851

*The lateral faces of a prism all have to be rectangular; they are all rectangles--all of these lateral faces here.*0862

*For the pyramid, all of them have to be triangles; so this one here, we know, is "pyramid."*0870

*And the next one: There is exactly one base.*0881

*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."*0884

*The next one: Find the lateral area of the regular polygon.*0902

*Lateral area is 1/2 times the perimeter times the slant height.*0907

*If you ever forget the formula, just think of it as finding the area of all four triangles.*0919

*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.*0927

*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.*0938

*They are not always going to be 4; it depends on the base.*0947

*Find the lateral area: 1/2 the perimeter--if this is 12, then this is 12, and this is 12, and this is 12.*0952

*The perimeter is 12 times 4, which is 48; and then, the slant height...*0963

*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.*0978

*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.*0995

*Be careful: this is not the slant height, so you have to look for it.*1012

*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,*1018

*it is going to be...using the Pythagorean theorem...l ^{2} + 6^{2} = 10^{2}.*1031

*This is l ^{2} + 36 = 100; then l^{2} is going to be 64, which makes l 8;*1040

*so we know that the slant height, that right there, is 8.*1055

*And that is the measure that I need, 8; so the lateral area is going to be 48 times 8, divided by 2.*1068

*Or you can just do 24 (because I cut this in half), times 8; so on your calculator, 24 times 8 equals 192.*1081

*And I don't see any units, so it will just be units squared.*1098

*The next example: Find the surface area of the pyramid.*1110

*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,*1114

*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...*1133

*we don't even need this; this is a side, and it is a square; so length times width is just side squared.*1143

*1/2...the perimeter would be...oh, do we know the side?*1156

*Well, we are not given this, but since we know that this is the center, and we are given half of that--*1163

*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.*1173

*So, the perimeter would be 12 times 4, which is 48.*1184

*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.*1192

*So again, we have to solve for it; now, slant height, we know, is from the vertex;*1204

*and just look at one of the triangles, one of the lateral faces, and then find the height of that.*1214

*If you are looking at a triangle, the lateral face, then it has to be that right there; that is the slant height.*1222

*OK, so then, here, how would we find the slant height?*1237

*Here we have a triangle; this is a right triangle; so this is 8, 6, and then the slant height would be the hypotenuse.*1242

*Using the Pythagorean theorem, 8 ^{2} + 6^{2} = the slant height, squared; as you can see, that is the right triangle.*1254

*This is 64 + 36 = l ^{2}; this gives you 100, which is the slant height squared;*1264

*and then, if you take the square root of that, then you get 10; so I know that this slant height is 10.*1275

*It is 10, plus...the area of the base is 12 times 12, which is 144; then, you just solve it out; so 24...*1284

*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.*1299

*And the fourth example: Find the lateral area and the surface area of the cone.*1330

*My lateral area: the formula is π times the radius times the slant height--that is the lateral area.*1339

*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.*1354

*First, let's look for the lateral area: LA is π times radius (is 5), and the slant height would be this right here.*1373

*So, on your calculator, you are going to multiply out π times 5 times 13; and I get 204 and 20 hundredths inches squared.*1392

*It is area, so make sure that it is units squared.*1418

*Then, to find surface area, let's first find the area of this base right here.*1425

*The area of the circle is B for area of the base; it is πr ^{2}; that is π, and then the radius is 5, squared, which gives me 78.6.*1433

*And we'll just do π times 5 squared; make sure that you square this first, and then multiply it.*1453

*Because of Order of Operations, you have to do the exponent before you multiply: 25 times π is 78.54, or 78.6;*1464

*so, my surface area is going to be my lateral area, that number right there, added to my area of the circle.*1478

*And then, when you add it together, we should get 282.74; and then here, it is inches squared.*1498

*Now, if you remember from the prism, remember: when you find surface area, you are not just adding the base.*1516

*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.*1527

*You have to take that into consideration.*1533

*I only have one base, so whatever my lateral area is, I can just add this number to that.*1536

*But if I have two bases, then I have to multiply that base times 2 and then add it.*1543

*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.*1552

*That is it for this lesson; thank you for watching Educator.com.*1567

## Start Learning Now

Our free lessons will get you started (Adobe Flash

Sign up for Educator.com^{®}required).Get immediate access to our entire library.

## Membership Overview

Unlimited access to our entire library of courses.Learn at your own pace... anytime, anywhere!