Volumes of Pyramids and Cones
A pyramid or cone holds one-third of the prism or cylinder with the same base and height: \(V = \dfrac{1}{3}Ah\). This NSW Year 9 Advanced maths subtopic finds volumes of pyramids, cones and solids made from them (NESA outcome MA5‑VOL‑P‑01).
You will find volumes of square and triangular pyramids and cones, and work back to a base length, height or radius. You will use a slant height to find a pyramid's height, find the volume of a silo, a tent and a pile of sand, fill paper cups from a bottle, see why six pyramids fill a cube, and find the volume of water in a cone held point down. It is part of Year 9 Advanced Surface Area and Volume.
Theory
A pyramid or cone holds exactly one third of the prism or cylinder with the same base and height: \(V = \dfrac{1}{3}Ah\), and for a cone \(V = \dfrac{1}{3}\pi r^2 h\). This Year 9 Advanced maths page is part of the path towards Standard and Advanced maths in Years 11 and 12.
A pyramid has a polygon base and triangular faces that meet at a point, the apex. It is named after its base: square pyramid, rectangular pyramid, triangular pyramid. A cone has a circular base and a curved surface meeting at its apex.
In a right pyramid or cone the apex is directly above the centre of the base. The perpendicular height \(h\) is measured straight down from the apex to the base. The slant height runs along a face (or the curved surface) and is longer.
Fill a pyramid with water and pour it into the prism with the same base and height: it takes exactly three pyramids to fill it. So a pyramid's volume is one third of the prism's, and the same holds for a cone and its cylinder.
The height, the slant height and a horizontal distance in the base form a right-angled triangle, so Pythagoras' theorem connects them.
\(A\) is the base area (a square, rectangle or triangle) and \(h\) the perpendicular height. For a cone with slant height \(l\):
How to find the volume of a pyramid or cone
- Find the base area \(A\): \(s^2\) for a square, \(lw\) for a rectangle, \(\dfrac{1}{2}bh\) for a triangle, \(\pi r^2\) for a circle (halve any diameter).
- Find the perpendicular height. If only a slant height is given, use Pythagoras' theorem.
- Multiply \(A \times h\) and take one third. Put all lengths in the same unit first.
- Answer exactly (in terms of \(\pi\)) or rounded, and in mL or L if a capacity is asked for.
Harder problems
- Composite solids: add the parts (a silo is a cylinder plus a cone; a tent can be a prism plus a pyramid) or subtract a part cut out (a pyramid carved out of a cube). Use the height of each part, not of the whole solid.
- Working backwards: substitute \(V\) and solve for the base area, a side, the height or the radius. A height equal to a multiple of \(r\) leads to \(r^3\) and a cube root.
- Counting: how many cups can be filled completely means round down; how many truck loads are needed means round up.
- Comparing solids: write both volumes, set them equal (or divide one by the other) and solve; common factors such as \(\pi r^2\) cancel.
- Partly filled cones: the liquid forms a similar cone. Use the ratio of radius to height of the full cone to find the radius of the liquid surface.
The radius, the perpendicular height and the slant height form a right-angled triangle; the slant height is the hypotenuse.
| \(h\) | \(=\) | \(\sqrt{13^2 - 5^2} = \sqrt{144} = 12~\text{cm}\) |
| \(V\) | \(=\) | \(\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \pi \times 5^2 \times 12\) |
| \(\) | \(=\) | \(\boldsymbol{100\pi} \approx \mathbf{314.16}~\text{cm}^3\) |
Both parts have radius \(0.4\) cm. Find each volume and add.
| \(V_{\text{cyl}}\) | \(=\) | \(\pi \times 0.4^2 \times 15 = 2.4\pi\) |
| \(V_{\text{cone}}\) | \(=\) | \(\dfrac{1}{3} \times \pi \times 0.4^2 \times 1.2 = 0.064\pi\) |
| \(V\) | \(=\) | \(2.464\pi \approx \mathbf{7.74}~\text{cm}^3\) |
\(250\) mL \(= 250~\text{cm}^3\) and \(r = 5\) cm. Substitute, then make \(h\) the subject.
| \(\dfrac{1}{3}\pi \times 5^2 \times h\) | \(=\) | \(250\) |
| \(\dfrac{25\pi}{3}h\) | \(=\) | \(250\) |
| \(h\) | \(=\) | \(\dfrac{3 \times 250}{25\pi} = \dfrac{30}{\pi}\) |
| \(\) | \(=\) | \(9.5492\ldots \approx \mathbf{9.55}~\text{cm}\) |
Check: \(\dfrac{1}{3}\pi \times 25 \times 9.5492\ldots = 250.0\) \(\checkmark\)
(i) Find the radius of the water surface.
(ii) Find the volume of water exactly.
(iii) What fraction of the cone is full?
(i) The water is a smaller cone with the same shape, so radius and height are in the same ratio, \(9 : 15\).
| \(\dfrac{r}{5}\) | \(=\) | \(\dfrac{9}{15}\) |
| \(r\) | \(=\) | \(\mathbf{3}~\text{cm}\) |
| \(V_{\text{water}}\) | \(=\) | \(\dfrac{1}{3}\pi \times 3^2 \times 5 = \boldsymbol{15\pi}~\text{cm}^3\) |
| \(V_{\text{cone}}\) | \(=\) | \(\dfrac{1}{3}\pi \times 9^2 \times 15 = 405\pi\) |
| \(\text{fraction}\) | \(=\) | \(\dfrac{15\pi}{405\pi} = \mathbf{\dfrac{1}{27}}\) |
A third of the depth holds only \(\left(\dfrac{1}{3}\right)^3 = \dfrac{1}{27}\) of the volume, because every length of the small cone is a third of the big one.
Common pitfalls
Frequently asked questions
What is the formula for the volume of a pyramid?
\(V = \dfrac{1}{3}Ah\): one third of the base area times the perpendicular height. For a square base of side \(s\), \(V = \dfrac{1}{3}s^2 h\).
What is the formula for the volume of a cone?
\(V = \dfrac{1}{3}\pi r^2 h\), where \(r\) is the base radius and \(h\) is the perpendicular height.
Why is the volume of a cone one third of a cylinder?
A cone tapers to a point, so it holds less than the cylinder with the same base and height. Filling experiments show it holds exactly one third; the same is true for a pyramid and its prism.
How do you find the height of a cone from the slant height?
Use Pythagoras' theorem: the slant height is the hypotenuse, so \(h^2 = l^2 - r^2\).
How do you find the radius of a cone from its volume?
Rearrange \(V = \dfrac{1}{3}\pi r^2 h\): multiply the volume by \(3\), divide by \(\pi h\), then take the square root.
What is the difference between slant height and perpendicular height?
The perpendicular height goes straight down from the apex to the centre of the base. The slant height goes down the sloping surface. Volume uses the perpendicular height.