9.07 Further composite solids

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:

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

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 h	Write the formula
	\displaystyle =	\displaystyle \dfrac{1}{3}\times\pi\times 2^2 \times 6	Substitute the values
	\displaystyle =	\displaystyle 8\pi \text{ unit}^3	Evaluate

Watch question walkthrough

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 h	Write the formula
	\displaystyle =	\displaystyle \dfrac{1}{3}\times\pi\times 3^2 \times 9	Substitute the values
	\displaystyle =	\displaystyle 27\pi \text{ unit}^3	Evaluate

Watch question walkthrough

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}^3	Evaluate

Watch question walkthrough

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

A truncated cone

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

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

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+10^2	Substitute the values
\displaystyle c^2	\displaystyle =	\displaystyle 104	Evaluate 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+5^2	Substitute the values
\displaystyle c^2	\displaystyle =	\displaystyle 26	Evaluate 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

Watch question walkthrough

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.

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

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 5.1	Multiply the trapezium area by 4
	\displaystyle =	\displaystyle 61.2\text{ cm}^2	Evaluate

\displaystyle SA	\displaystyle =	\displaystyle 20+61.2	Add the areas
	\displaystyle =	\displaystyle 81.2\text{ cm}^2	Evaluate and round

Watch question walkthrough

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

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.

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^2	Write the formula
	\displaystyle =	\displaystyle 4^2+10^2	Substitute 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 s	Write the formula
	\displaystyle =	\displaystyle \pi \times 4 \times \sqrt{116}	Subsitute the dimensions
	\displaystyle =	\displaystyle 4 \pi \sqrt{116} \text{ cm}^2	Evaluate

\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}^2	Evaluate

\displaystyle \text{Surface area}	\displaystyle =	\displaystyle 32\pi + 4 \pi \sqrt{116}	Add the areas
	\displaystyle =	\displaystyle 235.87 \text{ cm}^2	Evaluate and round

Watch question walkthrough