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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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.
Come back to this at the end of the lesson.
Wrong: Rounding 4.5 down gives 4.
Right: Standard rounding rounds 4.5 up to 5; rounding to the nearest even number (banker's rounding) is different.
Pyramids — $V = \tfrac{1}{3}Ah$
Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.
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:
A square pyramid has a base side length of 9 m and perpendicular height 8 m. Find its volume.
Cones — $V = \tfrac{1}{3}\pi r^2 h$
A cone is a pyramid with a circular base. Its volume is one-third that of the cylinder with the same radius and height.
A cone has base radius 5 cm and vertical height 12 cm. Find its volume correct to 2 decimal places.
A cone has a base radius of 6 cm and a volume of $168\pi$ cm³. Find its perpendicular height.
Spheres — $V = \tfrac{4}{3}\pi r^3$
A sphere of radius $r$ has volume $V = \frac{4}{3}\pi r^3$. A hemisphere is exactly half a sphere, giving $V = \frac{2}{3}\pi r^3$.
A spherical water balloon has diameter 18 cm. Find its volume, leaving in terms of $\pi$.
Section A — Pyramids
Section B — Cones
Section C — Spheres
Section D — Composite Solids
$\frac{1}{3}(36)(10) = \mathbf{120 \text{ m}^3}$
$\frac{1}{3}(40)(9) = \mathbf{120 \text{ cm}^3}$
$200 = \frac{1}{3}(100)h \Rightarrow h = \frac{600}{100} = \mathbf{6 \text{ cm}}$
$\frac{1}{3}\pi(49)(15) = \frac{735\pi}{3} = \mathbf{245\pi \text{ cm}^3}$
$h = \sqrt{13^2-5^2} = \sqrt{144} = 12$ cm; $V = \frac{1}{3}\pi(25)(12) = \mathbf{100\pi \text{ cm}^3}$
$48\pi = \frac{1}{3}\pi(16)h = \frac{16\pi h}{3} \Rightarrow h = \frac{144}{16} = \mathbf{9 \text{ cm}}$
$r=3$; $V = \frac{1}{3}\pi(9)(10) = 30\pi \approx \mathbf{94.25 \text{ cm}^3}$
$\frac{4}{3}\pi(216) = \mathbf{288\pi \text{ cm}^3}$
$r=5$; $\frac{4}{3}\pi(125) = \frac{500\pi}{3} \approx \mathbf{523.60 \text{ m}^3}$
$\frac{2}{3}\pi(729) = \mathbf{486\pi \text{ cm}^3}$
$\frac{500\pi}{3} = \frac{4}{3}\pi r^3 \Rightarrow r^3 = 125 \Rightarrow r = \mathbf{5 \text{ cm}}$
Cylinder: $\pi(16)(10) = 160\pi$; Cone: $\frac{1}{3}\pi(16)(6) = 32\pi$; Total $= \mathbf{192\pi \text{ cm}^3}$
Cylinder: $\pi(25)(8) = 200\pi$; Hemisphere: $\frac{2}{3}\pi(125) = \frac{250\pi}{3}$; Total $= 200\pi + \frac{250\pi}{3} = \frac{850\pi}{3} \approx \mathbf{890.1 \text{ cm}^3}$
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Multiple Choice
1 A cone has base radius 6 cm and vertical height 9 cm. Its volume in terms of $\pi$ is:
? Regarding this topic, 1 A cone has base radius 6 cm and vertical height 9 cm. Its volume in terms of $\pi$ is:
2 A sphere has volume $\frac{256\pi}{3}$ cm³. Its radius is:
? Regarding this topic, 2 A sphere has volume $\frac{256\pi}{3}$ cm³. Its radius is:
3 A square pyramid has base side 12 cm and volume 576 cm³. Its perpendicular height is:
? Regarding this topic, 3 A square pyramid has base side 12 cm and volume 576 cm³. Its perpendicular height is:
Short Answer
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)
$h = \sqrt{10^2 - 6^2} = \sqrt{100-36} = \sqrt{64} = \mathbf{8 \text{ m}}$
$V = \frac{1}{3}\pi(36)(8) = \frac{288\pi}{3} = \mathbf{96\pi \text{ m}^3}$
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)
$V_{\text{hemi}} = \frac{2}{3}\pi(64) = \mathbf{\frac{128\pi}{3} \text{ cm}^3}$
$V_{\text{cone}} = \frac{1}{3}\pi(16)(9) = \mathbf{48\pi \text{ cm}^3}$
Total $= \frac{128\pi}{3} + 48\pi = \frac{128\pi + 144\pi}{3} = \frac{272\pi}{3} \approx \mathbf{285 \text{ cm}^3}$
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)
$r=3$; $V_{\text{cyl}} = \pi(9)(8) = \mathbf{72\pi \text{ m}^3}$
$V_{\text{cone}} = \frac{1}{3}\pi(9)(3) = \mathbf{9\pi \text{ m}^3}$
Total $= 81\pi \approx \mathbf{254.5 \text{ m}^3}$
$254.5 \times 750 = 190\,875 \text{ kg} \approx \mathbf{191 \text{ t}}$
Challenge the boss with your knowledge of volumes of pyramids, cones and spheres. Pool: lessons 1–10.