Pyramids and Cones

A pyramid in its box

A pyramid has a flat base and triangular faces that rise from the edges of the base and meet at one point, called the apex. Its height is the distance straight up from the base to the apex, measured at right angles to the base.

Put a pyramid inside the box that has the same base and the same height. The pyramid narrows as it rises, so it takes up less room than the box. The question is how much less.

The pyramid stands on the bottom face of the box, so the two share a base, and its apex touches the middle of the top face, so they share a height.

Three pyramids make a cube

A cube can be cut into three pyramids that are exactly the same. Choose one corner of the cube to be the apex of all three. The three faces of the cube that do not touch that corner are their bases. Each pyramid has a face of the cube as its base, s by s, and its height is s, the distance from that face to the far corner.

The three identical pyramids fill the cube, so each one is a third of it: $\frac{1}{3}\times s^3$⅓ × s³. And $s^3=s^2\times s$s³ = s² × s is the base times the height, so the volume of each pyramid is ⅓ × base × height.

Drag to pull the cube apart. The three pyramids are identical, each with a base of $s^2$s² and a height of s, so each is $\frac{1}{3}{s}^3$⅓ s³.

Any box, and the apex anywhere

A cube is a special box. Stretch it upward to twice its height. Every thin layer of the cube becomes twice as thick, and so does every layer of each pyramid inside it, so every volume doubles, and each pyramid is still a third of the box. The rule still gives it: the pyramid on the bottom face now has base $s^2$s² and height 2s, and $\frac{1}{3}\times s^2\times 2s=\frac{2}{3}{s}^3$⅓ × s² × 2s = ⅔ s³, which is a third of the box’s $2s^3$2s³. A stretch along any edge of the box works the same way, and stretches like these turn a cube into any box, so the rule holds for these pyramids in any box.

The apex does not have to sit above a corner of the base either. Slide it sideways, keeping it at the same height. Cut the pyramid into thin layers parallel to the base: sliding the apex slides each layer sideways, but no layer changes its shape or its size, so the volume does not change. It is the same argument that keeps a triangle’s area at ½ × base × height wherever its top corner is.

The cone

A cone is like a pyramid with a circular base: a curved surface rises from the edge of the circle to the apex. It fits inside the cylinder with the same base and the same height, the way a pyramid fits inside its box.

To see why a cone is also a third, stand the cone inside a square pyramid with the same apex, whose base is the square just around the cone’s circle, and stand the cylinder inside the matching box. Cut all four solids across at any height. The cone’s layer is a circle sitting just inside the pyramid’s square layer. A circle of radius r has area $\pi r^2$πr², and the square around it has side 2r and area $4r^2$4r², so the circle is $\frac{\pi }{4}$π/4 of the square, at every height. In the same way, every layer of the cylinder is $\frac{\pi }{4}$π/4 of the layer of the box.

Layer by layer, then, the cone is $\frac{\pi }{4}$π/4 of the pyramid, and the cylinder is $\frac{\pi }{4}$π/4 of the box. The pyramid is a third of the box, so the cone is a third of the cylinder. The same reasoning works for a base of any shape: at every height, its layer is the same fraction of the layer of the pyramid on the rectangle around it.

Water gives a check: fill the cone and pour it into the cylinder. Three cone-fulls fill the cylinder.

The cone stands on the bottom of the cylinder and its apex touches the middle of the top, so they share a circular base and a height. Three cone-fulls of water fill the cylinder.

One third of base times height

Either way, a pyramid or a cone holds ⅓ of its container: volume = ⅓ × base area × height, whatever the shape of the base. For a cone the base is a circle of area $\pi r^2$πr², so its volume is $\frac{1}{3}\pi r^{2}h$⅓πr²h.

A pyramid on a square base 6 cm by 6 cm, with a height of 5 cm, holds ⅓ × 36 × 5 = 60 cm³, a third of the 180 cm³ box around it.

A cone with a radius of 3 cm and a height of 4 cm holds $\frac{1}{3}\times \pi \times 3^2\times 4=\frac{1}{3}\times 36\pi =12\pi$⅓ × π × 3² × 4 = ⅓ × 36π = 12π cm³, which is about 37.7 cm³. The cylinder around it holds $\pi \times 3^2\times 4=36\pi$π × 3² × 4 = 36π cm³, three times as much.

The cone holds ⅓ × base area × height, a third of the cylinder’s base area × height.

The height, not a slanted edge

The height in the formula is the perpendicular height, straight up from the base to the apex. A sloping edge is longer than that, and using it gives too much.

A slanted length is needed for a different job: covering the sloping faces. Each face of a pyramid is a triangle, and its height is the slant height, measured from the apex straight down the middle of the face to the middle of a base edge. The pyramid’s height, the slant height and the line from the center of the base to the middle of that edge make a right-angled triangle, with the slant height as its hypotenuse. So Pythagoras’ theorem gives the slant height from the other two.

Two slips

Leaving out the ⅓ gives the container. A cone with a base area of 12 cm² and a height of 5 cm holds ⅓ × 12 × 5 = 20 cm³; the 12 × 5 = 60 cm³ is the cylinder around it.

Halving instead, 60 ÷ 2 = 30 cm³, is the usual guess from looking at the shape. The cube splits into three pyramids, and the cylinder takes three cone-fulls, so divide by 3.