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# 7.3: Base, Lateral and Surface Areas of Prisms

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use nets to represent prisms.
• Find the surface area of a prism.

## Prisms

A prism is a 3-dimensional figure with a pair of parallel and congruent ends, or bases. The sides of a prism are parallelograms. Prisms are identified by their bases.

Look at the names of the prisms above. Each type of prism is named after its base. In the figure on the left, the base is a rectangle, so it is called a rectangular prism. In the middle, the base is a triangle, so the shape is a triangular prism. On the right, the base is a hexagon (6-sided figure), so the prism is called a hexagonal prism.

As you can see, the base is not always on the “bottom” of the prism!

Prisms are named by their ______________________.

If a prism is called a pentagonal prism, then its base is a __________________________.

## Surface Area of a Prism Using Nets

The prisms in the picture above are right prisms. In a right prism, the lateral sides are perpendicular to the bases of prism. In the picture below, compare a right prism to an oblique prism, in which sides and bases are not perpendicular:

Reading Check:

1. Fill in the blanks:

A prism is a _____-dimensional figure with parallel and congruent ________________ and sides that are __________________________________.

2. In a right prism, what is the relationship between the sides and the bases?

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

3. What is the difference between a right prism and an oblique prism?

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## Surface Area

Surface area is the exposed area of a 3-dimensional or solid figure. This means that surface area is a measurement of the outside area of an object. For example, the surface area of a soccer ball is the outside part of the ball that touches the air. The inside of the ball is not included in surface area. Surface area is, as its name describes, the area of the surface of a solid object.

Surface area is the area of the _________________________ of a 3-dimensional object.

Area Addition Postulate

The surface area of a 3-dimensional figure is the sum of the areas of all of its non-overlapping parts.

This means that we find the surface area by adding up all of the face areas of the figure. Since the faces are on the surface of the object, if we add up all of the faces, we will have the entire surface area!

To find the surface area of a shape, you add the areas of its _______________________.

You can use a net and the Area Addition Postulate to find the surface area of a right prism:

From the net of the prism, you can see that that the surface area of the entire prism equals the sum of the shapes that make up the net. Since there are 6 faces of the prism and 6 shapes in the net, there are 6 areas that you add together:

Total surface area=area A+area B+area C+area D+area E+area F\begin{align*}\text{Total surface area} = \text{area}\ A + \text{area}\ B + \text{area}\ C + \text{area}\ D + \text{area}\ E + \text{area}\ F\end{align*}

\begin{align*}\rightarrow\end{align*} You may notice that the faces labeled E\begin{align*}E\end{align*} and ____ are called the bases of the prism, and the faces labeled A,B,\begin{align*}A, B,\end{align*} ____ , and ____ are called the sides of the prism. The surface area is the sum of the areas of the bases and the areas of the sides.

To find the areas of shapes in the net, we must use the formula for the area of a rectangle:

Area=lengthwidthArea of shape A=105=50 square units\begin{align*}&\text{Area} = \text{length} \cdot \text{width}\\ &\text{Area of shape}\ A = 10 \cdot 5 = 50 \ \text{square units}\end{align*}

Find the areas of the other rectangles in the net:

Area of shape B=310=\begin{align*}B = 3 \cdot 10 = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*} square units

Area of shape C==\begin{align*}C = \underline{\;\;\;\;\;\;\;\;\;\;} \cdot \underline{\;\;\;\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*} square units

Area of shape D==\begin{align*}D = \underline{\;\;\;\;\;\;\;\;\;\;} \cdot \underline{\;\;\;\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*} square units

Area of shape E=35=\begin{align*}E = 3 \cdot 5 = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*} square units

Area of shape F==\begin{align*}F = \underline{\;\;\;\;\;\;\;\;\;\;} \cdot \underline{\;\;\;\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*} square units

Then insert these areas back into the surface area equation above:

Total surface area=area A+area B+area C+area D+area E+area FTotal surface area=50+++++Total surface area=190 square units\begin{align*}&\text{Total surface area} = \text{area}\ A + \text{area}\ B + \text{area}\ C + \text{area}\ D + \text{area}\ E + \text{area}\ F\\ &\text{Total surface area} = 50 + \underline{\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;} \\ &\text{Total surface area} = 190 \ \text{square units}\end{align*}

If 2 polygons (or plane figures) are congruent, then their areas are congruent.

This means that if 2 shapes are the same, then they have the same area! This may seem simple, and it is!

Example 1

Use a net to find the surface area of the prism.

The area of the net is equal to the surface area of the figure:

Surface area=area A+area B+area C+area D+area E\begin{align*}\text{Surface area} = \text{area}\ A + \text{area}\ B + \text{area}\ C + \text{area}\ D + \text{area}\ E\end{align*}

To find the area of the triangles (shapes A\begin{align*}A\end{align*} and E\begin{align*}E\end{align*}), we use the formula:

Area=12 bh\begin{align*}\text{Area} = \frac{1}{2} \ bh\end{align*}, where b\begin{align*}b\end{align*} is the base of the triangle and h\begin{align*}h\end{align*} is its height

Since triangles A\begin{align*}A\end{align*} and E\begin{align*}E\end{align*} are congruent, their areas are the same:

Area of shape A=12(129)=12(108)=54\begin{align*}\text{Area of shape}\ A = \frac{1}{2} (12 \cdot 9) = \frac{1}{2} (108) = 54\end{align*}

So the area of shapes A\begin{align*}A\end{align*} is 54 square units and the area of shape E\begin{align*}E\end{align*} is 54 square units.

For the areas of shapes B,C,\begin{align*}B, C,\end{align*} and D\begin{align*}D\end{align*}, we use the formula for the area of a rectangle.

Area of shape B=69=54 square unitsArea of shape C== square unitsArea of shape D== square units\begin{align*}&\text{Area of shape}\ B = 6 \cdot 9 = 54 \ \text{square units}\\ &\text{Area of shape}\ C = \underline{\;\;\;\;\;\;\;\;\;\;} \cdot \underline{\;\;\;\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;\;} \ \text{square units}\\ &\text{Area of shape}\ D = \underline{\;\;\;\;\;\;\;\;\;\;} \cdot \underline{\;\;\;\;\;\;\;\;\;\;} = \underline{\;\;\;\;\;\;\;\;\;\;} \ \text{square units}\end{align*}

Now insert the areas you found back into the surface area equation on the previous page:

Surface area=area A+area B+area C+area D+area ESurface area=54+54+++54Surface area=324 square units\begin{align*}&\text{Surface area} = \text{area}\ A + \text{area}\ B + \text{area}\ C + \text{area}\ D + \text{area}\ E\\ &\text{Surface area} = 54 + 54 + \underline{\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;} + 54\\ &\text{Surface area} = 324 \ \text{square units}\end{align*}

Reading Check:

1. True/False: The surface area of a 3-D figure is the same as the area of its net.

2. True/False: You can figure out the area of a net by finding the area of each of the shapes in it (rectangles, triangles, etc.) and multiplying them together.

\begin{align*}{\;}\end{align*}

## Surface Area of a Prism Using Perimeter

The hexagonal prism below has 2 regular hexagons for bases and 6 sides. Since all sides of the hexagon are congruent, all of the rectangles that make up the lateral sides of the 3-dimensional figure are also congruent.

You can break down the figure like this:

The word “lateral” means “on the side.”

For example, in football, a lateral pass is when the quarterback throws the ball sideways to a receiver: the ball is passed to the side instead of forwards.

Lateral sides of a solid are the faces on the sides of the shape (not the bases).

Lateral area, which you will learn next, is the area of the sides of the shape.

The surface area of the rectangular sides of the figure is called the lateral area of the figure. The lateral area does not include the area of the bases. To find the lateral area, you can add up all of the areas of the 6 rectangles:

Lateral area=6(area of one rectangle)=6(sh)=6sh\begin{align*}\text{Lateral area} &= 6 \cdot \text{(area of one rectangle)}\\ &= 6 \cdot (s \cdot h)\\ &= 6sh\end{align*}

You can also see that the perimeter of the base of the hexagonal prism on the previous page is (s+s+s+s+s+s)\begin{align*}(s + s + s + s + s + s)\end{align*} or 6s\begin{align*}6s\end{align*}.

Another way to find the lateral area of the figure is to multiply the perimeter of the base by h\begin{align*}h\end{align*}, which is the height of the figure:

Lateral area=6sh=(6s)h=(perimeter of base)h=Ph\begin{align*}\text{Lateral area} &= 6sh\\ &= (6s) \cdot h\\ &= \text{(perimeter of base)} \cdot h\\ &= Ph\end{align*}

Substituting P\begin{align*}P\end{align*}, the perimeter of the base, for 6s\begin{align*}6s\end{align*}, we get the formula for any lateral area of a right prism:

Lateral area of a prism=Ph\begin{align*}\text{Lateral area of a prism} = Ph\end{align*}

The lateral area is the surface area of the rectangular ____________________ of a right prism.

The lateral area does not include the area of the _____________________.

Find the lateral area of a right prism by multiplying the ________________________ of the base by the ________________________ of the prism.

We can use the lateral area formula to calculate the total surface area of the prism. Remember, lateral area does not include the area of the bases, but we must include the bases to find the total surface area of the prism!

Using P\begin{align*}P\end{align*} for the perimeter of the base and B\begin{align*}B\end{align*} for the area of the base:

Total surface area=Lateral area+area of 2 bases=(perimeter of base  height)+2(area of base)=Ph+2B\begin{align*}\text{Total surface area} &= \text{Lateral area} + \text{area of 2 bases}\\ &= \text{(perimeter of base}\ \cdot \ \text{height}) + 2 \cdot \text{(area of base)}\\ &= Ph + 2B\end{align*}

You can use this formula A=Ph+2B\begin{align*}A = Ph + 2B\end{align*} to find the surface area of any right prism.

Reading Check:

1. True/False: The lateral area includes the area of the sides and the area of the bases.

2. True/False: The perimeter of the base and the area of the base are the same thing.

3. How could you find the total surface area of a 3-dimensional shape if you know the height of the shape and the perimeter of its base? What other information would you need to know? Explain.

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Example 2

Use the formula to find the total surface area of the trapezoidal prism below.

The dimensions of the trapezoidal base are shown in the diagram, but we should list our information so it is easy to see:

Height of the entire prism (we will call this H\begin{align*}H\end{align*}) =21\begin{align*}= 21\end{align*}

Base 1 of the trapezoid \begin{align*}(b_1) = 4\end{align*}

Base 2 of the trapezoid \begin{align*}(b_2) = 10\end{align*}

height of the trapezoid (call this \begin{align*}h\end{align*} so it is not confused with \begin{align*}H\end{align*} above) \begin{align*}= 2.64\end{align*}

Use the formula for total surface area from the previous page:

\begin{align*}\text{Total surface area} = PH + 2B\end{align*}

(where \begin{align*}P\end{align*} is perimeter of the base, \begin{align*}H\end{align*} is height, and \begin{align*}B\end{align*} is area of the base)

Find the area of each trapezoidal base. Do this with the formula for area of a trapezoid. Remember that the height of the trapezoid is small \begin{align*}h\end{align*}: \begin{align*}A_B = \frac{1}{2} h (b_1 + b_2)\end{align*}

\begin{align*}A_B &= \frac{1}{2} \cdot 2.64 ( \underline{\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;} )\\ A_B & = 18.48 \ \text{square units}\end{align*}

Now find the perimeter of the base:

\begin{align*}P &= 10 + 4 + 4 + 4\\ &= 22 \end{align*}

Use these values to find the total surface area of the solid:

\begin{align*}\text{Total surface area} &= PH + 2B\\ &=( \underline{\;\;\;\;\;\;\;\;\;\;\;\;} )(21) + 2( \underline{\;\;\;\;\;\;\;\;\;\;\;\;} )\\ &=462 + 36.96\\ &=498.96 \ \text{square units}\end{align*}

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