Volume of Pyramids and Cones
Theory
The volume of a pyramid or cone is one third of the base area times the perpendicular height: \(V = \dfrac{1}{3}Ah\), and for a cone \(V = \dfrac{1}{3}\pi r^2 h\). This Year 10A maths page covers finding the perpendicular height with Pythagoras' theorem, working backwards to a radius, height or base edge, capacities, solids with the top cut off, and hollow and joined solids.
A pyramid has a polygon base and triangular faces meeting at the apex. A cone has a circular base. In a right pyramid or right cone the apex is directly above the centre of the base.
A pyramid holds exactly one third of the prism with the same base and height, and a cone holds one third of the matching cylinder. That is where the \(\dfrac{1}{3}\) comes from.
The \(h\) in the formula is the perpendicular height: straight up from the centre of the base to the apex. It is not the slant height or a sloping edge. If you are given one of those, use Pythagoras' theorem to find \(h\).
Capacities follow the usual links: \(1\text{ cm}^3 = 1\) mL and \(1\text{ m}^3 = 1000\) L.
| Solid | Volume | Base area \(A\) |
|---|---|---|
| Any pyramid | \(\dfrac{1}{3}Ah\) | area of the base polygon |
| Square pyramid | \(\dfrac{1}{3}b^2 h\) | \(b^2\) |
| Rectangular pyramid | \(\dfrac{1}{3}lwh\) | \(lw\) |
| Cone | \(\dfrac{1}{3}\pi r^2 h\) | \(\pi r^2\) |
How to find the volume
- Find the base area \(A\) (a triangle, square, rectangle or circle).
- Find the perpendicular height, using Pythagoras' theorem if needed.
- Multiply \(A \times h\) and divide by \(3\).
- Convert to a capacity if asked, rounding once at the end.
Other questions
- Radius from the circumference or a sector: \(2\pi r = C\). When a sector is rolled into a cone, its radius is the slant height and its arc is \(2\pi r\).
- Joined solids: add the volumes. Two cones on the same base use the sum of their heights.
- Top cut off, or hollow: subtract the small solid from the big one. Use similar triangles to find the size of the piece removed.
- Equal volumes: set the two volume expressions equal and solve for the unknown.
- Mass and counting: volume = mass \(\div\) density. Round down for how many whole objects can be made.
- Percentage changes: multiply the factors, squaring the radius factor: \(r\) up \(20\%\) and \(h\) down \(30\%\) gives \(1.2^2 \times 0.7\).
- Unknown in an expression: form the equation, rearrange to a quadratic equal to zero and keep the positive solution.
The volume needs the perpendicular height. It forms a right-angled triangle with half a base edge (\(9\) cm) and the slant height (the hypotenuse):
| \(h^2\) | \(=\) | \(15^2 - 9^2 = 144\) |
| \(h\) | \(=\) | \(12\text{ cm}\) |
| \(V\) | \(=\) | \(\dfrac{1}{3}Ah = \dfrac{1}{3} \times 18^2 \times 12\) |
| \(\) | \(=\) | \(\mathbf{1296}\text{ cm}^3\) |
\(150\) mL is \(150\text{ cm}^3\). Substitute into \(V = \dfrac{1}{3}\pi r^2 h\) and make \(r^2\) the subject:
| \(\dfrac{1}{3}\pi r^2 \times 9\) | \(=\) | \(150\) |
| \(3\pi r^2\) | \(=\) | \(150\) |
| \(r^2\) | \(=\) | \(\dfrac{150}{3\pi} = 15.9154\ldots\) |
| \(r\) | \(=\) | \(3.9894\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{4.0}\text{ cm}\) |
Take the square root of the full value of \(r^2\), not a rounded one.
The piece cut off is a small pyramid, similar to the full one. Its height is \(\dfrac{1}{4}\) of the full height, so its base edge is \(\dfrac{1}{4} \times 3 = 0.75\) m.
| \(V_{\text{full}}\) | \(=\) | \(\dfrac{1}{3} \times 3^2 \times 4 = 12\text{ m}^3\) |
| \(V_{\text{top}}\) | \(=\) | \(\dfrac{1}{3} \times 0.75^2 \times 1 = 0.1875\text{ m}^3\) |
| \(V\) | \(=\) | \(12 - 0.1875\) |
| \(\) | \(=\) | \(\mathbf{11.8125}\text{ m}^3\) |
The same subtraction works for a hollow solid: outside volume minus the space inside.
The apex is above the centre of the base, where the diagonals cross. Half a diagonal, the height and a sloping edge make a right-angled triangle.
| \(\text{diagonal}\) | \(=\) | \(\sqrt{8^2 + 6^2} = 10\) |
| \(\text{half}\) | \(=\) | \(5\text{ cm}\) |
| \(h^2\) | \(=\) | \(13^2 - 5^2 = 144\) |
| \(h\) | \(=\) | \(\mathbf{12}\text{ cm}\) |
| \(V\) | \(=\) | \(\dfrac{1}{3} \times 8 \times 6 \times 12\) |
| \(\) | \(=\) | \(\mathbf{192}\text{ cm}^3\) |
Common pitfalls
Frequently asked questions
What is the formula for the volume of a pyramid?
\(V = \dfrac{1}{3}Ah\), where \(A\) is the base area and \(h\) the perpendicular height. For a square pyramid, \(V = \dfrac{1}{3}b^2 h\).
What is the formula for the volume of a cone?
\(V = \dfrac{1}{3}\pi r^2 h\), where \(r\) is the radius of the base and \(h\) is the perpendicular height.
Why is the volume of a cone one third of a cylinder?
A cone fits inside the cylinder with the same base and height, and it takes exactly three cones of water to fill the cylinder. The same is true of a pyramid and its prism.
How do you find the height of a cone from its slant height?
Use Pythagoras' theorem with the slant height as the hypotenuse: \(h^2 = l^2 - r^2\).
How do you find the volume of a cone with the top cut off?
Find the volume of the full cone, find the volume of the small cone that was removed, and subtract. Similar triangles give the size of the small cone if it is not stated.