Size of Surface of Regular Dodecahedron

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Theorem

In the words of Hypsicles of Alexandria:

If there be an equilateral and equiangular pentagon and a circle circumscribed about it, and if a perpendicular be drawn from the centre to one side, then
$30$ times the rectangle contained by the side and the perpendicular is equal to the surface of the dodecahedron.

(The Elements: Book $\text{XIV}$: Proposition $3$)


Proof

Euclid-XIV-3.png

Let $ABCDE$ be the regular pentagon which is the face of a regular dodecahedron.

Let the circle $ABCDE$ be circumscribed around the pentagon $ABCDE$.

Let $F$ be the center of the circle $ABCDE$.

Let $FG$ be the perpendicular dropped from $F$ to $CD$.


Let $CF$ and $FD$ be joined.

Then:

\(\ds CD \cdot FG\) \(=\) \(\ds 2 \cdot \triangle CDF\) Area of Triangle in Terms of Side and Altitude
\(\ds \therefore \ \ \) \(\ds 30 \cdot CD \cdot FG\) \(=\) \(\ds 60 \cdot \triangle CDF\)
\(\ds \) \(=\) \(\ds 12 \cdot \map \AA {ABCDE}\)

where $\map \AA {ABCDE}$ denotes the area of the regular pentagon $ABCDE$.

The result follows from the definition of a regular dodecahedron as having $12$ such faces.

$\blacksquare$


Historical Note

This proof is Proposition $3$ of Book $\text{XIV}$ of Euclid's The Elements.


Sources