Volume of Pyramids, Cones, and Spheres
A pyramid or cone holds exactly one-third as much as the prism or cylinder enclosing it. That factor of ⅓ is the key — and for spheres, $\frac{4}{3}\pi r^3$ captures the whole solid.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
If you filled a cone-shaped funnel with water, then poured it into a cylinder of the same radius and height, how much of the cylinder do you think would be filled? Why? What does this tell you about the relationship between their volumes?
Pyramids and cones share the one-third rule: they always hold exactly one-third the volume of the prism or cylinder that encloses them. Spheres have their own formula involving $r^3$.
$V = \tfrac{1}{3}Ah$ — for any pyramid or cone, where $A$ is the base area and $h$ is the perpendicular height. Always use $h$, never the slant height $\ell$.
$V = \tfrac{4}{3}\pi r^3$ — for a sphere of radius $r$. A hemisphere uses $V = \tfrac{2}{3}\pi r^3$.
Key facts
- $V = \frac{1}{3}Ah$ for any pyramid or cone
- $V = \frac{4}{3}\pi r^3$ for a sphere; hemisphere is half
- The factor $\frac{1}{3}$ compared to prism/cylinder
- Use perpendicular height, not slant height, in cone formula
Concepts
- Why a pyramid is $\frac{1}{3}$ of the enclosing prism
- Why the cone formula uses $h$ not $\ell$
- How to work backwards (find a dimension from volume)
Skills
- Calculate volume of square pyramids, rectangular pyramids, cones
- Calculate volume of spheres and hemispheres
- Work backwards to find an unknown dimension from a given volume
- Solve composite solid volume problems
A pyramid always has exactly one-third the volume of the prism with the same base and height. This is true regardless of the shape of the base.
To calculate pyramid volume:
- Find the area $A$ of the base (rectangle, square, triangle, etc.)
- Identify the perpendicular height $h$ (vertical, from base to apex)
- Apply: $V = \frac{1}{3}Ah$
What to write in your book
- The one-third rule: $V = \tfrac{1}{3}Ah$ for any pyramid or cone. The base area $A$ depends on the shape of the base.
- Always use perpendicular height $h$ — the vertical distance from base to apex, not the slant height.
- Cone: if only slant height $\ell$ and radius $r$ are given, find $h$ first using Pythagoras: $h = \sqrt{\ell^2 - r^2}$.
- Sphere: $V = \tfrac{4}{3}\pi r^3$. The exponent is 3. Hemisphere: $V = \tfrac{2}{3}\pi r^3$.
True or false: A cone has exactly one-third the volume of the cylinder with the same base radius and height.
Worked examples · 3 in a row, reveal as you go
A square pyramid has a base side length of 9 m and perpendicular height 8 m. Find its volume.
A cone has base radius 5 cm and vertical height 12 cm. Find its volume correct to 2 decimal places.
A spherical water balloon has diameter 18 cm. Find its volume, leaving in terms of $\pi$.
What to write in your book
- Working backwards: substitute the known volume into the formula, then solve for the unknown dimension. E.g. given $V$ and $r$, find $h$ for a cone: $h = \dfrac{3V}{\pi r^2}$.
- Composite solids: add volumes of component shapes (cylinder + cone, hemisphere + cylinder).
- Leave answers in terms of $\pi$ when exact form is required — only evaluate when asked for decimal.
- Hemisphere: $V = \tfrac{2}{3}\pi r^3$ — exactly half of the sphere formula.
Quick check: A cone has base radius 6 cm and vertical height 9 cm. Its volume in terms of $\pi$ is:
Common errors · the 3 traps that cost marks
What to write in your book
- Always check: is this a prism/cylinder ($V = Ah$) or a pyramid/cone ($V = \tfrac{1}{3}Ah$)?
- In the cone formula, $h$ is the vertical height — draw the right-triangle if in doubt: $h^2 + r^2 = \ell^2$.
- Sphere: $V = \tfrac{4}{3}\pi r^3$. Exponent is 3. Hemisphere = exactly half.
- Composite: identify each part, calculate separately, then combine.
Fill the gap: A cone has base radius 4 cm and volume $48\pi$ cm³. Setting up $48\pi = \tfrac{1}{3}\pi(16)h$ gives $48\pi = \dfrac{16\pi h}{3}$, so $h =$ cm.
Quick-fire practice · 5 calculations
A square pyramid has base side 6 m and perpendicular height 10 m. Find its volume.
A rectangular pyramid has base 8 cm × 5 cm and perpendicular height 9 cm. Find its volume.
An ice cream cone has diameter 6 cm and vertical height 10 cm. Find its volume in cm³ (to 2 d.p.).
A sphere has diameter 10 m. Find its volume correct to 2 d.p.
A solid cylinder (radius 4 cm, height 10 cm) has a cone on top (same radius, height 6 cm). Find the total volume in terms of $\pi$.
Odd one out: Three of these statements about the sphere formula $V = \tfrac{4}{3}\pi r^3$ are correct. Which one is wrong?
Earlier you predicted how much of the cylinder a full cone would fill. The answer is exactly one-third:
Cone: $V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h$ · Cylinder: $V_{\text{cyl}} = \pi r^2 h$
Ratio $= \dfrac{\tfrac{1}{3}\pi r^2 h}{\pi r^2 h} = \dfrac{1}{3}$
So if you poured three full cones into the cylinder, it would be exactly full. This is not an approximation — it is a mathematical fact. The factor of $\tfrac{1}{3}$ is why pyramids and cones always need special formulas, separate from their enclosing prism or cylinder.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A cone has slant height 10 m and base radius 6 m. (a) Find the perpendicular height of the cone. (b) Find the volume of the cone, leaving your answer in terms of $\pi$. (3 marks)
Q2. A decorative paperweight consists of a hemisphere sitting on top of a cone. Both have radius 4 cm. The cone has perpendicular height 9 cm. (a) Find the volume of the hemisphere. Leave in terms of $\pi$. (b) Find the volume of the cone. Leave in terms of $\pi$. (c) Find the total volume of the paperweight, correct to the nearest cm³. (3 marks)
Q3. A grain silo is modelled as a cylinder (diameter 6 m, height 8 m) with a cone on top (same diameter, perpendicular height 3 m). (a) Find the volume of the cylindrical section in terms of $\pi$. (b) Find the volume of the conical section in terms of $\pi$. (c) Find the total volume in m³, correct to 1 decimal place. (d) Grain has a density of 750 kg/m³. Find the maximum mass of grain the silo can hold (to the nearest tonne). (4 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $\tfrac{1}{3}(36)(10) = 120\text{ m}^3$ · 2: $\tfrac{1}{3}(40)(9) = 120\text{ cm}^3$ · 3: $r = 3$; $\tfrac{1}{3}\pi(9)(10) = 30\pi \approx 94.25\text{ cm}^3$ · 4: $r = 5$; $\tfrac{4}{3}\pi(125) = \tfrac{500\pi}{3} \approx 523.60\text{ m}^3$ · 5: $\pi(16)(10) + \tfrac{1}{3}\pi(16)(6) = 160\pi + 32\pi = 192\pi\text{ cm}^3$
Q1 (3 marks): (a) $h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\text{ m}$ [1]. (b) $V = \tfrac{1}{3}\pi(36)(8) = \tfrac{288\pi}{3} = 96\pi\text{ m}^3$ [2].
Q2 (3 marks): (a) $V_{\text{hemi}} = \tfrac{2}{3}\pi(64) = \tfrac{128\pi}{3}\text{ cm}^3$ [1]. (b) $V_{\text{cone}} = \tfrac{1}{3}\pi(16)(9) = 48\pi\text{ cm}^3$ [1]. (c) Total $= \tfrac{128\pi}{3} + 48\pi = \tfrac{272\pi}{3} \approx 285\text{ cm}^3$ [1].
Q3 (4 marks): (a) $r = 3$; $V_{\text{cyl}} = \pi(9)(8) = 72\pi\text{ m}^3$ [1]. (b) $V_{\text{cone}} = \tfrac{1}{3}\pi(9)(3) = 9\pi\text{ m}^3$ [1]. (c) Total $= 81\pi \approx 254.5\text{ m}^3$ [1]. (d) $254.5 \times 750 = 190\,875\text{ kg} \approx 191\text{ t}$ [1].
Five timed questions on volumes of pyramids, cones and spheres. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Pool: lessons 1–10. Replays welcome.
⚔ Enter the arenaClimb platforms by answering pyramid, cone and sphere questions. Pool: lessons 1–10.
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