How to calculate the Volume of a Frustum

Learning how to calculate the volume of a frustum comes down to one idea: complete the cone or pyramid, then subtract the part that was cut off. Edexcel IGCSE Maths calls a frustum a truncated cone or a truncated pyramid, and this page covers both, with clear diagrams and worked examples. When you are ready, test yourself on the randomly generated, auto-marked practice questions below.

Prior Knowledge This page requires confidence with Volume of Prisms and Cylinders, Similar Shapes and Scale Factors and Circles, Arcs and Sectors.
⇩ Jump to Practice Questions ⇩

What is a Frustum?

A frustum is the solid left behind when the top of a cone or a pyramid is sliced off by a cut parallel to the base. Edexcel IGCSE Maths questions usually say truncated cone or truncated pyramid, which means exactly the same thing: truncated is just the word for "shortened by having its top cut off". A plant pot, a lampshade and a bucket are all everyday frustums.

Truncated cone

Truncated pyramid

Slice parallel to the base, keep the bottom. The grey dashed part is the small cone or pyramid that was removed; the orange solid that remains is the frustum.

How to Calculate the Volume of a Frustum

There is no new formula to learn: a frustum is one solid subtracted from another. Every question uses the same three steps.

  1. Complete the solid. Extend the sloping sides until they meet, rebuilding the whole cone or pyramid. Sketch it: the frustum sits at the bottom, the removed top sits above the cut.
  2. Find the full height with similar triangles. The whole solid and the removed top are similar shapes, so matching measurements are in the same ratio. Compare the two radii (or the two base edges) to find the heights you were not given.
  3. Subtract the volumes. Work out the volume of the whole solid, then the volume of the removed top, and subtract.
The frustum formula \[ V_{\text{frustum}} = V_{\text{whole}} - V_{\text{top}} \]

volume of the whole solid, minus the volume of the removed top

Cone:  \( V = \dfrac{1}{3}\pi r^2 h \) Pyramid:  \( V = \dfrac{1}{3} \times \text{base area} \times h \)
Complete the Solid Rebuild the whole cone or pyramid first. The frustum is what is left when the top is subtracted, so you always work with two complete solids.
Similar Triangles Find the Height The removed top is a scaled-down copy of the whole solid. The ratio of the radii (or base edges) equals the ratio of the heights.
Subtract, Never Shortcut Volume of the whole solid minus volume of the removed top. Never subtract the lengths first: volume does not scale in a straight line.

Worked Examples

💡 Example 1: both cones given

A cone with base radius 9 cm and height 12 cm has its top removed. The removed cone has radius 3 cm and height 4 cm. Find the volume of the frustum that remains, to 3 significant figures.

\[ \begin{array}{rcl} V_{\text{whole}} &=& \frac{1}{3}\pi \times 9^2 \times 12 \\ &=& 324\pi \\ V_{\text{top}} &=& \frac{1}{3}\pi \times 3^2 \times 4 \\ &=& 12\pi \\ V_{\text{frustum}} &=& 324\pi - 12\pi \\ &=& 312\pi \\ &=& 980\ \text{cm}^3 \end{array} \]
What's happening?

Both cones are fully described, so no similar triangles are needed: find each volume with \( \dfrac{1}{3}\pi r^2 h \) and subtract.

Keep the answer in terms of \( \pi \) until the last line, then round once.

💡 Example 2: find the full height first

A frustum of a cone has base radius 10 cm, top radius 4 cm and height 9 cm. Find its volume, to 3 significant figures.

Let the full cone have height \( H \):

\[ \begin{array}{rcl} \dfrac{H-9}{H} &=& \dfrac{4}{10} \\ 10(H-9) &=& 4H \\ 10H - 90 &=& 4H \\ 6H &=& 90 \\ H &=& 15 \end{array} \] \[ \begin{array}{rcl} V &=& \frac{1}{3}\pi \times 10^2 \times 15 - \frac{1}{3}\pi \times 4^2 \times 6 \\ &=& 500\pi - 32\pi \\ &=& 468\pi \\ &=& 1470\ \text{cm}^3 \end{array} \]
What's happening?

Only the frustum is described, so first rebuild the whole cone. The removed top is similar to the whole cone: its radius is \( \dfrac{4}{10} \) of the base radius, so its height \( H - 9 \) is \( \dfrac{4}{10} \) of \( H \).

Solving gives a full height of 15 cm, so the removed cone has height \( 15 - 9 = 6 \) cm. Then subtract as usual.

💡 Example 3: truncated pyramid

A truncated pyramid has a square base of edge 12 cm, a square top of edge 4 cm, and height 6 cm. Find its volume.

Let the full pyramid have height \( H \):

\[ \begin{array}{rcl} \dfrac{H-6}{H} &=& \dfrac{4}{12} \\ 12(H-6) &=& 4H \\ 12H - 72 &=& 4H \\ 8H &=& 72 \\ H &=& 9 \end{array} \] \[ \begin{array}{rcl} V &=& \frac{1}{3} \times 12^2 \times 9 - \frac{1}{3} \times 4^2 \times 3 \\ &=& 432 - 16 \\ &=& 416\ \text{cm}^3 \end{array} \]
What's happening?

Exactly the same method as a cone, with \( V = \dfrac{1}{3} \times \text{base area} \times h \) instead. The base edges play the role the radii played.

The removed pyramid has height \( 9 - 6 = 3 \) cm and base edge 4 cm. Here everything is exact, so no rounding is needed.

💡 Example 4: a frustum in context

A plant pot is a truncated cone, 20 cm deep, with a top diameter of 24 cm and a base diameter of 16 cm. Find the volume of the pot, to 3 significant figures.

Radii: \( 24 \div 2 = 12 \) and \( 16 \div 2 = 8 \). Let the full cone have height \( H \):

\[ \begin{array}{rcl} \dfrac{H-20}{H} &=& \dfrac{8}{12} \\ 12(H-20) &=& 8H \\ 12H - 240 &=& 8H \\ 4H &=& 240 \\ H &=& 60 \end{array} \] \[ \begin{array}{rcl} V &=& \frac{1}{3}\pi \times 12^2 \times 60 - \frac{1}{3}\pi \times 8^2 \times 40 \\ &=& 2880\pi - \frac{2560\pi}{3} \\ &=& 6366.9\ldots \\ &=& 6370\ \text{cm}^3 \end{array} \]
What's happening?

Two traps in one question. First, these are diameters: halve them before doing anything else.

Second, the pot is widest at the top, so the "base" of the imaginary cone is the pot's top rim. Picture the cone point-down: the removed tip is the similar cone below the pot's base, with height \( 60 - 20 = 40 \) cm.

🔑 Key Points

  • A frustum (a truncated cone or pyramid) is what remains when the top is cut off parallel to the base.
  • One method every time: complete the solid, then \( V_{\text{frustum}} = V_{\text{whole}} - V_{\text{top}} \).
  • The removed top is similar to the whole solid: the ratio of radii or base edges equals the ratio of heights.
  • Both volume formulae carry the \( \dfrac{1}{3} \), and every answer is in cubic units.

⚠️ Pitfalls

  • Using the frustum's height as the height of the whole cone or pyramid. The whole solid is taller: find its height with similar triangles first.
  • Subtracting lengths and then cubing, or scaling volume by the length ratio. Volume scales with the cube of the length ratio, so always subtract the two volumes.
  • Mixing up diameter and radius: halve any diameter before it goes anywhere near \( \dfrac{1}{3}\pi r^2 h \).
  • Rounding too early. Keep values exact (in terms of \( \pi \) where you can) and round once, at the end.
⇩ Practice Questions ⇩

The similar-shapes idea behind the frustum method has a whole topic of its own: length, area and volume scale factors.

Similar Shapes →

Volume of a Frustum: Practice Room

Practise how to calculate the volume of a frustum with randomly generated, auto-marked questions covering both truncated cones and truncated pyramids. Every question uses the same plan: complete the whole solid, find its full height with similar triangles if it is not given, then subtract the removed top. Room 0 warms up the cone and pyramid volume formulae if you need them first. Give each answer to 3 significant figures (exact answers are accepted too) and type the number only; the units are in the question. Diagrams are not drawn to scale. Questions are randomly generated, so refresh a room for a new set. Difficulty rises down each column.

✓ Correct 0 ✗ Re-attempts 0 🔥 Streak 0 🏆 Best 0

Video tutorial on how to calculate the Volume of a Frustum

The following video goes into much more detail and will explain the process of calculating the volume of a frustum visually.