Pyramids and cones are three-dimensional shapes that taper from a flat base to a point. The volume of a pyramid equals one-third of the base area multiplied by the perpendicular height; the volume of a cone is ⅓πr²h. Both formulae require the perpendicular height, not the slant height.
What is the formula for the volume of a pyramid?
A pyramid has a flat polygonal base and triangular faces that slope up to meet at a single point called the apex. The perpendicular height h is the shortest distance from the apex straight down to the base — measured at 90° to the base, not along any sloping edge.
V = ⅓ × base area × h
where A is the area of the base and h is the perpendicular height.
The table below places the pyramid formula alongside the corresponding prism and cylinder formulae so that the one-third pattern is immediately clear.
| Shape | Volume formula |
|---|---|
| Prism (any polygon base) | Base area × height |
| Pyramid (any polygon base) | ⅓ × base area × height |
| Cylinder | πr²h |
| Cone | ⅓πr²h |
A pyramid always holds exactly one-third of the volume of a prism with the same base and the same perpendicular height. A cone holds exactly one-third of the volume of a cylinder with the same radius and the same perpendicular height.
Why is there a factor of one-third?
The cleanest physical explanation: a cube can be dissected into exactly three congruent square pyramids, each sharing the same square base area and the same height as the cube. So one pyramid occupies one-third of the cube, giving V = ⅓ × base area × height.
A rigorous proof for an arbitrary pyramid uses calculus — integrating the areas of infinitely thin horizontal slices as they shrink towards the apex — but at GCSE you take the result as given. What you do carry into every question is the habit of writing ⅓ as the very first factor before substituting any numbers. That single habit prevents the most common error in this topic.
How do I calculate the volume of a pyramid or cone?
Work through the following three worked examples in order. They build from straightforward to slightly more involved.
Worked example 1 — square-based pyramid
A square-based pyramid has base side length 6 cm and perpendicular height 10 cm. Find its volume.
- Base area: A = 6 × 6 = 36 cm²
- Apply the formula: V = ⅓ × 36 × 10
- V = ⅓ × 360 = 120 cm³
Worked example 2 — triangular pyramid
A pyramid has a triangular base with base 8 cm and perpendicular height of the triangle 5 cm. The perpendicular height of the pyramid is 9 cm. Find the volume.
- Base area: A = ½ × 8 × 5 = 20 cm²
- Apply the formula: V = ⅓ × 20 × 9
- V = ⅓ × 180 = 60 cm³
Worked example 3 — cone
A cone has radius 4 cm and perpendicular height 9 cm. Find the volume. Give your answer to 1 decimal place.
- Apply V = ⅓ × π × r² × h
- V = ⅓ × π × 4² × 9 = ⅓ × π × 16 × 9 = ⅓ × 144π = 48π
- V = 48 × 3.14159… ≈ 150.8 cm³
Leave the answer as 48π if the question asks for an exact value. Convert to a decimal only when specified.
How do I find the perpendicular height when only the slant height is given?
The slant height l runs along the curved surface of a cone from the apex down to the circumference of the base. It is always greater than the perpendicular height h. Because the radius, the perpendicular height, and the slant height form a right-angled triangle with l as the hypotenuse, Pythagoras' theorem gives:
l² = r² + h² so h = √(l² − r²)
Worked example 4 — cone with slant height given
A cone has radius 5 cm and slant height 13 cm. Find its volume. Give your answer to 1 decimal place.
- Find the perpendicular height: h = √(13² − 5²) = √(169 − 25) = √144 = 12 cm
- V = ⅓ × π × 5² × 12 = ⅓ × π × 25 × 12 = ⅓ × 300π = 100π
- V = 100 × 3.14159… ≈ 314.2 cm³
Always sketch the cone with r, h, and l labelled before writing any equation — making the right-angled triangle visible removes the temptation to substitute the wrong measurement.
How do I find the height or radius when the volume is given?
Substitute the known values into the formula and then isolate the unknown. Substituting first is far safer than rearranging in the abstract.
Worked example 5 — finding the perpendicular height
A cone has volume 200π cm³ and radius 10 cm. Find its perpendicular height.
- Write the formula: V = ⅓ × π × r² × h
- Substitute: 200π = ⅓ × π × 100 × h
- Divide both sides by π: 200 = ⅓ × 100 × h = 100h ÷ 3
- Multiply both sides by 3: 600 = 100h
- Divide by 100: h = 6 cm
Dividing by π at step 3 clears it from both sides and keeps the arithmetic straightforward.
What are the most common mistakes on these questions?
Three errors account for the majority of lost marks on GCSE cone and pyramid volume questions.
1. Omitting the factor of ⅓. Using V = base area × height instead of V = ⅓ × base area × height gives an answer exactly three times too large. Write ⅓ first — before any substitution. If your method shows it, you earn method marks even if arithmetic goes wrong later.
2. Using the slant height instead of the perpendicular height. When a diagram labels the sloping edge of a cone rather than the vertical height, you must apply Pythagoras (h = √(l² − r²)) before using the volume formula. The slant height and the perpendicular height are never equal.
3. Forgetting to calculate the base area. For non-circular pyramids, the base area must be worked out as a separate step. For a square base of side a, write A = a². For a triangular base, write A = ½bh_base. Show this step explicitly — rushing it mentally is where arithmetic slips occur.
Frequently asked questions
Are the cone and pyramid volume formulae given on the GCSE exam paper?
Yes. AQA, Edexcel, and OCR all supply a formulae sheet with Higher-tier papers that includes V = ⅓πr²h for a cone. The general pyramid formula V = ⅓ × base area × height is also typically provided. Even so, you need to identify the correct formula, choose the right height, and evaluate the base area yourself — the sheet does not do those steps.
What is the difference between a cone and a pyramid?
A pyramid has a flat polygonal base (square, triangle, hexagon, and so on) and straight triangular lateral faces. A cone has a circular base and a smoothly curved lateral surface. Mathematically, a cone is a special case of a pyramid whose base is a circle of area πr² — which is why both volumes share the same ⅓ × base area × height structure.
Can I use the diameter instead of the radius in the cone formula?
No. The formula uses the radius r, which is half the diameter d. If a question gives the diameter, halve it before substituting: r = d ÷ 2. Substituting the full diameter instead of the radius makes r² four times too large and inflates the volume by a factor of four.
What does "leave your answer in terms of π" mean?
It means do not multiply by 3.14159. Simply write the numerical coefficient followed by π — for example, 48π cm³. This is an exact answer with no rounding error. Convert to a decimal only when the question asks for one, and always round to the number of significant figures or decimal places specified.
For Socratic GCSE geometry tutoring on cones, pyramids, and all Higher-tier topics, try Professor Pi at aitutors.me.