Printable Worksheets
Print or save as PDF — or build a custom worksheet from any module's questions.
1
Learn
2
Questions
3
Game
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.
55–60 min
MS-M1 — MEDIUM
3 MC
3 SA
Lesson 10 of 22
Free
🔺
Choose how you work: type answers on screen, or work in your book.
Type on screen
Work in book
Think First
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?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Write your initial thinking in your book
Saved
Come back to this at the end of the lesson.
Pyramid
h
A = base area
V = ⅓Ah
Cone
h
r
V = ⅓πr²h
Sphere
r
V = 4/3 πr³
🧠 Know
$V = \frac{1}{3}Ah$ 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
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
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
Pyramid
A solid with a polygonal base and triangular faces meeting at a single apex; volume is one-third of the enclosing prism
Cone
A solid with a circular base tapering to an apex; volume is one-third of the enclosing cylinder
Perpendicular height
The height $h$ measured at right angles from the base to the apex — always use this in $V = \frac{1}{3}Ah$
Hemisphere
Half a sphere; volume $= \frac{2}{3}\pi r^3$, surface area $= 3\pi r^2$ (curved face + flat circle)
Misconceptions to Fix
✗
Wrong: The volume of a cone is the same as a cylinder with the same base and height
✗
Right: V(cone) = ⅓ × base area × height. A cone holds exactly one-third the volume of the corresponding cylinder.
Pyramids — $V = \tfrac{1}{3}Ah$
Key Point
Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.
Key Terms
Formula A rule showing the relationship between variables using symbols.
Substitution Replacing variables with their known values in an equation.
Unit Conversion Changing a measurement from one unit to another.
Capacity The amount of liquid a container can hold, measured in litres or millilitres.
Perimeter The total distance around the outside of a shape.
Area The amount of space inside a two-dimensional shape.
The One-Third Rule
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: $V = \frac{1}{3}Ah$
Pyramid
V = ⅓Ah
÷3
h
Prism (same base & h)
V = Ah
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.
Problem
A square pyramid has a base side length of 9 m and perpendicular height 8 m. Find its volume.
Solution
1
$A = 9^2 = 81 \text{ m}^2$
Base is a square; area = side²
2
$V = \dfrac{1}{3} \times 81 \times 8 = \dfrac{648}{3} = 216 \text{ m}^3$
Apply $V = \frac{1}{3}Ah$; $\frac{1}{3}$ of 648
Show next step
Cones — $V = \tfrac{1}{3}\pi r^2 h$
Volume of a Cone
A cone is a pyramid with a circular base. Its volume is one-third that of the cylinder with the same radius and height.
Use vertical height $h$ (not slant height $\ell$)
If given diameter, halve to find $r$ first
If $h$ is unknown but $\ell$ and $r$ are known: use Pythagoras — $h = \sqrt{\ell^2 - r^2}$
Problem
A cone has base radius 5 cm and vertical height 12 cm. Find its volume correct to 2 decimal places.
Solution
1
$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \pi \times 25 \times 12$
Substitute $r = 5$, $h = 12$ directly
2
$V = \dfrac{300\pi}{3} = 100\pi \approx 314.16 \text{ cm}^3$
$\frac{1}{3} \times 300\pi = 100\pi$; evaluate with calculator
Show next step
Problem
A cone has a base radius of 6 cm and a volume of $168\pi$ cm³. Find its perpendicular height.
Solution
1
$168\pi = \dfrac{1}{3}\pi(6)^2 h = \dfrac{1}{3}\pi(36)h = 12\pi h$
Write the formula with known values substituted
2
$h = \dfrac{168\pi}{12\pi} = 14 \text{ cm}$
Divide both sides by $12\pi$; the $\pi$ cancels
Show next step
Spheres — $V = \tfrac{4}{3}\pi r^3$
Volume of a Sphere and Hemisphere
A sphere of radius $r$ has volume $V = \frac{4}{3}\pi r^3$. A hemisphere is exactly half a sphere, giving $V = \frac{2}{3}\pi r^3$.
Always halve diameter to get $r$ before substituting
The exponent is 3 (cubic) — a common error is squaring instead
Leave answers in terms of $\pi$ when exact form is required
Problem
A spherical water balloon has diameter 18 cm. Find its volume, leaving in terms of $\pi$.
Solution
1
$r = 18 \div 2 = 9 \text{ cm}$
Halve the diameter to find radius
2
$V = \dfrac{4}{3}\pi(9)^3 = \dfrac{4}{3}\pi(729) = 972\pi \text{ cm}^3$
$\frac{4}{3} \times 729 = \frac{2916}{3} = 972$
Show next step
Practice Questions
Show all working. Exact answers (in terms of $\pi$) are preferred unless told otherwise.
Section A — Pyramids
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.
A pyramid has volume 200 cm³ and square base side 10 cm. Find its perpendicular height.
Section B — Cones
A cone has base radius 7 cm and vertical height 15 cm. Find its volume, leaving in terms of $\pi$.
A cone has slant height 13 cm and base radius 5 cm. Find the vertical height, then calculate the volume (in terms of $\pi$).
A cone has volume $48\pi$ cm³ and base radius 4 cm. Find its vertical height.
An ice cream cone has diameter 6 cm and vertical height 10 cm. Find its volume in cm³ (to 2 d.p.).
Section C — Spheres
A sphere has radius 6 cm. Find its volume in terms of $\pi$.
A sphere has diameter 10 m. Find its volume correct to 2 d.p.
A hemisphere has radius 9 cm. Find its volume in terms of $\pi$.
A sphere has volume $\frac{500\pi}{3}$ cm³. Find its radius.
Section D — Composite Solids
A solid consists of a cylinder (radius 4 cm, height 10 cm) with a cone on top (same radius, height 6 cm). Find the total volume in terms of $\pi$.
A solid hemisphere (radius 5 cm) sits on top of a cylinder (radius 5 cm, height 8 cm). Find the total volume in terms of $\pi$.
Show Answers ▼
Q1 $\frac{1}{3}(36)(10) = \mathbf{120 \text{ m}^3}$
Q2 $\frac{1}{3}(40)(9) = \mathbf{120 \text{ cm}^3}$
Q3 $200 = \frac{1}{3}(100)h \Rightarrow h = \frac{600}{100} = \mathbf{6 \text{ cm}}$
Q4 $\frac{1}{3}\pi(49)(15) = \frac{735\pi}{3} = \mathbf{245\pi \text{ cm}^3}$
Q5 $h = \sqrt{13^2-5^2} = \sqrt{144} = 12$ cm; $V = \frac{1}{3}\pi(25)(12) = \mathbf{100\pi \text{ cm}^3}$
Q6 $48\pi = \frac{1}{3}\pi(16)h = \frac{16\pi h}{3} \Rightarrow h = \frac{144}{16} = \mathbf{9 \text{ cm}}$
Q7 $r=3$; $V = \frac{1}{3}\pi(9)(10) = 30\pi \approx \mathbf{94.25 \text{ cm}^3}$
Q8 $\frac{4}{3}\pi(216) = \mathbf{288\pi \text{ cm}^3}$
Q9 $r=5$; $\frac{4}{3}\pi(125) = \frac{500\pi}{3} \approx \mathbf{523.60 \text{ m}^3}$
Q10 $\frac{2}{3}\pi(729) = \mathbf{486\pi \text{ cm}^3}$
Q11 $\frac{500\pi}{3} = \frac{4}{3}\pi r^3 \Rightarrow r^3 = 125 \Rightarrow r = \mathbf{5 \text{ cm}}$
Q12 Cylinder: $\pi(16)(10) = 160\pi$; Cone: $\frac{1}{3}\pi(16)(6) = 32\pi$; Total $= \mathbf{192\pi \text{ cm}^3}$
Q13 Cylinder: $\pi(25)(8) = 200\pi$; Hemisphere: $\frac{2}{3}\pi(125) = \frac{250\pi}{3}$; Total $= 200\pi + \frac{250\pi}{3} = \frac{850\pi}{3} \approx \mathbf{890.1 \text{ cm}^3}$
Revisit Your Initial Thinking
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
← Back to Lesson
Multiple Choice
1 A cone has base radius 6 cm and vertical height 9 cm. Its volume in terms of $\pi$ is:
A $108\pi$ cm³
B $324\pi$ cm³
C $162\pi$ cm³
D $54\pi$ cm³
A — $V = \frac{1}{3}\pi(36)(9) = \frac{324\pi}{3} = 108\pi$ cm³.
2 A sphere has volume $\frac{256\pi}{3}$ cm³. Its radius is:
A 2 cm
B 4 cm
C 6 cm
D 8 cm
B — $\frac{256\pi}{3} = \frac{4}{3}\pi r^3 \Rightarrow r^3 = 64 \Rightarrow r = 4$ cm.
3 A square pyramid has base side 12 cm and volume 576 cm³. Its perpendicular height is:
A 4 cm
B 8 cm
C 12 cm
D 16 cm
C — $576 = \frac{1}{3}(144)h \Rightarrow 576 = 48h \Rightarrow h = 12$ cm.
Retry Questions
Continue to Game ->
Short Answer
01
SA 4 3 marks
A cone has slant height 10 m and base radius 6 m.
(a) Find the perpendicular height of the cone. (1 mark)
(b) Find the volume of the cone, leaving your answer in terms of $\pi$. (2 marks)
Show Answer ▼
(a) $h = \sqrt{10^2 - 6^2} = \sqrt{100-36} = \sqrt{64} = \mathbf{8 \text{ m}}$
(b) $V = \frac{1}{3}\pi(36)(8) = \frac{288\pi}{3} = \mathbf{96\pi \text{ m}^3}$
02
SA 5 3 marks
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 $\pi$. (1 mark)
(b) Find the volume of the cone. Leave in terms of $\pi$. (1 mark)
(c) Find the total volume of the paperweight, correct to the nearest cm³. (1 mark)
Show Answer ▼
(a) $V_{\text{hemi}} = \frac{2}{3}\pi(64) = \mathbf{\frac{128\pi}{3} \text{ cm}^3}$
(b) $V_{\text{cone}} = \frac{1}{3}\pi(16)(9) = \mathbf{48\pi \text{ cm}^3}$
(c) Total $= \frac{128\pi}{3} + 48\pi = \frac{128\pi + 144\pi}{3} = \frac{272\pi}{3} \approx \mathbf{285 \text{ cm}^3}$
03
SA 6 4 marks
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. Give your answer in terms of $\pi$. (1 mark)
(b) Find the volume of the conical section. Give your answer in terms of $\pi$. (1 mark)
(c) Find the total volume in m³, correct to 1 decimal place. (1 mark)
(d) Grain has a density of 750 kg/m³. Find the maximum mass of grain the silo can hold (to the nearest tonne). (1 mark)
Show Answer ▼
(a) $r=3$; $V_{\text{cyl}} = \pi(9)(8) = \mathbf{72\pi \text{ m}^3}$
(b) $V_{\text{cone}} = \frac{1}{3}\pi(9)(3) = \mathbf{9\pi \text{ m}^3}$
(c) Total $= 81\pi \approx \mathbf{254.5 \text{ m}^3}$
(d) $254.5 \times 750 = 190\,875 \text{ kg} \approx \mathbf{191 \text{ t}}$