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Volume of a Pyramid, Cone and Sphere

Pointed and round solids each have their own volume rule, but they share a friendly pattern. Learn the three formulas, watch which height goes in, and these questions become quick wins.

MYP 5Standard MathsMensurationCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

Volume of a pyramid

A pyramid rises from a flat base to a single point, the apex. Its volume is exactly one third of the prism that would sit on the same base with the same height. The rule is:

\[ V = \tfrac13 \times \text{base area} \times h \]

Here \(h\) is the perpendicular height, the straight-up distance from the base to the apex, not the slanting edge. Work out the base area first (a square base of side \(a\) has area \(a^2\)), then take a third of base times height.

The one third links them all

A cone is just a pyramid with a circular base, so it carries the same one third. Spotting this means you really only have two new ideas here: the factor of a third for anything that comes to a point, and the separate rule for a sphere.

Volume of a cone

A cone has a circular base of radius \(r\) and rises to a point. Since the base area is \(\pi r^2\), slot that into the pyramid rule:

\[ V = \tfrac13 \pi r^2 h \]

Again \(h\) is the perpendicular height from the centre of the base up to the apex. It is easy to confuse this with the slant height along the surface, so check which one a question gives you.

Volume of a sphere

A sphere is a perfectly round ball with every surface point the same distance \(r\) from the centre. Its volume depends only on the radius:

\[ V = \tfrac43 \pi r^3 \]

Note the radius is cubed, not squared, so a small change in \(r\) makes a large change in volume. If you are given the diameter, halve it to get \(r\) before cubing.

A worked example

Worked example

Find the volume of a cone with radius 6 cm and perpendicular height 10 cm. Give your answer to 1 decimal place.

1
Write the rule: \(V = \tfrac13 \pi r^2 h\).
2
Substitute \(r = 6\) and \(h = 10\): \(V = \tfrac13 \pi \times 6^2 \times 10 = \tfrac13 \pi \times 36 \times 10\).
3
Simplify inside: \(36 \times 10 = 360\), and \(\tfrac13 \times 360 = 120\), so \(V = 120\pi\).
4
Evaluate: \(120\pi = 376.99\ldots\)
\(V \approx 377.0\) cm\(^3\)

Where this is assessed

Recalling and using the right formula is Criterion A (Knowing and understanding). Clear, ordered working with the third handled correctly is Criterion C (Communicating). Estimating how much a conical cup or a ball of ice holds is Criterion D (Applying maths in context). Read carefully whether the height given is perpendicular or slant.

Check yourself

1. Find the volume of a square-based pyramid with base side 6 cm and perpendicular height 10 cm. +

Base area \(= 6^2 = 36\) cm\(^2\). \(V = \tfrac13 \times 36 \times 10 = \tfrac13 \times 360 = \mathbf{120}\) cm\(^3\).

2. Find the volume of a sphere with radius 3 cm (to 1 d.p.). +

\(V = \tfrac43 \pi r^3 = \tfrac43 \pi \times 3^3 = \tfrac43 \pi \times 27 = 36\pi = 113.09\ldots \approx \mathbf{113.1}\) cm\(^3\).

3. Find the volume of a cone with radius 5 cm and perpendicular height 12 cm (to 1 d.p.). +

\(V = \tfrac13 \pi r^2 h = \tfrac13 \pi \times 5^2 \times 12 = \tfrac13 \pi \times 25 \times 12 = \tfrac13 \pi \times 300 = 100\pi = 314.15\ldots \approx \mathbf{314.2}\) cm\(^3\).

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