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
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
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): The So-Called Book $\text{XIV}$, by Hypsicles