# 7.3: Base, Lateral and Surface Areas of Prisms

**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?*

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

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

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

## 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:

\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 \begin{align*}E\end{align*} and ____ are called the **bases** of the prism, and the faces labeled \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:

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

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

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

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

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

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

\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:

\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 \begin{align*}A\end{align*} and \begin{align*}E\end{align*}), we use the formula:

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

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

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

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

\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:

\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:

\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 \begin{align*}(s + s + s + s + s + s)\end{align*} or \begin{align*}6s\end{align*}.

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

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

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

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

\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 \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.*

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

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

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

**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 \begin{align*}H\end{align*}) \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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