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Stage 5.1-3

11.11 Volume of pyramids and cones

Lesson

Volume of a pyramid

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.

This image shows a triangular pyramid, a square pyramid and a hexagonal pyramid.

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.

2 cubes. 1 cube has a pyramid inside with its base equal to a face of the cube and the vertex is at the centre of the cube.

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.

Two rectangular prisms on which they are divided into 2 simple pyramids. Ask your teacher for more information.

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}.

\displaystyle \dfrac{s^3}{6}\displaystyle =\displaystyle \dfrac{1}{3}\times \dfrac{s^3}{2}Breaking \dfrac{s^3}{2} into two factors
\displaystyle \text{Volume of pyramid}\displaystyle =\displaystyle \dfrac{1}{3}\times \text{Volume of rectangular prism}Using what we found in the diagrams
\displaystyle =\displaystyle \frac{1}{3}\times \text{Area of base}\times \text{Height }Previously shown

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.

A square pyramid with square side length of 9 centimetres and a height of 3 centimetres.
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 hWrite the formula
\displaystyle =\displaystyle \frac{1}{3}\times 9^2\times 3Subsitute the values
\displaystyle =\displaystyle 81 \text{ cm}^3Evaluate
Idea summary

The volume of a pyramid is given by:

\displaystyle V=\dfrac{1}{3} Ah
\bm{A}
is the area of the base of the pyramid
\bm{h}
is the height of the pyramid

Volume of a cone

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.

This image shows a cone with a diameter of 32 metres and slant height of 34 metres.
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^2Write the formula
\displaystyle 34^2\displaystyle =\displaystyle h^2+16^2Substitute the values
\displaystyle 1156\displaystyle =\displaystyle h^2+256Evaluate the powers
\displaystyle 900\displaystyle =\displaystyle h^2Subtract 256 from both sides
\displaystyle h\displaystyle =\displaystyle \sqrt{900}Square root both sides
\displaystyle h\displaystyle =\displaystyle 30 \text{ m}Evaluate
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 hWrite the formula
\displaystyle =\displaystyle \dfrac{1}{3}\times \pi \times 16^2 \times 30Substitute the values
\displaystyle =\displaystyle 8042 \text{ m}^3Evaluate
Idea summary

The volume of a cone is given by:

\displaystyle V=\dfrac{1}{3}\pi r^2h
\bm{r}
is the radius of the cone's base
\bm{h}
is the perpendicular height of the cone

Outcomes

MA5.3-14MG

applies formulas to find the volumes of right pyramids, right cones, spheres and related composite solids

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