Truncated pyramids and cones are made from the difference between similar figures, so we can calculate their volumes using the scale factor between the solids.

Examples

Example 1

A small square pyramid has a height of x \text{ m} and a base side length of y \text{ m}. A large pyramid has dimensions triple that of the small pyramid.

A square pyramid of base length y and height x.

What are the dimensions of the large pyramid?

Worked Solution

Create a strategy

Multiply the dimensions of the small pyramid by the scale factor 3.

Apply the idea

Height of the large pyramid is 3x \text{ m}.

Base side length of the large pyramid is 3y \text{ m.}

What is the volume of the large pyramid?

Worked Solution

Create a strategy

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

Apply the idea

A square pyramid of base length 3y and height 3x.

From part (a), we know that the large pyramid has a height h=3x and a base side length s=3y.

\begin{aligned} V &=\dfrac{1}{3}s^2 h \\ &=\dfrac{1}{3}\times (3y)^2 \times 3x \\ &= 9xy^2 \text{ m}^3 \end{aligned}

How many times can the volume of the small pyramid go into the volume of the large pyramid?

Worked Solution

Create a strategy

Cube the length scale factor.

Apply the idea

The dimensions of the small pyramid were multiplied by a scale factor of 3. So the volume scale factor is 3^3=27.

So the volume of the small pyramid can go into the volume of the large pyramid 27 times.

If the small pyramid has a volume of 46 \text{ m}^3, what is the volume of the large pyramid?

Worked Solution

Create a strategy

Multiply the given volume by the volume scale factor.

Apply the idea

We know from part (c) that the volume of the large pyramid is 27 times that of the small pyramid. So we have:

\displaystyle \text{Volume}	\displaystyle =	\displaystyle 46 \times 27	Multiply 46 by 27
	\displaystyle =	\displaystyle 1242 \text{ m}^3	Evaluate

Example 2

Lucy makes a truncated cone by cutting off a smaller cone halfway from the top, as shown in the diagram below:

A truncated cone of height x and radius 2 y. Ask your teacher for more information.

Find the exact volume of the original cone.

Worked Solution

Create a strategy

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

Apply the idea

We can see from the diagram that the original cone had a base radius of 2y and a height of x+x=2x.

\displaystyle V=	\displaystyle =	\displaystyle \dfrac{1}{3}\pi r^2 h	Write the formula
	\displaystyle =	\displaystyle \dfrac{1}{3}\times\pi\times (2y)^2 \times 2x	Substitute the values
	\displaystyle =	\displaystyle \dfrac{1}{3}\times\pi\times 4y^2 \times 2x	Evaluate (2y)^2
	\displaystyle =	\displaystyle \dfrac{8 \pi xy^2}{3} \text{ unit}^3	Simplify

Find the exact volume of the cone section that was cut from the original cone.

Worked Solution

Create a strategy

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

Apply the idea

We can see from the diagram that the removed cone section had a height of x, which means that the original height was multiplied by a scale factor of \dfrac{1}{2} since \dfrac{1}{2} \times 2x = x.

So the volume of the cut cone would be \left(\dfrac{1}{2}\right)^3 times the volume of the original cone.

\displaystyle V	\displaystyle =	\displaystyle \left(\dfrac{1}{2}\right)^3 \times \dfrac{8 \pi xy^2}{3}	Multiply the original volume by \left(\dfrac{1}{2}\right)^3
	\displaystyle =	\displaystyle \dfrac{1}{8} \times \dfrac{8 \pi xy^2}{3}	Evaluate \left(\dfrac{1}{2}\right)^3
	\displaystyle =	\displaystyle \dfrac{\pi xy^2}{3} \text{ unit}^3	Simplify

What fraction of the original cone did Lucy cut off?

Worked Solution

Create a strategy

Divide the volume of the removed section by the volume of the original cone.

Apply the idea

In part (b), the volume of the removed section was \dfrac{\pi xy^2}{3}, and in part (a), the volume of the original cone was \dfrac{8 \pi xy^2}{3}.

\displaystyle \text{Fraction}	\displaystyle =	\displaystyle \dfrac{\frac{\pi xy^2}{3}}{\frac{8 \pi xy^2}{3}}	Divide the two volumes
	\displaystyle =	\displaystyle \dfrac{\frac{\pi xy^2}{3}\times 3}{\frac{8 \pi xy^2}{3}\times 3}	Multiply both parts by 3
	\displaystyle =	\displaystyle \dfrac{\pi xy^2}{8 \pi xy^2}	Evaluate
	\displaystyle =	\displaystyle \dfrac{1}{8}	Simplify

If the original cone had a volume of 184 \text{ cm}^3, what is the exact volume of Lucy's truncated cone?

Worked Solution

Create a strategy

Multiply the volume of the original cone by the fraction of the remaining part of the truncated cone.

Apply the idea

From part (c), the cut off part of the cone was \dfrac{1}{8} of the original cone. So what is left is 1-\dfrac{1}{8} =\dfrac{7}{8} of the original cone.

\displaystyle \text{Volume remaining}	\displaystyle =	\displaystyle 184 \times \dfrac{7}{8}	Multiply the volume by \dfrac{7}{8}
	\displaystyle =	\displaystyle 161 \text{ cm}^3	Evaluate

Idea summary

For two similar figures with a scale factor of s, the volume of the larger figure will be equal to s^3 times greater than the volume of the smaller figure.

For a truncated cone or pyramid, if we know that the top part of the solid that was cut off has a height of \dfrac{1}{x} times the height of the original solid, then the cut off volume will be \dfrac{1}{x^3} times the original solid's volume.

Outcomes

MA5.3-13MG

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

MA5.3-14MG

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

MA5.3-16MG

proves triangles are similar, and uses formal geometric reasoning to establish properties of triangles and quadrilaterals

7.09 Further composite solids

Introduction

Ideas

Volumes of truncated pyramids and cones

Examples

Example 1

Example 2

Outcomes

MA5.3-13MG

MA5.3-14MG

MA5.3-16MG

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