---
title: "On the Volume of the Platonic Solids"
description: |
  How big are the Platonic solids in relation to one another?
format:
  html:
    html-math-method: katex
date: "TODO"
date-modified: "2025-06-03"
categories:
  - geometry
---


The Platonic solids have been known for millennia.
They bear the name of Plato, who spoke of them in his dialogue *Timaeus*.
He describes their "construction" (sans the dodecahedron) from the most basic
  "isosceles and scalene" triangles, or in modern parlance, "45-45-90 and 30-60-90" triangles, respectively.
However, the construction was not mathematical, and to my knowledge, each solid was first
  rigorously described from first principles in Book XIII of Euclid's Elements.

In my teenage years, I recall viewing articles on the solids with their volumes
  proudly displayed next to their surface area.
While surface area may be troublesome in the case of the dodecahedron
  (as the geometry of regular pentagons is not widely taught),
  it is easy enough for grade schoolers to calculate for cubes,
  and for trigonometry students to calculate for the solids composed of equilateral triangles.
On the other hand, the volume is somewhat mystical.

The volume itself is a meaningless quantity for comparison unless put in ratio with another volume.
Fortunately, edge length is the only free variable in a Platonic solid,
  meaning their volumes can be parametrized by this value alone.
Further, since the cube has such simple expression for its volume (the edge length cubed),
  it is a natural choice as a base for the comparison[^1].
Therefore, I will derive this ratio for the solids in question.

[^1]: Though this is a pure mathematical concept, empirical units use the same convention.
For instance, a cubic centimeter is defined as the volume occupied by a cube which is a
  centimeter long in each dimension, despite being applicable to volumes of any shape.

This post will calculate the volume without using any trigonometric functions (sine, cosine, tangent),
  and instead opts for a more compass-and-straightedge approach.
Consequently, it becomes more natural to calculate the *square* of the volume to better cooperate
  with the Pythagorean theorem.


A Recap of Geometry
-------------------

For those with only a vague recollection (or perhaps none at all) of geometry,
  this section is intended as a refresher.


### Planar Geometry

There are [many centers of a triangle](https://faculty.evansville.edu/ck6/encyclopedia/etc.html),
  but for us, two are of primary interest:

- The *circumcenter* is equidistant from every vertex.
  In other words, it is the center of a circle containing all three vertices.
    - It can be constructed by finding the intersection of the edges' perpendicular bisectors.
    - The distance from a vertex to the circumcenter is called the *circumradius* (*c*).
- The *incenter* is equidistant from every edge.
  It is the center of a circle which lies tangent to every edge
  (i.e., radii can be drawn which are are perpendicular to the edge).
    - It is constructed by finding the intersection of the lines which bisect each angle.
    - The perpendicular distance from an edge to the incenter is called the *inradius* (*a*).

::: {}
![](index_files/triangle centers.png)

Constructing the circumcenter and incenter.
Angle bisectors in blue, perpendicular bisectors in red, in- and circumradii in green.
:::

The inradius is special because it is also an altitude for a triangle formed by the inradius and an edge of the larger triangle.
This means that the area of the larger triangle is the sum of these smaller triangles.

![](index_files/incenter area.png)

$$
\begin{align*}
  A &= \left ({e_1 a \over 2} + {e_2 a \over 2} + {e_3 a \over 2} \right)
    = \left ({a \over 2} \right ) (e_1 + e_2 + e_3)
  \\
  &= {Pa \over 2}
\end{align*}
$$

This gives an expression for the area.
For an equilateral triangle, these two centers coincide.
This is because the perpendicular bisectors of the edges *are* the angle bisectors.
In fact, the bisection of an angle involves constructing a rhombus, which is made up of
  two isosceles triangles (of which the equilateral triangle is a special case).
In this case, the inradius is also called the *apothem*, and the difference between it and the circumradius
  is immediately apparent and called the *sagitta* (*s*).

![](index_files/equilateral center.png)

This idea of incenters and circumcenters can be extended to other 2D figures
  such as the square and regular pentagon.
For a square, the center is simply the intersection of the diagonals (i.e., the diagonals' common midpoint).
The pentagon is trickier and will be discussed later.
Regardless, the expression for the area ${Pa \over 2}$ still works,
  since the polygon can be triangulated through the center in a similar way.

![](index_files/square pentagon circles.png)


### Cubes, Prisms, and Pyramids

Now we speak of 3D geometry.
The volume of a prism is equal to the height times the area of the base,
  where the "height" is orthogonal to the plane of the base.
Pyramids with the same height and base have one-third this area.

$$
  V_\text{prism} = Bh,~~ V_\text{pyramid} = {Bh \over 3}
$$

This volume formula can be made more intuitive by considering the cube.
The pyramid formed by one of the faces and an edge perpendicular to it will contain
  one square and two half-squares, or two squares in total.
Therefore three pyramids are needed to recreate all six faces of the cube.

For a slightly more detailed explanation, consider a point inside the face on top of the cube.
Its (perpendicular) distance from one edge is *x* and its distance to an edge adjacent to that is *y*.
Connecting all other bases to this point produces five pyramids, whose bases all have the same area.
Designate these pyramids as "bottom", "left", "right", "front", and "back",
  where left and right correspond to *x* and front and back correspond to *y*.

![](index_files/cube pyramids.png)

$$
\begin{align*}
  V_\text{cube}
    &= Bh = V_\text{bottom} + V_\text{left} + V_\text{right} V_\text{front} + V_\text{back}
  \\
  &= rBh + rBx + rB(h-x) + rBy + rB(h-y)
  \\
  &= rBh + rBh + rBh \implies 1 = 3r
  \\
  r &= {1 \over 3}
\end{align*}
$$

This can be generalized to a pyramid based on any prism, where the top point lies
  in the plane of one of the bases.
However, this is beyond the scope of this post.


Simple Solids: the Octahedron and the Tetrahedron
-------------------------------------------------

While "simple" is a bit of a misnomer, their volumes are easiest to appreciate,
  since they do not need regular pentagons.


### Octahedron

The octahedron can be thought of as two square pyramids joined end-on-end,
  with uniform edge length throughout.
Since the base is a square, its center is equidistant from the vertices of the base.
An alternative, congruent square can be noticed by the symmetry of the octahedron,
  meaning the center is also equidistant from the top of the square pyramid,
  and that the segment connecting the two is an altitude of the pyramid.

::: {}
![](index_files/octahedron squares.png)

Primary square in blue, secondary square in red. Diagonals of both squares shown.
:::

The length of this altitude is simply half of the diagonal of the square.
Therefore, the volume of an octahedron (calculated using edge length 1) is:

$$
\begin{align*}
  B^2
    &= (1^2)^2
  \\
  (2h)^2
    &= 4h^2 = 1^2 + 1^2 = 2
  \\
  V_\text{sq.pyr.}^2
    &= {B^2 h^2 \over 3^2} = {1 \cdot {2 / 4} \over 3^2}
  \\
  (1^3 \cdot V_\text{oct})^2
    &= (2V_\text{sq.pyr})^2 = 4V_\text{sq.pyr}^2
    = 4 \cdot {2 / 4 \over 3^2} = {2 \over 3^2}
  \\ \\
  V_\text{oct} &= {\sqrt{2} \over 3}
\end{align*}
$$


### Tetrahedron

From here on out, it becomes convenient to specify the edge length of every solid to be 2,
  since this simplifies the bisection of edges.
This happens to coincide with Plato's description, where the equilateral triangle is described as
  being formed from two 30-60-90 triangles.
That is, the edge length of the equilateral triangle was twice the "unit" length: the shortest side of the 30-60-90 triangle.

However, we must remember that volumes need to be put in ratio with a cube volume of 8:

$$
\begin{align*}
  { V_\text{solid[2]} \over V_\text{cube[2]} }
    &= { V_\text{solid[1]} \over V_\text{cube[1]} }
  \\
  \implies V_\text{solid[2]}
    &= { V_\text{cube[2]} \over V_\text{cube[1]} } V_\text{solid[1]}
  \\
  &= { 2^3 \over 1 } V_\text{solid[1]}
$$

Since the tetrahedron is itself a pyramid, its volume follows from the earlier formula.
First, we must calculate the (square of the) area of the base of an equilateral triangle.

:::: {layout-ncol="2"}

::: {.column width="49%"}
![](index_files/equilateral triangle area.png)
:::

::: {.column width="49%"}
$$
\begin{align*}
  d_\text{altitude}^2
    &= \textcolor{orange}{2}^2 -\ \textcolor{green}{1}^2 = 3
  \\
  B^2
    &= \left ( {2 \cdot d_\text{altitude} \over 2} \right )^2 = 3
\end{align*}
$$
:::

::::

Next, bisect this triangle through any edge, then use it to bisect the tetrahedron through
  the plane containing this line and the remaining edge.
This forms an isosceles triangle containing an edge and the altitudes of two faces.
Bisecting the angle where the two alittudes meet (perpendicularly) bisects the edge.

:::: {layout-ncol="2"}

::: {.column width="49%"}
![](index_files/tetrahedron bisect.png)
:::

::: {.column width="49%"}
$$
\begin{align*}
  \textcolor{blue}{d_\text{length}}^2
    &= d_\text{altitude}^2 -\ \textcolor{green}{1}^2 = 2
  \\
  (2A_\text{center})^2
    &= (2d_\text{length})^2 = (\textcolor{red}{h} d_\text{altitude})^2
  \\
  &= 4 \cdot 2 = 3h^2
  \\
  h^2 &= 8 / 3
\end{align*}
$$
:::

::::

Since *h* is known, we can calculate the volume.

$$
  ({ 2^3 \cdot V_\text{tet[1]} })^2
    &= {B^2 h^2 \over 3^2} = {3 \cdot (8/3) \over 3^2}
  \\
  V_\text{tet[1]}^2
    &= {8 \over 2^6 \cdot 3^2} = {1 \over 2^3 \cdot 3^2} = {1 \over 2 \cdot 6^2}
  \\
  V_\text{tet[1]}
    &= \sqrt{1 \over 6^2 \cdot 2} = {1 \over 6\sqrt 2}
\end{align*}
$$


### Returning to 2D: Regular Pentagons

Both of the icosahedron and dodecahedron contain regular pentagons.
Thus, it is necessary to examine them in detail.

The regular pentagon has five diagonals, which form a pentagram.
Since all angles in a regular pentagon are equal, the trapezoid formed by three consecutive edges
  and one diagonal is isosceles.
This means the diagonal is parallel to one of the edges, which applies to all diagonals by symmetry.

Since the diagonal is parallel to one of the sides, a parallelogram can be formed from
  two sides and segements from two diagonals.
More specifically, this parallelogram is a rhombus, since the segments must have equal lengths
  to the sides.

::: {}
![](index_files/pentagram rhombus.png)

Left: Pentagram in regular pentagon;
Middle: Isosceles trapezoid, with parallel lines marked in blue;
Right: Rhombus in regular pentagon
:::

Bisect the pentagon vertically and let the length of half of the diagonal of a pentagon be *d*,
  half the length of the other diagonal of a rhombus be *h*,
  and the remaining height of the pentagon be *g*.

![](index_files/pentagon measurements.png)

$$
\begin{align*}
  \textcolor{orange}{d}^2 + \textcolor{red}{h}^2
    &= \textcolor{blue}{2}^2
  \\
  2\textcolor{darkblue}{A_\text{blue}}
    &= 2\textcolor{green}{g} = h(\textcolor{magenta}{d -\ (2 -\ d)}) = h (2d -\ 2)
  \\
  \implies g &= {h(2d -\ 2) \over 2} = h(d -\ 1)
\end{align*}
$$

Notice that the center of a pentagram contains a regular pentagon.
This means that the ratio of its height to the side is equal to the ratio of
  the larger pentagon's height to its side.
This is enough information to deduce *d*:

$$
\begin{align*}
  {\textcolor{red}{h} \over 2(\textcolor{brown}{2 -\ d})}
    &= {2\textcolor{red}{h} + \textcolor{green}{g} \over \textcolor{blue}{2}}
    = {2h + h(d-1) \over 2} = {h(1 + d) \over 2 }
  \\
  2h &= 2h(1 + d)(2 -\ d)
  \\
  1 &= (1 + d)(2 -\ d) = 2 -\ d + 2d -\ d^2
  \\
  0 &= d^2 -\ d -\ 1
\end{align*}
$$

This is the minimal polynomial of the golden ratio *φ*.
It is half the length of the diagonal, so the ratio of a diagonal to a side is also *φ*.

To make calculations easier, some conversions will be made to base *φ*, or phinary.
If you are not familiar already with phinary, I have already written at length about it [here](
  /posts/polycount/1
).
Finally, the apothem *a* and height *l* can be calculated by similar triangles.

::: {}
![](index_files/pentagon apothem.png)
:::

$$
\begin{align*}
  \textcolor{blue}{c \over a}
    &= \textcolor{brown}{2 \over \phi},~ a^2 + 1^2 = c^2
    \implies 1 = c^2 -\ a^2 = (c + a)(c -\ a)
  \\
  l
    &= c + a = {2a \over \phi} + a = a{2 + \phi \over \phi}
    = a{12_\phi \over 10_\phi}
    = a{2\bar{1}0_\phi \over 10_\phi} = a(2\bar{1}_\phi)
  \\ \\
  s
    &= c -\ a = {2a \over \phi} -\ a
    = a{2 -\ \phi \over \phi} = a{\bar{1}2_\phi \over 10_\phi}
    = a{2\bar{3}0_\phi \over 10_\phi} = a(2\bar{3}_\phi)
  \\ \\ \\
  1
    &= ls = a^2(2\bar{1}_\phi)(2\bar{3}_\phi)
    = a^2(4\bar{8}3_\phi)
    = a^2(\bar{4}7_\phi) = a^2(3.\bar{4}_\phi)
  \\
  a^2
    &= {1 \over 3.\bar{4}_\phi} \cdot {43_\phi \over 43_\phi}
    = {43_\phi \over [12]\bar{7}.[\bar{12}]_\phi}
    = {3 + 4\phi \over 5}
  \\ \\
  \implies l^2
    &= a^2(2\bar{1}_\phi)^2
    = {3 + 4\phi \over 5} \cdot (4\bar{4}1_\phi)
    = {3 + 4\phi \over 5} \cdot 5 = 3 + 4\phi
\end{align*}
$$

The last few steps in solving for $a^2$ are somewhat tricky.
The conjugate of *φ* is $-{1 \over \phi}$.
Since the digit in the *φ*^^-1^^ place value is negative, its conjugate has a positive value in the *φ* place value,
  i.e., $3.\bar{4}_{\phi^*} = 43_\phi$.
Multiplying a quadratic root by its conjugate produces an integer value,
  which means that the scary quantity $[12]\bar{7}.[\bar{12}]_\phi$ resolves cleanly to 5.

The division can also be done explicitly in phinary:

$$
\begin{align*}
  {1 \over 3.\bar{4}_\phi}
    &= {1 \over 0.\bar{1}3_\phi}
    = {500_\phi \over 5 (\bar{1}3_\phi)}
    = {233_\phi + (0 = \textcolor{red}{4\bar{4}}\bar{4}0_\phi
    = \textcolor{red}{26\bar{2}}\bar{4}0_\phi) \over 5 (\bar{1}3_\phi)}
  \\
  &= {\bar{2}60\bar{1}3_\phi \over 5 (\bar{1}3_\phi)}
    = {2001_\phi \over 5}
    = {221_\phi \over 5}
    = {43_\phi \over 5}
    = {3 + 4\phi \over 5}
\end{align*}
$$


The Remaining Solids
--------------------

With the diagonal length and apothem of a regular pentagon in tow, the geometry of
  the final two solids may be explored.
As a reminder, these solids will have edge length 2, so they will be put in ratio with a cube volume of 8.

The icosahedron and dodecahedron are easiest to dissect as many pyramids joined to a single center.
This is reminiscent of the area formula which uses the triangulation of a regular polygon its center.

The altitude (*h*) of any one pyramid is the radius of the *insphere* of the solid,
  which is tangent to the plane of every face.
Similarly, the *circumsphere* (circumradius, *r*) contains all vertices,
  and the *midsphere* (midradius, *ρ*) is tangent to every edge.
These will become important shortly.


### The Icosahedron: an Antiprism in Profile

The icosahedron may also be thought of as two pentagonal pyramids connected to
  either base of a pentagonal *antiprism*.
An antiprism is a figure similar to a prism, but with the one of the bases twisted relative
  to the other and with (equilateral) triangles joining them.

:::: {layout-ncol="2"}

::: {.column width="49%"}
![](index_files/icosahedron with antiprism.png)

Icosahedron with pentagonal antiprism in blue
:::

::: {.column width="49%"}
![](index_files/equilateral triangles.png)

Construction showing $2a = c$
:::

::::

A segment connecting the centers of two antipodal faces is a diameter of the insphere.
The altitude of one of these faces will be cut into circumradius and inradius.
By similar triangles, the circumradius of an equilateral triangle is
  exactly twice the length of the inradius.
This means the inradius is 1/3 of the altitude, or 1/9 of the square of the altitude.
With edge length 2, the square of the altitude is 3, so the square of the inradius is
  ${3 \over 9} = {1 \over 3}$ .

The pentagonal antiprism may be bisected bisected along the plane containing the altitudes
  of two triangles opposite one another.
This forms a parallelogram with side lengths of the altitude of an equilateral triangle
  and height of a pentagon.

![](index_files/antiprism measures.png)

$$
\begin{align*}
  (\textcolor{green}{2h})^2
    &= \textcolor{red}{l}^2 -\ {1 \over 3}
    = 3 + 4\phi -\ {1 \over 3}
    = {3(3 + 4\phi) -\ 1 \over 3}
  \\
  h^2
    &= {8 + 12\phi \over 3 \cdot 4}
    = {2 + 3\phi \over 3}
    = {32_\phi \over 3}
  \\
  32_\phi
    &= 210_\phi = 1100_\phi = 10000_\phi
    = \phi^4
  \\ \\
  (2^3 \cdot V_\text{ico[1]})^2
    &= \left ( 20 \cdot {Bh \over 3} \right )^2
    = {20^2 B^2 h^2 \over 3^2} = {5^2 \cdot 4^2 \cdot 3 \cdot {\phi^4 \over 3} \over 3^2}
    = {5^2 \cdot 2^4 \cdot \phi^4 \over 3^2}
  \\
  V^2
    &= {5^2 \cdot 2^4 \cdot \phi^4 \over 2^6 \cdot 3^2}
    = {5^2 \cdot \phi^4 \over 2^2 \cdot 3^2}
  \\
  V &= {5 \phi^2 \over 6}
\end{align*}
$$


### The Dodecahedron

The dodecahedron is a bit trickier.
It belongs to a class of polyhedra known as *truncated trapezohedra*.
However, the bisection trick from the icosahedron still works.

Begin by bisecting the solid through antipodal altitudes.
This produces an oblong hexagon made up of four pentagon heights and two edges.

![](index_files/dodecahedron with hemisphere.png)

The segment connecting the antipodal midpoints bisects the hexagon into two (isosceles) trapezoids,
  and is a diameter of the midsphere.
Additionally, it is parallel to the two edges. A second midradius is perpendicular to this one,
  bisecting the trapezoid.

![](index_files/dodecahedron measures.png)

The inradius is the altitude of a triangle formed by the length of a pentagon (its base),
  a midradius, and a circumradius.
However, the altitude with respect to the midradius is another midradius.
This means that the height can be found by equating areas and completing the square.

$$
\begin{align*}
  a^2 + \textcolor{green}{h}^2
    &= \textcolor{blue}{\rho}^2,~~
    \textcolor{orange}{l}\textcolor{green}{h}
    = \textcolor{blue}{\rho \rho} \implies \textcolor{orange}{l}^2\textcolor{green}{h}^2
    = (\textcolor{orange}{(2\bar{1}_\phi) a})^2 h^2
    = 5a^2h^2
    = \textcolor{blue}{\rho}^4
  \\
  5a^2h^2
    &= (a^2 + h^2)^2
    = a^4 + 2a^2h^2 + h^4
  \\ \\
  0
    &= a^4 -\ 3a^2h^2 + h^4
    = (h^2 -\ x)^2 + y
    = h^4 -\ 2xh^2 + x^2 + y
  \\
  -2x
    &= -3a^2
    \implies x = {3a^2 \over 2},~~
    x^2 + y = {9a^4 \over 4} + y = a^4
  \\
  y
    &= {4a^4 \over 4} -\ {9a^4 \over 4} = -{5a^4 \over 4}
    = -{(2\bar{1}_\phi)^2 a^4 \over 4}
    = -\left ( {(2\bar{1}_\phi) a^2 \over 2} \right )^2
  \\ \\
  (h^2 -\ {3a^2 \over 2})^2
    &= -y
    = \left ( {(2\bar{1}_\phi) a^2 \over 2} \right )^2
  \\
  h^2 -\ {3a^2 \over 2}
    &= {(2\bar{1}_\phi) a^2 \over 2}
  \\
  h^2
    &= {(2\bar{1}_\phi) a^2 \over 2} + {3a^2 \over 2}
    = {(22_\phi) a^2 \over 2} = (11_\phi) a^2
  \\
  &= {(11_\phi)(43_\phi) \over 5}
    = {473_\phi \over 5}
    = {[11]7_\phi \over 5} = {7 + 11\phi \over 5}
\end{align*}
$$

With the square of the height known, all that is left to do is find the volume.

$$
\begin{align*}
  B^2
    &= \left( {Pa \over 2} \right)^2
    = (5a)^2 = 25a^2 = 5(43_\phi)
  \\
  5h^2
    &= [11]7_\phi = 740_\phi = 4300_\phi
    = (43_\phi)(100_\phi)
  \\
  (2^3 \cdot V_\text{dodec[1]})^2
    &= \left (12 \cdot {Bh \over 3} \right )^2 = 4^2 B^2 h^2
    = 2^4 \cdot 5(43_\phi) \cdot {(43_\phi)(100_\phi) \over 5}
  \\
  V^2
    &= {2^4 \cdot (43_\phi)^2 \cdot (100_\phi) \over 2^6}
    = {(43_\phi)^2(10_\phi)^2 \over 2^2}
  \\
  V
    &= {(43_\phi)(10_\phi) \over 2} = {(430_\phi) \over 2} = {(74_\phi) \over 2}
    = {4 + 7\phi \over 2}
\end{align*}
$$


Closing
-------

Since each of these volumes has been calculated algebraically, there have been no approximate decimal forms.
Ordered by size, the volumes of each of the solids are:

| Solid        | Volume                | Approximation   | Length of Side with Unit Volume |
|--------------|-----------------------|-----------------|---------------------------------|
| Tetrahedron  | ${1 \over 6\sqrt 2}$  | 0.1178511302... | 2.039648903...                  |
| Octahedron   | ${\sqrt{2} \over 3}$  | 0.4714045208... | 1.284898293...                  |
| Cube         | $1^3$                 | 1               | 1                               |
| Icosahedron  | ${5\phi^2 \over 6}$   | 2.181694991...  | 0.7710253465...                 |
| Dodecahedron | ${4 + 7\phi \over 2}$ | 7.663118961...  | 0.5072220724...                 |

The dodecahedron being so much larger than the icosahedron surprised me, to be honest.
When one glances at a set of dice (as one does), it seems like the d20 is larger than the d12,
  albeit with smaller edges.
However, at least in one of my sets, the edges of the d20 are in fact about 1.5 times as long
  as those of the d12, implying their volumes are roughly equal.

***

I tried to use as much coordinate-free geometry as I could in producing these diagrams.
GeoGebra lacks a tool for producing Platonic solids other than cubes and tetrahedra,
  so I ended up cheating in coordinates for the octahedron and icosahedron diagrams.
On the other hand, the hexagon I described in the dodecahedron is of such importance to
  its construction that I ended up constructing it from scratch.
I am rather proud of this because I did so without looking up someone else's.
After having written this post, I feel much more competent with compass-and-straightedge constructions.

All diagrams made with GeoGebra.
