Lesson

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 $s$`s` units. Now lets divide the cube up into 6 simple pyramids by joining all the vertices to the midpoint of the cube.

This creates $6$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.

$\text{Volume of the cube }=s^3$Volume of the cube =`s`3

$\text{Volume of one of the pyramids }=\frac{s^3}{6}$Volume of one of the pyramids =`s`36

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 $l\times b\times h=s\times s\times\frac{s}{2}$`l`×`b`×`h`=`s`×`s`×`s`2= $\frac{s^3}{2}$`s`32.

We know that the volume of the pyramid is $\frac{s^3}{6}$`s`36 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 $\frac{s^3}{2}$`s`32.

$\frac{s^3}{6}$s36 |
$=$= | $\frac{1}{3}\times\frac{s^3}{2}$13×s32 |
Breaking $\frac{s^3}{2}$ |

$\text{Volume of pyramid}$Volume of pyramid | $=$= | $\frac{1}{3}\times\text{Volume of rectangular prism}$13×Volume of rectangular prism |
Using what we found in the diagrams |

$=$= | $\frac{1}{3}\times\text{Area of base}\times\text{height }$13×Area of base×height |
Previously shown |

So what we can see here is that the volume of the pyramid is $\frac{1}{3}$13 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!), so the volume for any pyramid is given by the formula:

Volume of a pyramid

$\text{Volume of a pyramid }=\frac{1}{3}\times\text{Base area }\times\text{Height }$Volume of a pyramid =13×Base area ×Height

$V=\frac{1}{3}Ah$`V`=13`A``h`

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:

Volume of Right Cone

$\text{Volume of a cone }=\frac{1}{3}\times\text{Base area }\times Height$Volume of a cone =13×Base area ×`H``e``i``g``h``t`

$V=\frac{1}{3}\pi r^2h$`V`=13π`r`2`h`

Where $r$`r` is the radius of the cone's base and $h$`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.

Find the volume of the square pyramid shown.

Find the volume of the following pyramid.

Give your answer in cubic centimetres.

Consider the cone below.

Find the perpendicular height $h$

`h`of the cone.Enter each line of working as an equation.

What is the volume of the cone in cubic metres?

Round your answer to three significant figures.

Solve problems involving surface area and volume of right pyramids, right cones, spheres and related composite solids.