Skip to content

Volume of pyramids and cones

Measurement · level 6, lesson 1 of 6

A question to start

Take a paper cone (like the ones beside a water cooler) and a cylinder-shaped cup with exactly the same circular top and exactly the same height. Fill the cone with water and pour it into the cup. How many cone-fulls does it take to fill the cup?

Most people guess two: the cone looks like "half" of the cup, a triangle inside a rectangle. But a cone is pointed in every direction, not only in the side view, and it holds less than you think. The answer is three, every time, for every cone and its matching cylinder. Pyramids do the same with their matching prisms.

In this lesson you will see three different reasons why the magic number is 3, so that you do not just remember the formula but know it has to be true.

The idea

Pyramids and cones

A pyramid is a solid with a flat polygon as its base and triangular sides that all meet at one point, the apex (the tip). It is named after its base: a square pyramid, a triangular pyramid (also called a tetrahedron), a hexagonal pyramid.

A cone is the same idea with a circle for its base: a curved surface sweeps from the edge of the circle up to the apex.

The height \(h\) of a pyramid or cone is the straight distance from the apex down to the plane of the base, meeting it at a right angle. It is not the length of a sloping edge or of the sloping side. That sloping distance is called the slant height, and it matters for surface area (lesson 3), not for volume.

If the apex is directly above the centre of the base, the pyramid or cone is called a right pyramid or cone. If the apex leans to one side, it is oblique. The volume rule is the same for both.

The rule

For every pyramid and every cone:

\[V = \tfrac13 \times \text{base area} \times \text{height}\]

For a cone, the base is a circle of area \(\pi r^2\), so \(V = \tfrac13 \pi r^2 h\). Compare it with the prism and cylinder rule from level 5, \(V = \text{base area} \times \text{height}\): the pointed solid is exactly one third of the straight solid with the same base and height.

r h cone = 1/3 of the cylinder

Reason 1: pour it and see

The water (or rice, or sand) experiment at the top of the page is a real experiment you can do. Make a cone and a cylinder from card with the same base and height, fill the cone and empty it into the cylinder. After three pours the cylinder is full. Try it with a square pyramid and a box of the same base and height, and it is three again.

An experiment is convincing, but it cannot tell the difference between 3 and 3.02. For that we need a reason.

Reason 2: three pyramids make a cube

Take a cube with edges of length \(s\). Pick one corner at the top, call it \(P\). Three faces of the cube do not touch \(P\): the bottom, the back and the right-hand face. Join \(P\) to every corner of each of those three faces. You have cut the cube into three pyramids, each with a square face of the cube as its base and \(P\) as its apex.

P cube cut from P one of 3 equal pieces

Why do they fill the cube with no gaps and no overlaps? Every point inside the cube can be joined to \(P\) by a straight line, and if you carry that line on beyond the point, it leaves the cube through exactly one of the three far faces (or along an edge where two of them meet). So every point belongs to the pyramid on that face.

The three pyramids are identical: turning the cube about its long diagonal through \(P\) carries each one onto the next. So each has one third of the cube's volume:

\[V = \tfrac13 s^3 = \tfrac13 \times s^2 \times s = \tfrac13 \times \text{base area} \times \text{height}\]

since each pyramid's base is a face (\(s^2\)) and its height is an edge of the cube (\(s\)).

Now stretch the cube into a box \(l\) by \(w\) by \(h\). Stretching multiplies every small piece of volume by the same amount, so the three pieces still have equal volumes, each one third of the box. The piece standing on the bottom face is a pyramid with base \(l \times w\) and height \(h\), so its volume is \(\tfrac13 lwh\): again one third of base times height.

Reason 3: slices, and Cavalieri's idea

In level 5 you met Cavalieri's principle: two solids with equal cross-section areas at every height have equal volumes (like a straight and a leaning stack of coins). Slice a pyramid of height \(h\) parallel to its base. Halfway up, the slice is a copy of the base shrunk to half its width in every direction, so its area is \(\left(\tfrac12\right)^2 = \tfrac14\) of the base. A quarter of the way down from the apex it is \(\left(\tfrac14\right)^2 = \tfrac1{16}\) of the base. In general the slice's area depends only on the base area and on how far up the slice is, not on the base's shape and not on where the apex sits.

So any two pyramids or cones with the same base area and the same height have the same slices at every level, and by Cavalieri's principle the same volume. A cone of base area 12 cm² and height 5 cm has the same volume as a square pyramid of base area 12 cm² and height 5 cm, and as a leaning one. Reason 2 showed that one particular kind of pyramid is one third of base times height, so all of them are.

Adding up thin slices like this is exactly how calculus finds volumes; with an integral the \(\tfrac13\) appears from adding up the squares \(t^2\), because \(\int_0^h t^2\,dt = \tfrac13 h^3\).

Worked examples

1. A square pyramid. A pyramid has a square base 6 m on each side and a height of 4 m.

\(V = \tfrac13 \times 6^2 \times 4 = \tfrac13 \times 36 \times 4 = 48\) m³.

2. An ice-cream cone. A cone has radius 3 cm and height 10 cm.

\(V = \tfrac13 \pi \times 3^2 \times 10 = \tfrac13 \times 90\pi = 30\pi \approx 94.2\) cm³. The matching cylinder holds \(90\pi\): three times as much.

3. The Great Pyramid. When it was built, the Great Pyramid of Giza had a square base about 230 m on each side and was about 146.6 m tall. Its volume was about

\(\tfrac13 \times 230^2 \times 146.6 = \tfrac13 \times 52\,900 \times 146.6 \approx 2\,585\,000\) m³, about 2.6 million cubic metres.

4. A pile of sand. Sand poured onto the ground forms a cone 4 m across and 1.5 m high. How much sand is there?

  1. Radius \(4 \div 2 = 2\) m.
  2. \(V = \tfrac13 \pi \times 2^2 \times 1.5 = \tfrac13 \times 6\pi = 2\pi \approx 6.3\) m³.

5. Using the slant height. A cone has radius 5 cm and a slant height of 13 cm. Find its volume.

  1. The height, the radius and the slant height make a right triangle, with the slant height as the hypotenuse. By Pythagoras (Geometry): \(h^2 + 5^2 = 13^2\), so \(h^2 = 169 - 25 = 144\) and \(h = 12\) cm.
  2. \(V = \tfrac13 \pi \times 25 \times 12 = 100\pi \approx 314.2\) cm³.

6. Working backwards. A conical paper cup must hold 100 cm³ and have a radius of 4 cm. How tall must it be?

\(\tfrac13 \pi \times 16 \times h = 100\), so \(h = \dfrac{300}{16\pi} \approx 5.97\) cm.

Try it

  1. A square pyramid has a base 5 cm on each side and a height of 9 cm. Find its volume.
  2. A cone has radius 6 cm and height 7 cm. Find its volume, exactly and to one decimal place.
  3. A pyramid has a rectangular base 8 m by 6 m and a height of 10 m. Find its volume.
  4. A triangular pyramid has a base that is a triangle with base 6 cm and height 4 cm. The pyramid is 10 cm tall. Find its volume.
  5. A cone is 10 cm across its base and its slant height is 13 cm. Find its volume.
  6. A conical glass holds 150 mL. Its rim has radius 4 cm. How deep is it?
  7. Olympus Mons on Mars is about 600 km across and about 22 km high. As a very rough model, treat it as a cone. Estimate its volume. (Real shield volcanoes are lower and flatter near the top than a cone, so this is only a first guess.)
  8. A cone-shaped glass (point down) is filled with water to half its height. What fraction of the glass is full? (Hint: the water is a smaller cone.)
Answers
  1. \(\tfrac13 \times 25 \times 9 = 75\) cm³.
  2. \(\tfrac13 \pi \times 36 \times 7 = 84\pi \approx 263.9\) cm³.
  3. \(\tfrac13 \times 48 \times 10 = 160\) m³.
  4. Base area \(\tfrac12 \times 6 \times 4 = 12\) cm². \(V = \tfrac13 \times 12 \times 10 = 40\) cm³.
  5. \(r = 5\), and \(h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12\). \(V = \tfrac13 \pi \times 25 \times 12 = 100\pi \approx 314.2\) cm³.
  6. 150 mL = 150 cm³. \(\tfrac13 \pi \times 16 \times h = 150\), so \(h = \dfrac{450}{16\pi} \approx 9.0\) cm.
  7. \(r = 300\) km, \(h = 22\) km. \(V = \tfrac13 \pi \times 300^2 \times 22 = 660\,000\pi \approx 2\,070\,000\) km³, about 2 million km³.
  8. The water is a cone with half the height and (because the sides are straight) half the radius. Its volume is \(\tfrac13 \pi \left(\tfrac r2\right)^2 \tfrac h2 = \tfrac18 \times \tfrac13 \pi r^2 h\). So only \(\tfrac18\) of the glass is full, even though the water reaches halfway up. To be half full, the water must reach about 79% of the height, since \(0.794^3 \approx 0.5\).

Common mistake

Forgetting the \(\tfrac13\). A cone or pyramid must always come out smaller than the cylinder or prism with the same base and height. If your cone holds as much as the can around it, you have left out the third.

Using the slant height as the height. The height goes straight down from the apex to the base, at a right angle. If you are given the slant height, use Pythagoras to find the true height first, as in example 5.

Where it leads