# 11.03 Volume of pyramids and cones

Lesson

### Pyramids

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.

#### Exploration

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.

$\text{Volume of Cube }=s^3$Volume of Cube =s3

$\text{Volume of one of the Pyramids }=\frac{s^3}{6}$Volume of one of the Pyramids =s36

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×s2= $\frac{s^3}{2}$s32

We know that the volume of the pyramid is $\frac{s^3}{6}$s36 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 $\frac{s^3}{2}$s32.

 $\frac{s^3}{6}$s36​ $=$= $\frac{1}{3}\times\frac{s^3}{2}$13​×s32​ Breaking $\frac{s^3}{2}$s32​ into two factors $\text{Volume of pyramid}$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.

Volume of Pyramid

$\text{Volume of Pyramid }=\frac{1}{3}\times\text{Area of base }\times\text{Height }$Volume of Pyramid =13×Area of base ×Height

#### Practice questions

##### question 1

Find the volume of the square pyramid shown.

##### question 2

A small square pyramid of height $4$4 cm was removed from the top of a large square pyramid of height $8$8 cm forming the solid shown. Find the exact volume of the solid.

##### question 3

A right square pyramid has a height of $24$24 cm and a volume $2592$2592 cm3. What is its base length of the pyramid?

### Cones

The volume of a cone has the same relationship to a cylinder as we just saw that a pyramid has with a prism.

That is:

Volume of Right Cone

$\text{Volume of Right Cone }=\frac{1}{3}\times\text{Area of Base }\times\text{Height of cylinder}$Volume of Right Cone =13×Area of Base ×Height of cylinder

$V=\frac{1}{3}\pi r^2h$V=13πr2h

The mathematical derivation of the formula for the volume of a cone is beyond this level of mathematics, so for now it is suffice to know the rule and how to use it.

#### Practice questions

##### QUESTIOn 4

Find the volume of the cone shown.

##### QUESTION 6

Find the radius of a cone that has a volume of $12441.02$12441.02 cm3 and a height of $30$30 cm.