Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Geometry
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share

Start Learning Now

Our free lessons will get you started (Adobe Flash® required).
Get immediate access to our entire library.

Sign up for Educator.com

Membership Overview

  • Unlimited access to our entire library of courses.
  • Search and jump to exactly what you want to learn.
  • *Ask questions and get answers from the community and our teachers!
  • Practice questions with step-by-step solutions.
  • Download lesson files for programming and software training practice.
  • Track your course viewing progress.
  • Download lecture slides for taking notes.
  • Learn at your own pace... anytime, anywhere!

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 = ½(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 = π × radius × slant height
  • Surface area of a right cone = lateral area + area of the circle

Surface Area of Pyramids and Cones

Determine whether the following statement is true or false.
All the faces of a pyramid intersect at the vertex.
  
False.
Determine whether the following statement is true or false.
All the faces of a pyramid are triangles.
  
False.
Determine whether the following statement is true or false.
All the lateral faces of a regular pyramid are congruent isosceles triangles.
  
True.
The base of a regular pyramid is a square, each side of the square is 4m, the slant height of the pyramid is 6m, find the lateral area of this pyramid.
  • LA = [1/2]Perimeter*slant height
  • LA = [1/2]*(4*4)*6
LA = 48m2
The base of a regular pyramid is a square, each side of the square is 7m, the slant height of the pyramid is 8m, find the surface area of this pyramid.
  • LA = [1/2]Perimeter*slant height
  • LA = [1/2]*(7*4)*8
  • LA = 112m2
  • SA = LA + (area of the base)
  • SA = 112 + 7*7
SA = 161m2
Determine whether the following statement is true or false.
The axis of a cone is the same as its altitude.
  
False.
Fill in the blank in the statement with sometimes, never or always.
The base of a cone is _____ a circle.
  
Always.
The radius of the base circle of a right cone is 5, the slant height is 8, find the lateral area of the cone.
  • LA = p*radius*slant height
LA = 3.14*5*8
The diameter of the base circle of a right cone is 12, the slant height is 9, find the surface area of this cone.
  • 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*radius2
  • SA = 169.56 + 3.14*6*6
SA = 282.6
Determine whether the following statement is true or false.
The slant height is always larger than the altitude of a cone.
  
True

*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 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

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 πr2.0751

So, it is πrl + πr2.0763

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...l2 + 62 = 102.1031

This is l2 + 36 = 100; then l2 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, 82 + 62 = the slant height, squared; as you can see, that is the right triangle.1254

This is 64 + 36 = l2; 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 πr2; 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