### Cones

A **cone** is a solid with a circular base and sides that taper up towards a vertex. A cone is generated from rotating a right triangle, around one leg. A cone has a **slant height**.

#### Surface Area

**Surface area** is a two-dimensional measurement that is the total area of all surfaces that bound a solid. The basic unit of area is the square unit. For the surface area of a cone we need the sum of the area of the base and the area of the sides.

**Surface Area of a Right Cone:** \begin{align*}SA=\pi r^2+\pi rl\end{align*}

Area of the base: \begin{align*}\pi r^2\end{align*}

Area of the sides: \begin{align*}\pi rl\end{align*}

#### Volume

To find the **volume** of any solid you must figure out how much space it occupies. The basic unit of volume is the cubic unit.

**Volume of a Cone:** \begin{align*}V=\frac{1}{3} \pi r^2 h\end{align*}

What if you were given a three-dimensional solid figure with a circular base and sides that taper up towards a vertex? How could you determine how much two-dimensional and three-dimensional space that figure occupies?

### Examples

#### Example 1

The surface area of a cone is \begin{align*}36 \pi\end{align*}

Plug what you know into the formula for the surface area of a cone and solve for \begin{align*}l\end{align*}

\begin{align*}36 \pi &= \pi 4^2+\pi 4l\\
36 &= 16+4l \qquad When \ each \ term \ has \ a \ \pi, \ they \ cancel \ out.\\
20 &= 4l\\
5 &= l\end{align*}

#### Example 2

The volume of a cone is \begin{align*}484 \pi \ cm^3\end{align*}

Plug what you know to the volume formula.

\begin{align*}484 \pi &= \frac{1}{3} \pi r^2 (12)\\
121 &= r^2\\
11 \ cm &= r\end{align*}

#### Example 3

What is the surface area of the cone?

First, we need to find the slant height. Use the Pythagorean Theorem.

\begin{align*}l^2 &= 9^2+21^2\\
&= 81+441\\
l &= \sqrt{522} \approx 22.85\end{align*}

The total surface area, then, is \begin{align*}SA=\pi 9^2+\pi (9)(22.85) \approx 900.54 \ units^2\end{align*}

#### Example 4

Find the volume of the cone.

First, we need the height. Use the Pythagorean Theorem.

\begin{align*}5^2+h^2 &=15^2\\ h &= \sqrt{200}=10\sqrt{2}\\ V &= \frac{1}{3}(5^2)\left(10\sqrt{2}\right) \pi \approx 370.24 \ units^3\end{align*}

#### Example 5

Find the volume of the cone.

We can use the same volume formula. Find the *radius.*

\begin{align*}V=\frac{1}{3} \pi (3^2)(6)=18 \pi \approx 56.55 \ units^3\end{align*}

### Review

Use the cone to fill in the blanks.

- \begin{align*}v\end{align*} is the ___________.
- The height of the cone is ______.
- \begin{align*}x\end{align*} is a __________ and it is the ___________ of the cone.
- \begin{align*}w\end{align*} is the _____________ ____________.

Sketch the following solid and answer the question. Your drawing should be to scale, but not one-to-one. Leave your answer in simplest radical form.

- Draw a right cone with a radius of 5 cm and a height of 15 cm. What is the slant height?

Find the slant height, \begin{align*}l\end{align*}, of one lateral face in the cone. Round your answer to the nearest hundredth.

Find the surface area and volume of the right cones. Round your answers to 2 decimal places.

- If the lateral surface area of a cone is \begin{align*}30 \pi \ cm^2\end{align*} and the radius is 5 cm, what is the slant height?
- If the surface area of a cone is \begin{align*}105 \pi \ cm^2\end{align*} and the slant height is 8 cm, what is the radius?
- If the volume of a cone is \begin{align*}30 \pi \ cm^3\end{align*} and the radius is 5 cm, what is the height?
- If the volume of a cone is \begin{align*}105 \pi \ cm^3\end{align*} and the height is 35 cm, what is the radius?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 11.6.