**Frustums** *GCSE Revision and Worksheets*

## What you need to know

The **frustum **of a cone or a pyramid is the 3D shape that is left-over after you cut the top off (see the pictures below for examples). The slice the you make when cutting the top off will always be parallel to the base of the shape.

You are expected to know how to calculate both the surface area and volume of a frustum, so before you go any further, make sure you’re familiar with the volume/surface area of normal, unsliced cones and pyramid. If you’re not sure, click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/surface-area-gcse-revision-and-worksheets/) and here (https://mathsmadeeasy.co.uk/gcse-maths-revision/volume-of-3d-shapes-gcse-maths-revision-and-worksheets/) for more information.

The best way to work out volumes and surface areas for frustums is to firstly calculate what the volume/surface area of the shape would be __without__ the top sliced off (here we will call this the **whole pyramid/cone**), and then subtract the volume/surface area of the bit which has been sliced off (here we will call this the **missing portion**). Since the slice that has been made to form the frustum must be parallel, the bit that you cut off will be the same type of shape as the bigger one (just smaller).

**Example: **Below is a frustum of a square-based pyramid. The height of the square-based pyramid is 15cm and the height of the frustum is 6cm. Calculate the volume of the frustum.

The formula for the volume of a pyramid is

\frac{1}{3}\times\text{ area of base }\times\text{ height}

The base of this shape is square, so that won’t be any trouble. Therefore, we get the volume of the whole pyramid (without the top chopped off) to be

\dfrac{1}{3} \times 5^2 \times 15 = 125\text{cm}^2

Next, we consider the missing portion. It is also a square-based pyramid, its base has side-length 3cm, and its height is the height of the whole pyramid subtract the height of the frustum. So, we get the volume of the missing portion to be

\dfrac{1}{3} \times 3^2 \times (15 - 6) = 27\text{cm}^2

Therefore, the volume of the frustum is 125-27=98\text{cm}^2.

Before this next example, recall that if two shapes are similar then their side-lengths are linked by a scale factor. If you’re unfamiliar with this, head over here (https://mathsmadeeasy.co.uk/gcse-maths-revision/similar-shapes-gcse-revision-and-worksheets/).

**Example: **Below is a frustum of a cone. The height of the cone was originally 36mm, and the height of the missing portion of the cone is 12mm. The radius of the cone is 15mm, and its slant height is 39mm. Work out the surface area of the frustum. Leave the answer in terms of \pi.

To work out the surface area of a cone, you need the area of the circular face \left(\pi r^2\right) and the area of the curved face (\pi rl, where l is the ‘slant height’ of the cone). For the frustum, we will need part of the whole cone’s curved face, and then we will need to add both the top circular face and the bottom circular face.

To answer this question, you will need to understand that the missing portion is not only a cone, it is also *similar *to the whole cone. This is **true of all frustums** and is incredibly useful – it means there is a scale factor between the dimensions of the whole cone and the missing portion. We’re told that the height of the whole cone is 36mm and the height of the missing portion is 12mm, therefore the scale factor is 36\div 12=3. From this, we can determine:

\text{Slant height of missing portion } = 39 \div 3 = 13\text{mm}

\text{Radius of missing portion } = 15 \div 3 = 5\text{mm}

Now we have all the information we need. So, we get

\text{Curved area of the whole cone } = \pi \times 15 \times 39 = 585\pi\text{ mm}^2

\text{Curved area of the missing portion } = \pi \times 5 \times 13 = 65\pi\text{ mm}^2

Therefore, the curved area of the frustum is 585\pi - 65\pi = 520\pi\text{ mm}^2. On top of this, the area of the bottom circular face is \pi \times 15^2 = 225\pi\text{ mm}^2 and the area of the top circular face is \pi \times 5^2 = 25\pi\text {mm}^2. Therefore,

\text{Total surface area } = 520\pi + 225\pi + 25\pi = 770\pi\text{ mm}^2.

## Example Questions

1) Below is the frustum of a cone. The height of the cone is 50cm, the radius of the base of the cone is 10cm, and the height of the frustum is 30cm. Work out the volume of the frustum to 3sf.

The volume of a cone is \dfrac{1}{3}\pi r^2 h. To find the volume of the frustum we need the volume of the whole cone and the volume of the missing portion. To calculate the latter, we will need the radius of the top of the frustum (aka the base of the missing portion), and to find this we will use similarity.

The whole cone and the missing portion are similar. To find the scale factor between the dimensions of these two shapes, we must divide the height of one by the height of the other. The height of the missing portion is 50-30=20\text{cm}, so the scale factor is 50\div20 = 2.5. Therefore, the radius of the top face of the frustum is 10\div 2.5 = 4\text{cm}. So, we get

\text{Volume of whole cone }=\dfrac{1}{3}\pi\times 10^2\times 50=\dfrac{5000}{3}\pi

\text{Volume of missing portion }=\dfrac{1}{3}\pi\times 4^2\times 20=\dfrac{320}{3}\pi

Finally, to 3 significant figures, we get

\text{Volume of frustum } = \dfrac{5000}{3}\pi - \dfrac{320}{3}\pi = 4,900\text{ cm}^3

2) Below is a frustum of a square-based pyramid. The pyramid has its apex directly above the centre of the base. The pyramid has height 7.5m and its base has side-length 8m. The top face of the frustum has side-length 4m. Work out the surface area of the frustum. (HINT: 3D Pythagoras)

To work out the surface area of this frustum we need the area of both the top square and the bottom square as well as the 4 trapezia that form the sides of the frustum. Since the apex is directly above the centre of the base, all these trapezia will be identical.

Working out the area of the squares is easy. The formula for the area of a trapezium is

\dfrac{1}{2}(a + b)h

where a and b are the two parallel sides. In this case, we know the two parallel sides are lengths 4m and 8m, but we don’t know the perpendicular height of each trapezium face. To do this, we’re going to need, as the question suggests, some 3D trig.

By constructing a right-angled triangle between the apex, the centre of the base, and the centre of one of the base’s sides, we can find the height of one of the triangular sides of the whole pyramid.

We know that the base of this triangle will be half the base of the whole pyramid, so 4m, and we know the height is 7.5m. So, using Pythagoras, we get

c^2 = 4^2 + 7.5^2 \text{ therefore } c = \sqrt{72.25} = 8.5\text{m}.

As is always the case, the whole pyramid and the missing portion are similar shapes. Given that the base of one has width 8m and the base of the other has 4m, the scale factor is clearly 2. Therefore, now we know that the perpendicular height of one of the triangular faces of the whole pyramid is 8.5, we also know that the perpendicular height of one of the trapezoidal faces of the frustum is

8.5 \div 2 = 4.25\text{m}.

Therefore, we get

\text{Area of trapezium } = \dfrac{1}{2}(4 + 8)\times 4.25 = 25.5\text{ m}^2

To get the whole area, we must add 4 of these areas to the areas of the top face and the bottom face. We get

\text{Total area } = (4 \times 25.5) + 4^2 + 8^2 = 182\text{ m}^2.

## Frustums Revision and Worksheets

## Frustums Teaching Resources

Frustum questions have appeared in many GCSE Maths papers over the past few years and they have often been high mark questions. They are still relevant to the 9-1 GCSE Maths course and there is no reason why they can’t continue coming up in the exams. Whether you are a GCSE Maths tutor in Harrogate looking for frustum questions for your pupils or a maths teacher in York, you will find these resources useful.