# 8.7: Cone: Base Area, Lateral Area, Surface Area and Volume

**At Grade**Created by: CK-12

## Learning Objectives

- Find the surface area of a
*cone*using a net or a formula. - Find the volume of a cone.

## Cones

A **cone** is a three-dimensional figure with a single curved base that tapers to a single point called an **apex.** The base of a cone can be a circle or an oval of some type. In this chapter, we will only use *circular* **cones.**

- The ______________________ is the point on top of a
**cone.** - The
*base*of the**cones**we will study is in the shape of a ____________________.

You can remember the name **“cone”** of this shape because it looks like an upside-down ice cream **cone.**

The **apex** of a **right cone** lies above the center of the cone’s circle. In an **oblique cone,** the **apex** is *not* in the center:

- The
**apex**of a**right cone**is the point directly above the _____________________ of the cone’s circular base.

The height of a **cone** \begin{align*}h\end{align*}*perpendicular distance* from the *center* of the cone’s *base* to its **apex.**

- The height of a cone is just like an altitude: a _______________________ line from the center of the circular base to the apex.

## Surface Area of a Cone Using Nets

Most three-dimensional figures are easy to deconstruct into a net. The cone is different in this regard. Can you predict what the net for a cone looks like? In fact, the net for a cone looks like a small circle and a **sector**, or part of a larger circle.

The diagram below shows how the half-circle sector folds to become a cone:

Note that the circle that the sector is cut from is much larger than the base of the cone.

- The net for a
**cone**is a circular base plus a _____________________.

**Example 1**

*Which sector will give you a taller cone—a half circle or a sector that covers three-quarters of a circle? Assume that both sectors are cut from congruent circles.*

Make a model of each sector:

1. The half circle makes a cone that has a height that is about equal to the radius of the semi-circle.

2. The three-quarters sector gives a cone that has a *wider base* (greater diameter) but its height as *not* as great as the half-circle cone.

**Example 2**

*Predict which will be greater in height—a cone made from a half-circle sector or a cone made from a one-third-circle sector. Again, assume that both sectors are cut from congruent circles.*

The relationship in Example #1 on the previous page holds true—the *greater* (in degrees) the **sector**, the *smaller* the *height* of the cone.

In other words, the fraction \begin{align*}\frac{1}{3}\end{align*}*greater height* than a one-half sector.

- The
*larger*the**sector,**the __________________________ the*height*of its**cone.**

**Example 3**

*Predict which will be greater in diameter—a cone made from a half-circle sector or a cone made from a one-third-circle sector. Assume that the sectors are cut from congruent circles.*

Here you have the opposite relationship—the *larger* (in degrees) the **sector**, the *greater* the *diameter* of the cone.

In other words, \begin{align*}\frac{1}{2}\end{align*} is greater than \begin{align*}\frac{1}{3}\end{align*}, so a one-half sector will create a cone with *greater diameter* than a one-third sector.

- The
*larger*the**sector,**the ________________________ the*diameter*of its**cone.**

## Surface Area of a Regular Cone

The **surface area** of a **regular pyramid** is given by:

\begin{align*}A = \left ( \frac{1}{2} l P \right ) + B\end{align*}

where \begin{align*}l\end{align*} is the *slant height* of the figure, \begin{align*}P\end{align*} is the *perimeter* of the *base*, and \begin{align*}B\end{align*} is the *area* of the *base.*

Imagine a series of **pyramids** in which \begin{align*}n\end{align*}, the number of *sides* of each figure’s *base,* increases.

As you can see, as \begin{align*}n\end{align*} increases, the figure more and more resembles a *circle.*

You can also think of this as: a *circle* is like a polygon with an *infinite* number of sides that are infinitely small.

Similarly, a **cone** is like a **pyramid** that has an *infinite* number of *sides* that are infinitely small in length.

As a result, the formula for finding the total **surface area** of a **cone** is similar to the **pyramid** formula. The only difference between the two is that the **pyramid** uses \begin{align*}P\end{align*}, the *perimeter* of the base, while a cone uses \begin{align*}C\end{align*}, the *circumference* of the base.

\begin{align*}A ( \text{pyramid} ) = \frac{1}{2} l P + B && and && A ( \text{cone}) = \frac{1}{2}l C + B\end{align*}

Since the *circumference* of a circle is \begin{align*}2 \pi r\end{align*}:

\begin{align*}A ( \text{cone} ) = \frac{1}{2} l C + B = \frac{1}{2} l (2 \pi r) + B = \pi rl + B\end{align*}

You can also express \begin{align*}B\end{align*} as \begin{align*}\pi r^2\end{align*} to get:

## Surface Area of a Right Cone

\begin{align*}A ( \text{cone} ) &= \pi rl + B\\ &= \pi rl + \pi r^2\\ &= \pi r ( l + r )\end{align*}

Any of these forms of the equation can be used to find the **surface area** of a **right cone.**

**Reading Check:**

*There are a few different formulas to find the surface area of a cone. Pick one formula and describe what every variable represents.*

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

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

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

**Example 4**

*Find the total surface area of a right cone with a radius of* 8 cm *and a slant height of* 10 cm.

Use the formula:

\begin{align*}A ( \text{cone} ) & = \pi r ( l + r )\\ & = \pi(8) \cdot (10 + 8)\\ & = 8 \pi \cdot 18\\ & = 144 \pi \ cm^2 \quad \text{or}\\ & \approx (144)(3.14) = 452.16 \ cm^2\end{align*}

The exact area of the cone is \begin{align*}144 \pi \ cm^2\end{align*} and the approximate area is \begin{align*}452.16 \ cm^2\end{align*}.

**Example 5**

*Find the total surface area of a right cone with a radius of* 3 feet *and an altitude (not slant height) of* 6 feet.

Use the Pythagorean Theorem to find the slant height \begin{align*}l\end{align*}:

\begin{align*}r^2 + h^2 &= l^2\\ 3^2 + 6^2 &= l^2\\ 9 + 36 &= l^2\\ 45 &= l^2\\ \sqrt{45} &= l\\ 3 \sqrt{5} &= l\end{align*}

Now use the area formula:

\begin{align*}A ( \text{cone} ) &= \pi r ( l + r )\\ &= \pi (3) \cdot ( 3 \sqrt{5} + 3)\\ &= 3 \pi ( 3 \sqrt{5} + 3)\end{align*}

If we leave this as an *exact* answer, we cannot simplify anymore. This would be an ideal time to use a decimal approximation with a calculator:

\begin{align*}3 \pi ( 3 \sqrt{5} + 3) \approx 3(3.14)( 3 \sqrt{5} + 3) = 91.45 \ cm^2\end{align*}

The surface area of the cone is approximately \begin{align*}91.45 \ cm^2\end{align*}.

## Volume of a Cone

Which has a *greater volume,* a **pyramid, cone,** or **cylinder** if the figures have bases with the same "diameter" (i.e., distance across the base) and the same altitude?

To find out, compare **pyramids, cylinders,** and **cones** that have *bases* with *equal diameters* and the *same altitude.*

Here are three figures that have the same dimensions—cylinder, a right regular hexagonal pyramid, and a right circular cone. Which figure appears to have a greater volume?

It seems obvious that the volume of the **cylinder** is *greater* than the other two figures. This is because the pyramid and cone taper off to a single point, while the cylinder’s sides stay the same width.

Determining whether the **pyramid** or the **cone** has a *greater volume* is not so obvious. If you look at the *bases* of each figure you see that the **apothem** of the hexagon is *congruent* to the **radius** of the circle. You can see the relative size of the two bases by superimposing one onto the other:

From the diagram you can see that the hexagon is slightly *larger* in *area* than the circle.

Therefore, the **volume** of the right hexagonal regular **pyramid** would be *greater* than the **volume** of a right circular **cone.** It is, but only because the *area* of the **base** of the *hexagon* is slightly *greater* than the *area* of the **base** of the *circular cone.*

- When comparing the
**volumes**of a**cylinder,**a**pyramid,**and a**cone,**the __________________________ has the*largest*volume and the __________________________ has the*smallest*volume. The __________________________ has a volume in between the other two shapes.

The formula for finding the **volume** of each figure is virtually identical. Both formulas follow the same basic form:

\begin{align*}V = \frac{1}{3} Bh\end{align*}

Since the *base* of a circular **cone** is, by definition, a **circle**, you can substitute the area of a circle, \begin{align*}\pi r^2\end{align*} for the *base* of the figure. This is expressed as a volume postulate for cones.

- Instead of using \begin{align*}B\end{align*} for base area, we use the area of a _____________________, \begin{align*}\pi r^2\end{align*}, in the formula for volume of a cone.

**Volume of a Right Circular Cone**

Given a right circular **cone** with height \begin{align*}h\end{align*} and a base that has radius \begin{align*}r\end{align*}:

\begin{align*}V &= \frac{1}{3} Bh\\ V &= \frac{1}{3} \pi r^2 h\end{align*}

**Example 6**

*Find the volume of a right cone with a radius of* 9 cm *and a height of* 16 cm.

Use the formula: \begin{align*}V = \frac{1}{3} \pi r^2 h\end{align*}

Substitute the values for \begin{align*}r =\end{align*} ___________ and \begin{align*}h =\end{align*} _____________ :

\begin{align*}V & = \frac{1}{3} \pi (9^2)(16)\\ V & = \frac{1296 \pi}{3} = 432 \pi \ cm^3 \quad \text{or}\\ & \approx (432)(3.14) = 1356.48 \ cm^3\end{align*}

The cone has an exact volume of \begin{align*}432 \pi\end{align*} cubic centimeters and an approximate volume of 1356.48 cubic centimeters.

*By now, you have seen the units \begin{align*}cm^2\end{align*} or \begin{align*}in^2\end{align*} and \begin{align*}cm^3\end{align*} or \begin{align*}in^3\end{align*} in the examples.*

*When we calculate area, we use a “square” unit, such as \begin{align*}cm^2\end{align*} (square centimeters) or \begin{align*}in^2\end{align*} (square inches)*

*When we calculate volume, we use a “cubic” unit, such as \begin{align*}cm^3\end{align*} (cubic centimeters) or \begin{align*}in^3\end{align*} (cubic inches)*

**Example 7**

*Find the volume of a right cone with a radius of* 10 feet *and a slant height of* 13 feet.

Use the Pythagorean theorem to find the height \begin{align*}h\end{align*}:

\begin{align*}r^2 + h^2 & =l2\\ 10^2 + h^2 & =132\\ 100 + h^2 & =169\\ h^2 & = 169 - 100 =69\\ h & = \sqrt{69} \approx 8.31 \ ft\end{align*}

Now use the volume formula: \begin{align*}V = \frac{1}{3} \pi r^2 h\end{align*}

Substitute the values for \begin{align*}r =\end{align*} ___________ and \begin{align*}h =\end{align*} _____________ :

\begin{align*}V &= \frac{1}{3} \pi (10^2)(8.31)\\ V & = \frac{831 \pi}{3} = 277 \pi \ ft^3 \quad \text{or}\\ & \approx (277)(3.14) = 869.78 \ ft^3\end{align*}

The cone’s volume can be written as \begin{align*}277 \pi \ ft^3\end{align*} or \begin{align*}869.78 \ ft^3\end{align*}.

**Reading Check:**

1. *What type of units are used to express volume? What type for area?*

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

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

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

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

2. *What shape is the base of a right circular cone?*

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

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

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

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

3. *When calculating the volume of a cone, what information do you need?*

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

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

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

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

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