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# 8.4: Surface Area of Prisms and Cylinders

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## Introduction

Mr. Samuels the custodian has been working overtime getting ready for the Olympics at Montgomery Middle School. To help the students, he has been creating long jump fields, bringing in tons of sand and offering to help out afterschool and on weekends.

“I think that we should do something nice for Mr. Samuels,” Crystal said at lunch.

“I agree. But what?” Kenneth responded.

“How about giving him one of our prism awards? I could wrap it up and we could present it to him at the Olympics,” Crystal suggested.

Kenneth, Marcy and Dylan all agreed. So that week, Crystal found a box for the prism award and began to wrap it. Because of the size of the award, she needed a good sized box.

How much wrapping paper will she need?

Surface area is the topic for this lesson, and Crystal’s wrapping paper is all about surface area. By the end of this lesson, you will know how much wrapping paper she will need to cover this box.

What You Will Learn

In this lesson you will learn how to complete the following skills.

• Recognize surface area of prisms and cylinders as the sum of areas of faces using nets.
• Find surface areas and linear dimensions of right prisms with any polygonal base.
• Find surface areas and linear dimensions of cylinders and truncated cylinders.
• Solve real – world problems involving surface area of prisms and cylinders.

Teaching Time

I. Recognize Surface Area of Prisms and Cylinders as the Sum of Areas of Faces using Nets

In the last lesson we learned about three-dimensional figures. Remember that a three–dimensional or solid figure has length, width and depth. It is not simply a flat two–dimensional plane figure. We learned about all of the different types of three–dimensional figures. In this lesson, you will learn how to calculate some very specific measurements when working with prisms and cylinders.

What is a prism?

A prism is a three–dimensional figure with two parallel congruent polygons as bases. The side faces of a prism are rectangular in shape.

What is a cylinder?

A cylinder has two parallel congruent circular bases with a curved rectangle as its side.

One of the things that we can measure when working with three–dimensional figures is called surface area. Surface area is the total of the areas of each face of a solid figure. Imagine you could wrap one of the figures above in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure represents its surface area. To find the surface area, we must be able to calculate the area of each face and then add these areas together.

There are several different ways to calculate surface area. One way is to use a net.

A net is a two-dimensional diagram of a three-dimensional figure. Imagine you could unfold a box so that it is completely flat. You would have something that looks like this.

If we folded this up, you could see that it would form a cube. A cube is made up of faces that are squares. If we wanted to figure out the surface area or measurement of the outer covering of this cube, then we could find the area of each surface of the cube and then add the products together.

We could also look at a net of a rectangular prism.

A rectangular prism is made up of rectangles. To find the surface area of a prism, we would need to calculate the area of each of the faces and then add them together.

We can look at the net of a cylinder too.

With the net of a cylinder, we would need to calculate the area of each circle and the area of the curved side of the cylinder. Then we could add these values together to find the surface area.

To find the area of a rectangle, we use the formula $A = lw$. If we know the length and width of each rectangle, we can calculate its area. Let’s put in the measurements for all the rectangles. Remember, each face represents two dimensions of the box: length and width, length and height, or width and height.

Let’s begin by calculating the area of a rectangular prism.

Now that we have all the information we need, we can calculate the area of each face and then add their areas together.

$& \mathbf{bottom \ face} && \mathbf{top \ face} && \mathbf{long \ side} && \mathbf{long \ side} && \mathbf{short \ side} && \mathbf{short \ side}\\& A=lw && A=lw && A=lh && A=lh && A=wh && A=wh\\& 12 \times 7 \quad + && 12 \times 7 \quad + && 12 \times 3 \quad + && 12 \times 3 \quad + && 7 \times 3 \quad + && 7 \times 3\\& 84 \qquad \ \ + && 84 \qquad \ \ + && 36 \qquad \ \ + && 36 \qquad \ \ + && 21 \qquad \ + && 21 \qquad \ \ = \ 282 \ in.^2$

We found the area of each rectangular face and then added all of these areas together. The total surface area of the rectangular prism is 282 square inches. Using a net helped us to locate all of the faces and find the measurements of each side.

We follow a similar process to find the surface area of a cylinder from a net.

To find the surface area, we need to calculate the area for each circle in the net. We use the formula $A =\pi r^2$ to find the area of a circle. If we know the radius or diameter of each circle, we can calculate its area. Look closely again at the cylinder above. The two circular faces are congruent, so they must have the same radius and diameter. Let’s calculate the area for each face.

$& \mathbf{bottom \ face} && \mathbf{top \ face}\\& A=\pi {r^2} && A=\pi {r^2}\\& A=\pi {(4^2)} && A=\pi {(4^2)}\\& A=16 \pi && A=16 \pi \\& A=50.24 \ cm^2 && A=50.24 \ cm^2$

The area of each circular face is 50.24 square centimeters when we use 3.14 to approximate pi.

Now we need to find the area of the side. The net shows us that, when we “unroll” the cylinder, the side is actually a rectangle. Recall that the formula we use to find the area of a rectangle is $A = lw$. For cylinders, the width of the rectangle is the same as the height of the cylinder. In this case, the height of the cylinder is 8 centimeters.

What about the length? The length is actually the same as the perimeter of the circle, which we call its circumference. When we “roll” up the side, it fits exactly once around the circle. To find the area of the cylinder’s side, then, we multiply the circumference of the circle by the height of the cylinder. We find the circumference of a circle with the formula $C = 2 \pi r$, and then we multiply it by the height. Let’s try it.

$C &=2 \pi r\\C &=2 \pi 4\\C &=8 \pi\\C &=25.12 \times 8 = 200.96 \ cm^2$

Now we know the area of both circular faces and the side. Let’s add them together to find the surface area of the cylinder.

$& \mathbf{bottom \ face} && \mathbf{top \ face} && \mathbf{side} && \mathbf{surface \ area}\\& 50.24 \ cm^2 \qquad + && 50.24 \ cm^2 \quad \ + && 200.96 \ cm^2 \quad =&& 301.44 \ cm^2$

The total surface area of the cylinder is 301.44 square centimeters. Using a net helped us to locate the faces and the find the measurements of the side.

We can use this same process to find the surface area of any prism or any net. Just be sure to find the area of each face and then add the areas together.

II. Find Surface Area and Linear Dimensions of Right Prisms with any Polygonal Base

Nets let us see each face so that we can calculate their area and then add them together. However, we can also use a formula to represent the faces as we find their area. The formula gives us a nice short cut that we can use for any kind of prism, no matter what shape its base is. Take a look at the formula below.

$SA=Ph+2B$

Let’s look at the first part of the formula. $P$ represents the perimeter of the base, and $h$ represents the height of the prism. By multiplying the perimeter and height, we are finding the area of all of the side faces at once. This will be very useful if the prism that we are working with isn’t just a cube or a rectangular prism.

The second part of the formula represents the area of the top and bottom faces. $B$ represents the area of one base, which we find using whichever area formula is appropriate for the shape of the base. Then we multiply it by 2 to show the area of the top and bottom faces at once. Let’s give it a try to see how this works.

Example

Find the surface area of this figure using a formula.

We have all the measurements we need. Let’s find the perimeter of the base first. It is a rectangle, so we add the lengths and widths: $21 + 21 + 14 + 14 = 70$. We can put this number in for $P$ in the formula. The height, we can see, is 5 centimeters.

Now let’s solve for $B$, the area of the base. The base of this prism is a rectangle, so we use the formula $A = lw$ to find its area.

$B &= lw\\B &=21 \times 14\\B &=294 \ cm^2$

Now we have all of the information we need to fill in the formula. Let’s put it in and solve for $SA$, surface area.

$SA &=Ph + 2B\\SA &= 70 (5) + 2 (294)\\SA &= 350 + 588\\SA &= 938 \ cm^2$

This rectangular prism has a surface area of 938 square centimeters.

We can use this method for all kinds of prisms. Let’s try another one.

Example

What is the surface area of the figure below?

Let’s look at the base first to find its perimeter. The triangle has two sides of 5 inches and one that is 8 inches: $5 + 5 + 8 = 18 \ inches$. This will be $P$ in the formula. The height of the prism is 15 inches. Be careful not to confuse the height of the prism with the height of the triangular base!

To find $B$, we need to use the area formula for triangles: $A =\frac{1}{2} bh$. The base of the triangle is 8 inches, and the height is 3 inches.

$A &=\frac{1}{2} bh\\A &= \frac{1}{2} (8) (3)\\A &= 4 (3)\\A &= 12 \ in.^2$

The area of the triangular base is 12 square inches, so we put this in for $B$ in the formula. Let’s put all of the values in and solve.

$SA &= Ph + 2B\\SA &= 18 (15) + 2 (12)\\SA &= 270 + 24\\SA &= 294 \ in.^2$

The surface area of this triangular prism is 294 square inches.

Write this formula for finding the surface area of a prism down in your notebook.

III. Find Surface Areas and Linear Dimensions of Cylinders and Truncated Cylinders

We use the same basic formula, $Ph + 2B$, to find the surface area of cylinders, too. As we know, the top and bottom faces of a cylinder are circles. The perimeter of a circle is its radius. We find circumference by using the formula $2 \pi r$. Then we multiply it by the height of the cylinder.

To find the area of the base, $B$, we use the area formula for circles: $\pi{r^2}$. We still multiply it by 2 because there is a circular top face and bottom face. This gives us the formula

$& SA =2 \pi {r^2} + 2 \pi rh\\& \qquad \ \ (2B) \quad \ (Ph)$

This formula may look long and intimidating, but all we need to do is put in the values for the radius of the circular faces and the height of the cylinder and solve.

Write this formula and the sentence on how to use it down in your notebooks.

Let’s give it a try.

Example

What is the surface area of the figure below? Use 3.14 to approximate pi.

We have all of the measurements we need. Let’s put them into the formula and solve for surface area, $SA$.

$SA &=2 \pi {r^2} + 2 \pi rh\\SA &= 2 \pi (3.5^2) + 2 \pi (3.5) (28)\\SA &= 2 \pi (12.25) + 2 \pi (98)\\SA &= 24.5 \pi + 196 \pi \\SA &= 220.5 \pi \\SA &= 692.37 \ {cm^2}$

This cylinder has a surface area of 692.37 square centimeters.

That wasn’t so bad! We just have to be careful to put each measurement in the right place in the formula and take it one step at a time. Let’s try another one.

Example

What is the surface area of the figure below?

Look carefully at the cylinder. This time we have been given the diameter, not the radius. Remember, the diameter of a circle is always twice the length of the radius. We can divide the diameter’s length by 2 to find the radius: $13 \div 2 = 6.5$. Now we have the radius and the height, so let’s put the numbers into the formula and solve.

$SA &= 2 \pi r^2 + 2 \pi rh\\SA &= 2 \pi (6.5^2) + 2 \pi (6.5) (11)\\SA &= 2 \pi (42.25) + 2 \pi (71.5)\\SA &= 84.6 \pi + 143 \pi\\ SA &= 227.6 \pi\\ SA &= 714.66 \ {ft^2}$

This cylinder has a surface area of 714.66 square feet when we approximate pi as 3.14.

Sometimes, we can have a cylinder that has been cut. We call it a truncated cylinder. This is where you only see a section of the cylinder and will need to figure out the surface area of what you see.

Now let’s say that one – half of the cylinder is pictured. We know that the radius of the circle will not change, so we can use that given measurement. The height of the cylinder will change because it has been cut in half. Therefore, we can figure out the surface area by using the given measurements and the same formula. We don’t have to do anything different because we are looking for the measurement of what is shown, the surface area of half of the cylinder.

$SA &=2 \pi {r^2} + 2 \pi rh\\SA &= 2 \pi (2^2) + 2 \pi (2) (4)\\SA &= 2 \pi (4) + 2 \pi (8)\\SA &=8 \pi + 16 \pi\\ SA &=75.36 \ {cm^2}$

We could also find the surface area of the whole cylinder by changing the height from 4 cm to 8 cm. Then we could use the same measurement for radius and calculate the surface area of the cylinder.

Now let’s look at applying what we have learned to some real – life examples.

IV. Solve Real – World Problems Involving Surface Area of Prisms and Cylinders

We have learned two ways to find surface area: drawing a net or using a formula. We can use either of these methods to solve word problems involving surface area. Nets may be especially useful if the problem does not provide an image of the figure. If you choose to use a formula, be sure you know whether the problem deals with a cylinder or a prism and what kind of prism it is. Let’s practice using what we have learned.

Example

Kevin is replacing the canvas on his tent. The tent has a front and back flap and two long sides. If the tent is 6 feet tall, 16 feet across the front, and 24 feet long, how many square feet of canvas will Kevin need?

First let’s decide what kind of solid figure the problem deals with. The front and back flaps of the tent are triangles, so this is a triangular prism.

Now read the problem carefully. What is it asking us to find? We need to find the surface area of only the canvas parts of the tent. The floor is not made of canvas, so we should not include this side in our calculations. This problem is a bit complicated, so it may help to draw a net. In your drawing, label the floor, the sides, and the front and back flap.

Look at this triangular prism. Notice that the center square is blue. That is the bottom of the tent. Since the bottom of the tent is not canvas, we don’t need to figure out the area of this space for our surface area. We will need to figure out the rest of the areas and add them together to find the surface area.

Now let’s fill in all of the measurements. Now we’re ready to find the area of each face.

$& \mathbf{front \ flap} && \mathbf{back \ flap} && \mathbf{side} && \mathbf{side}\\ & A =\frac{1}{2}bh && A =\frac{1}{2}bh && A = lw && A = lw\\& \frac{1}{2}(8)(6) && \frac{1}{2}(8)(6) && 24(10) && 24(10)\\& 24 \qquad \qquad + && 24 \qquad \qquad + && 240 \qquad \quad + && 240 \qquad = \ 528 \ ft^2$

Kevin will need 528 square feet of canvas to cover all of the sides of the tent, not including the floor.

We can also solve real – world problems involving cylinders. Take a look at this next problem.

Example

Mrs. Johnson is wrapping a cylindrical package in brown paper so that she can mail it to her son. The package is 22 centimeters tall and 11 centimeters across. How much paper will she need to cover the package?

The picture clearly shows us the height and diameter of the cylinder, so let’s use the formula for finding the surface area. But be careful—we have been given the diameter, not the radius. We need to divide it by 2 to find the radius: $11 \div 2 = 5.5$. Now we have the radius and height, so we can put these in for the appropriate variables in the formula.

$SA &= 2 \pi r^2 + 2 \pi rh\\SA &= 2 \pi (5.5^2) +2 \pi (5.5)(22)\\SA &= 2 \pi (30.25) + 2 \pi (121)\\SA &= 60.5 \pi + 242 \pi \\SA &= 302.5 \pi\\SA &= 949.85 \ cm^2$

Mrs. Johnson will need 949.85 square centimeters of brown paper in order to wrap the entire package.

Now let’s use what we have learned on the problem from the introduction.

## Real-Life Example Completed

Here is the original problem once again. Reread it and look for the given information. Then solve the problem.

Mr. Samuels the custodian has been working overtime getting ready for the Olympics at Montgomery Middle School. To help the students, he has been creating long jump fields, bringing in tons of sand and offering to help out afterschool and on weekends.

“I think that we should do something nice for Mr. Samuels,” Crystal said at lunch.

“I agree. But what?” Kenneth responded.

“How about giving him one of our prism awards? I could wrap it up and we could present it to him at the Olympics,” Crystal suggested.

Kenneth, Marcy and Dylan all agreed. So that week, Crystal found a box for the prism award and began to wrap it. Because of the size of the award, she needed a good sized box.

How much wrapping paper will she need?

Now you are ready to solve the problem.

Solution to Real – Life Example

First of all, what kind of solid figure is this? All of the faces are rectangles, including the base, so it is a rectangular prism. The picture clearly shows us what its length, width, and height are, so let’s use the formula for finding the surface area of prisms.

What is the perimeter of the base?

$12 + 12 + 9 + 9 = 42 \ inches$.

We’ll put this in for $P$.

We also need to find the area of the base, $B$. This base is a rectangle, so we use the formula $B = lw$.

$B &=lw \\B &=12 (9)\\B &=108 \ in.^2$

Now we have all of the measurements to put in for the appropriate variables in the formula.

$SA &=Ph + 2B\\SA &=42(6)+2(108)\\SA &=252 + 216\\SA &=468 \ in.^2$

Crystal will need 468 square inches of wrapping paper in order to cover the present.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Three Dimensional Figures
solid figures that have length, width and height.
Prisms
three – dimensional figures with parallel, congruent polygons as bases and rectangular side faces.
Cylinders
three – dimensional figures with circular parallel congruent bases and a curved rectangle as the side.
Surface Area
the measurement of the outer covering on a solid figure.
Net
the pattern of a solid figure-what a solid figure would look like if it were drawn out as a pattern.
Truncated Cylinder
a cylinder that is cut in part from a complete cylinder.

## Time to Practice

Directions: Look at each figure and then answer the following questions about each.

1. What is the name of the figure pictured above?
2. What is the surface area of this figure?

1. What is the name of this figure?
2. What is the shape of the base of this figure?
3. How many bases are there?
4. What is the surface area of this figure?

1. What is the name of this figure?
2. What is the surface area of this figure?

1. What is the name of this figure?
2. What is the shape of the base
3. How many bases does this figure have?
4. What is the shape of the side faces?
5. How many side faces are there?
6. What is the surface area of this figure?

1. What is the name of this figure?
2. Which measurement is given the radius or the diameter?
3. What is the surface area of the figure?

Directions: Use what you have learned to solve the following problems.

1. A cylindrical water tank is 35 long and 10 feet across. How much sheet metal is the tank made of?
2. Did you use area or surface area to solve this problem?
3. For a class project, Duane needs to cover three shoeboxes in aluminum foil. Each box has a height of 7 inches, a length of 13 inches, and a width of 8 inches. How much aluminum foil will Duane need to cover all three boxes?
4. Juan built a dog house entirely out of wood, even the floor. If he left a square opening measuring 2.5 on all sides for the door, how many square feet of wood did Juan use?

## Date Created:

Jan 14, 2013

Apr 29, 2014
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