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9.07 Further composite solids

Lesson

Introduction

Finding the volume and surface area of straight-edged composite solids is generally a case of adding together the area or volume of the basic shapes that make up the solid.

When our composite solids are also made from pyramids and curved surface solids like cones and spheres, there are a couple of new techniques that can be useful for calculating their surface area or volume.

Volume of truncated pyramids and cones

To truncate a solid is to cut off some part of it in order to shorten it. We can do this to pyramids and cones by cutting off the tops to get composite solids that look like this:

This image shows a truncated pyramid and a truncated cone where the tops have been removed.

A truncated pyramid is called a frustum and is made by subtracting a smaller pyramid from a larger pyramid.

Similarly, a truncated cone is made by subtracting a smaller cone from a larger cone.

It is worth noting that the smaller sections that are removed will always be similar to the larger, original solid.

We can find the volume of a frustum or truncated cone by subtracting the volume of the removed section from the original solid.

Examples

Example 1

The solid is a truncated cone, formed by cutting off a cone shaped section from the top of a larger, original, cone.

A truncated cone formed by cutting off a cone section from the top of the original. Ask your teacher for more information.
a

What was the volume of the cone section that was cut off? Give your answer as an exact value.

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 are given r=2 and h=6.

\displaystyle V=\displaystyle =\displaystyle \dfrac{1}{3}\pi r^2 hWrite the formula
\displaystyle =\displaystyle \dfrac{1}{3}\times\pi\times 2^2 \times 6Substitute the values
\displaystyle =\displaystyle 8\pi \text{ unit}^3Evaluate
b

What was the volume of the original cone? Give your answer as an exact value.

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 are given r=3 and h=6+3=9.

\displaystyle V=\displaystyle =\displaystyle \dfrac{1}{3}\pi r^2 hWrite the formula
\displaystyle =\displaystyle \dfrac{1}{3}\times\pi\times 3^2 \times 9Substitute the values
\displaystyle =\displaystyle 27\pi \text{ unit}^3Evaluate
c

Find the volume of the truncated cone. Round your answer to two decimal places.

Worked Solution
Create a strategy

Subtract the volume of the cut off cone from the original cone.

Apply the idea
\displaystyle V\displaystyle =\displaystyle V_{\text{original}}-V_{\text{cut off}}Subtract the two volumes
\displaystyle =\displaystyle 27\pi -8\pi Substitute the values
\displaystyle =\displaystyle 59.69 \text{ unit}^3Evaluate
Idea summary

To find the volume of a truncated pyramid, we subtract the volume of the cut off pyramid from the volume of the whole original pyramid.

Similarly, to find the volume of a truncated cone, we subtract the volume of the cut off cone from the volume of the whole original cone.

Surface area of truncated pyramids and cones

When we unfold a frustum for its net we get two base faces, the top and the bottom, and some trapeziums.

This image shows a square-based frustum and its net of two squares and four trapeziums.

This square-based frustum has a net consisting of two squares and four trapeziums.

To find the surface area of a frustum, we add the base areas to the areas of the various trapeziums. So what about the surface area of a truncated cone?

Since there isn't a curved equivalent to a trapezium, we will need a different method to find the surface area of a truncated cone. This method involves taking the difference between the curved surface areas of the original and removed cone sections.

Consider the truncated cone below. If we unfold it, we get the net on the right.

This image shows truncated cone. Ask your teacher for more information.
A truncated cone
The net of a truncated cone made of 2 circles and part of an annulus.
The net of a truncated cone

Notice that we have two circular bases, the top and the bottom, and a surface formed by subtracting a small sector from a larger sector.

In other words, the surface area of a truncated cone is equal to the surface area of the original cone, minus the curved surface area of the removed cone, plus the base area of the removed cone.

For the truncated cone above, we can calculate each of those areas to be: \begin{aligned} \text{Surface area of the original cone}&=\pi \times 5^2+\pi \times 5\times 20=125\pi \\ \text{Curved area of the removed cone}&=\pi \times 5^2+\pi \times 5\times 20=36\pi \\ \text{Base area of the removed cone}&=\pi \times 3^2=9\pi \end{aligned}

As such, the surface area of this truncated cone is:\text{Surface area of the truncated cone}=125\pi -36\pi +9\pi=100\pi

These types of calculations work particularly well with cones because the original and removed sections are always similar solids.

Examples

Example 2

A small square pyramid of height 5 cm was removed from the top of a large square pyramid of height 10 cm to form the solid shown.

A truncated square pyramid of base length 4 and height 5 centimetres. Ask your teacher for more information.
a

Find the length of the slant height of the sides of the new solid. Round your answer to two decimal places.

Worked Solution
Create a strategy

Use Pythagoras' theorem c^2=a^2+b^2 to find the slant height of the large pyramid and the smaller pyramid, then subtract them.

Apply the idea
A right angled triangle with base 2 and height 10. Inside at the top is a right angled triangle with base 1 and height 5.

We can visualise right-angled triangles to find the slant height.

To find the slant height of the original pyramid, we can use Pythagoras' theorem with a=2, \, b=10.

To find the slant height of the cut out pyramid, we can use Pythagoras' theorem with a=1, \, b=5.

To find the original slant height:

\displaystyle c^2\displaystyle =\displaystyle 2^2+10^2Substitute the values
\displaystyle c^2\displaystyle =\displaystyle 104Evaluate the powers
\displaystyle c\displaystyle =\displaystyle \sqrt{104}Take the square root of both sides

To find the cut out slant height:

\displaystyle c^2\displaystyle =\displaystyle 1^2+5^2Substitute the values
\displaystyle c^2\displaystyle =\displaystyle 26Evaluate the powers
\displaystyle c\displaystyle =\displaystyle \sqrt{26}Take the square root of both sides
\displaystyle \text{Slant height}\displaystyle =\displaystyle \sqrt{104}-\sqrt{26}Subtract the two slant heights
\displaystyle =\displaystyle 5.1 \text{ cm}Evaluate and round
b

Now find the surface area of the solid formed. Round your answer to one decimal place.

Worked Solution
Create a strategy

Add the areas of the four identical trapezium sides, the large square base and the small square top.

A large square plus a small square plus four identical trapeziums.

Use the area of the trapezium formula A=\dfrac{1}{2}\left(a+b\right)h and area of the square formula A=s^2.

Apply the idea

The side length of the bottom square is 4, and the side length of the top square is 2.

\displaystyle \text{Square areas}\displaystyle =\displaystyle 4^2+2^2Add the square areas
\displaystyle =\displaystyle 20 \text{ cm}^2Evaluate

Each trapezium has a height of h=5.1 and parallel sides of a=2,\, b=4.

\displaystyle 4\text{ trapeziums}\displaystyle =\displaystyle 4\times \dfrac{1}{2}\left(2+4\right)\times 5.1Multiply the trapezium area by 4
\displaystyle =\displaystyle 61.2\text{ cm}^2Evaluate
\displaystyle SA\displaystyle =\displaystyle 20+61.2Add the areas
\displaystyle =\displaystyle 81.2\text{ cm}^2Evaluate and round
Idea summary

To find the surface area of a truncated pyramid, we add the base areas to the areas of the side trapeziums.

To find the surface area of a truncated cone we find the surface area of the original cone, minus the curved surface area of the removed cone, plus the base area of the removed cone.

Other composite solids

In addition to truncated pyramids and cones, we can also make composite solids out of spheres, hemispheres, cylinders and other curved surface solids.

A composite solid made of a cone and cylinder and another composite solid made of a cone and hemisphere.

As we can see, while some dimensions will be common between the component solids, there is no guarantee that these parts will be similar in shape. As such, we can instead find the surface area and volume of these solids in the usual way - by adding and subtracting solids that we already know.

Examples

Example 3

Find the surface area of the composite figure shown, which consists of a cone and a hemisphere joined at their bases. Round your answer to two decimal places.

Composite figure consisting cone and hemisphere joined at their bases with radius of 4 centimetres and height 10 centimetres.
Worked Solution
Create a strategy

Add the curved surface area of the cone to the curved surface area of the hemisphere. The curved part of a cone has area A=\pi rs and the surface area of a sphere is 4\pi r^2.

Apply the idea

We use the Pythagora's theorem theorem to find the slope length s, where a=4 and b=10.

\displaystyle s^2\displaystyle =\displaystyle a^2+b^2Write the formula
\displaystyle =\displaystyle 4^2+10^2Substitute the values
\displaystyle s\displaystyle =\displaystyle \sqrt{4^2+10^2}Take the square root of both sides
\displaystyle =\displaystyle \sqrt{116}Evaluate
\displaystyle \text{Curved surface area of cone}\displaystyle =\displaystyle \pi r sWrite the formula
\displaystyle =\displaystyle \pi \times 4 \times \sqrt{116}Subsitute the dimensions
\displaystyle =\displaystyle 4 \pi \sqrt{116} \text{ cm}^2Evaluate
\displaystyle \text{Surface area of hemisphere}\displaystyle =\displaystyle 4\pi r^2 \times \dfrac{1}{2}Write the formula
\displaystyle =\displaystyle \dfrac{4\times \pi \times 4^2}{2}Subsitute the dimensions
\displaystyle =\displaystyle 32 \pi \text{ cm}^2Evaluate
\displaystyle \text{Surface area}\displaystyle =\displaystyle 32\pi + 4 \pi \sqrt{116}Add the areas
\displaystyle =\displaystyle 235.87 \text{ cm}^2Evaluate and round
Idea summary

We can find the surface area and volume of composite solids made of prisms, pyramids, cones, spheres and cylinders by adding and subtracting solids that we already know.

Outcomes

VCMMG365 (10a)

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

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