11.11 Volume of pyramids and cones

A pyramid is formed when the vertices of a polygon are projected up to a common point (called a vertex). A right pyramid is formed when the apex is perpendicular to the midpoint of the base.

We want to be able to calculate the volume of a pyramid. Let's start by thinking about the square based pyramid.

Think about a cube, with side length s units. We can divide the cube up into 6 simple pyramids by joining all the vertices to the midpoint of the cube. One of these pyramids is shown below.

This would create 6 square based pyramids with the base equal to the face of one of the sides of the cube, and height, equal to half the length of the side. \begin{aligned} \text{Volume of the cube }&=s^3\\ \text{Volume of one of the pyramids }&=\frac{s^3}{6} \end{aligned}

Now lets think about the rectangular prism, that is half the cube. This rectangular prism has the same base as the pyramid and the same height as the pyramid.

Now the volume of this rectangular prism is l\times b\times h=s\times s\times \dfrac{s}{2}=\dfrac{s^3}{2}.

We know that the volume of the pyramid is \dfrac{s^3}{6} and the volume of the prism with base equal to the base of the pyramid and height equal to the height of the pyramid is \dfrac{s^3}{2}.

So what we can see here is that the volume of the pyramid is \dfrac{1}{3} of the volume of the prism with base and height of the pyramid.

Of course this is just a simple example so we can get the idea of what is happening. However, if we performed the same exercise with a rectangular prism we would actually get the same result (try it out). In fact, the volume for any pyramid is given by V =\dfrac{1}{3}Ah, where A is the area of the base of the pyramid and h is the height of the pyramid.

Examples

Example 1

Find the volume of the square pyramid shown.

Worked Solution

Create a strategy

Use the formula for the volume of a pyramid given by V=\dfrac{1}{3} A h, where the area of a square is given by A=s^2.

Apply the idea

We are given s=9 and h=3.

\displaystyle V	\displaystyle =	\displaystyle \dfrac{1}{3} s^2 h	Write the formula
	\displaystyle =	\displaystyle \frac{1}{3}\times 9^2\times 3	Subsitute the values
	\displaystyle =	\displaystyle 81 \text{ cm}^3	Evaluate

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The volume of a cone has the same relationship to a cylinder as we just saw that a pyramid has with a prism. As such, the volume of a cone can be found using the formula V=\dfrac{1}{3}\pi r^2h where r is the radius of the cone's base and h is the perpendicular height of the cone.

Due to its circular base and curved surface, the tools required to derive this formula lie outside the scope of this level of mathematics. Regardless, it suffices to know this formula and how to apply it when solving problems involving the volume of a cone.

Examples

Example 2

Consider the cone below.

a

Find the perpendicular height h of the cone.

Worked Solution

Create a strategy

Use Pythagoras' theorem.

Apply the idea

Based on the diagram, we are given a=h, c=34, and b=\dfrac{32}{2}=16.

\displaystyle c^2	\displaystyle =	\displaystyle a^2+b^2	Write the formula
\displaystyle 34^2	\displaystyle =	\displaystyle h^2+16^2	Substitute the values
\displaystyle 1156	\displaystyle =	\displaystyle h^2+256	Evaluate the powers
\displaystyle 900	\displaystyle =	\displaystyle h^2	Subtract 256 from both sides
\displaystyle h	\displaystyle =	\displaystyle \sqrt{900}	Square root both sides
\displaystyle h	\displaystyle =	\displaystyle 30 \text{ m}	Evaluate

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b

Determine the volume of the cone, to the nearest cubic metre.

Worked Solution

Create a strategy

Use the volume of the cone given by V=\dfrac{1}{3} \pi r^2 h.

Apply the idea

Based on the given diagram, we are given r=16.

\displaystyle V	\displaystyle =	\displaystyle \dfrac{1}{3} \pi r^2 h	Write the formula
	\displaystyle =	\displaystyle \dfrac{1}{3}\times \pi \times 16^2 \times 30	Substitute the values
	\displaystyle =	\displaystyle 8042 \text{ m}^3	Evaluate

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