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hscscience Maths Std · Y11
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Module 2 · L10 of 22 ~55 min MS-M1 ⚡ +95 XP available

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.

Today's hook, 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 would be filled? Why? What does this tell you about the relationship between their volumes?
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Recall, your gut answer first

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

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?

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The three formulas you need to own
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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 $V = \tfrac{2}{3}\pi r^3$.

Volume formulas for the curved and pointed solids.
Pyramid = $\tfrac{1}{3}$ of enclosing prism, always
Pyramid
$V = \tfrac{1}{3}Ah$. Find base area first (square, rectangle, or triangle), then multiply by $\tfrac{1}{3}h$.
Cone
$V = \tfrac{1}{3}\pi r^2 h$. Use perpendicular height $h$, not slant height $\ell$. If only $\ell$ and $r$ given, find $h = \sqrt{\ell^2 - r^2}$.
Sphere
$V = \tfrac{4}{3}\pi r^3$. The exponent is 3, cube, not square. Halve diameter to get $r$ first.
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What you'll master
Know

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
Understand

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)
Can do

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
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Key terms
PyramidA solid with a polygonal base and triangular faces meeting at a single apex; volume is one-third of the enclosing prism.
ConeA solid with a circular base tapering to an apex; volume is one-third of the enclosing cylinder.
Perpendicular heightThe height $h$ measured at right angles from the base to the apex, always use this in $V = \frac{1}{3}Ah$.
HemisphereHalf a sphere; volume $= \frac{2}{3}\pi r^3$, surface area $= 3\pi r^2$ (curved face + flat circle).
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The one-third rule

A pyramid always has exactly one-third the volume of the prism with the same base and height.

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The one-third rule
core concept

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:

  1. Find the area $A$ of the base (rectangle, square, triangle, etc.)
  2. Identify the perpendicular height $h$ (vertical, from base to apex)
  3. Apply: $V = \frac{1}{3}Ah$
Same base and height: the pyramid holds one-third of the prism.
Same base area A and same height h, pyramid volume is always exactly ⅓ of the enclosing prism
Common error: Using slant height $\ell$ instead of perpendicular height $h$. These are different, check your diagram carefully.

Pyramid or cone volume = ⅓ × base area × height. The one-third factor always applies regardless of base shape. Use vertical height (not slant height). Sphere: V = (4/3)πr³.

Pause, copy the one-third pyramid/cone volume rule V = ⅓Ah (using vertical height), the sphere formula V = (4/3)πr³, and the reminder to use vertical height, not slant height, for both into your book.

True or false: A cone has exactly one-third the volume of the cylinder with the same base radius and height.

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · SQUARE PYRAMID

A square pyramid has a base side length of 9 m and perpendicular height 8 m. Find its volume.

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$A = 9^2 = 81 \text{ m}^2$
Base is a square; area = side².
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PROBLEM 2 · CONE

A cone has base radius 5 cm and vertical height 12 cm. Find its volume correct to 2 decimal places.

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$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \pi \times 25 \times 12$
Substitute $r = 5$, $h = 12$ directly.
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PROBLEM 3 · SPHERE

A spherical water balloon has diameter 18 cm. Find its volume, leaving in terms of $\pi$.

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$r = 18 \div 2 = 9 \text{ cm}$
Halve the diameter to find radius.
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Quick check: A cone has base radius 6 cm and vertical height 9 cm. Its volume in terms of $\pi$ is:

One-Third Rule Reference: Pyramid, Cone, Sphere Pyramid h A = base area V = ⅓Ah Cone h r V = ⅓πr²h Sphere r V = 4/3 πr³
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Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Trap 01
Using slant height instead of perpendicular height
In a cone, $\ell$ (slant height) and $h$ (perpendicular height) are different. Always use $h$ in the formula. If only $\ell$ and $r$ are given, use Pythagoras: $h = \sqrt{\ell^2 - r^2}$.
Trap 02
Squaring instead of cubing in the sphere formula
$V = \tfrac{4}{3}\pi r^3$, the exponent is 3 (cubic), not 2. Writing $\pi r^2$ gives the area of a circle, not the volume of a sphere. Check every time.
Trap 03
Forgetting the ⅓ in the pyramid/cone formula
Students often write $V = Ah$ or $V = \pi r^2 h$ (the prism/cylinder formula) instead of including $\tfrac{1}{3}$. The one-third is the entire point, leaving it out triples your answer.

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.

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Quick-fire practice · 5 calculations

Quick-fire practice · 5 calculations

Quick-fire practice · 5 calculations

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A square pyramid has base side 6 m and perpendicular height 10 m. Find its volume.

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A rectangular pyramid has base 8 cm × 5 cm and perpendicular height 9 cm. Find its volume.

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An ice cream cone has diameter 6 cm and vertical height 10 cm. Find its volume in cm³ (to 2 d.p.).

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A sphere has diameter 10 m. Find its volume correct to 2 d.p.

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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?

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Revisit your thinking

Earlier you predicted how much of the cylinder a full cone would fill.

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Revisit your thinking

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.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 43 marks

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)

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ApplyBand 43 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)

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AnalyseBand 54 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)

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📖 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].