Pyramid Geometry

ancient-egypt style geometry for square pyramids

Starting values — Pyramid Geometry

Base side length (s)	230.4
Vertical height (h)	146.6

Change any control above and the results recompute from these.

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The one third is exact, but you cannot prove it by cutting 🖖

A pyramid holds exactly one third of the prism sharing its base and height — not approximately, exactly. In two dimensions the matching fact is easy to see: a triangle is half its rectangle, and you can demonstrate it by cutting the rectangle up and rearranging the pieces. In three dimensions that route is closed. Hilbert asked in 1900 whether any two solids of equal volume can be cut into finitely many pieces and reassembled into each other, and Max Dehn proved in 1901 that they cannot — a regular tetrahedron and a cube of the same volume are not reachable from one another by any finite set of cuts. The factor of 1/3 is therefore genuinely a result about limits, which is why it arrives with calculus rather than with scissors.

Two numbers define the whole pyramid 🖖

A square pyramid is fully fixed by just two measurements: the base side s and the vertical height h. Everything else follows from one right triangle running from the center of the base to the midpoint of an edge. The slant height is its hypotenuse, l = √(h² + (s/2)²), and once you have l, the surface area and volume drop right out. Change one input and every result recomputes together.

How Egyptians measured slope backwards 🖖

The ancient Egyptians used no angles. In the Rhind Mathematical Papyrus they recorded a pyramid's slope as its seked: the horizontal run, in palms, for every royal cubit (7 palms) of vertical rise — essentially the cotangent, the inverse of our modern slope. The Great Pyramid's seked is about 5½ palms, giving a face angle near 51.84°. This tool's seked output revives that 3,600-year-old convention.

Problem solved in full

Two Great Pyramid coincidences, settled with one number 6 steps
The Great Pyramid: base 230.4 m, height 146.6 m. Get the slant, the areas, the volume and the face angle — then settle the two famous coincidences with one number.
1. Slant height is the hypotenuse of the triangle from the centre of the base to the mid-edge and up to the apex — half the base, not the full base, which is the commonest slip here.
  
  $ℓ = h^{2} + (\frac{s}{2})^{2} = 146. 6^{2} + 115. 2^{2} = 34 760.2 = 186.4473 m$
2. The lateral surface is four triangles, each with base s and height ℓ, so 4 × ½sℓ collapses to 2sℓ. Add the base for the total.
  
  $A_{b} = s^{2} = 53 084.16, A_{l} = 2 s ℓ = 85 914.92, A = 138 999.08 m^{2}$
3. Volume is one third of the prism, and the third is not an approximation — it is exact for any pyramid on any base.
  
  $V = \frac{1}{3} s^{2} h = \frac{1}{3} (53 084.16) (146.6) = 2594 045.95 m^{3}$
4. The face angle comes from rise over half-run. The seked is its reciprocal, and that is the number the builders actually worked in.
  
  $θ = arctan \frac{2 h}{s} = arctan \frac{293.2}{230.4} = 51.839 3^{\circ}, k = \frac{s /2}{h} = 0.7858$
5. Compare the measured seked with the simplest fraction near it. Five and a half palms per cubit of seven palms is 11/14, and it matches to one part in eight thousand.
  
  $k = \frac{5 \frac{1}{2} palms}{7 palms} = \frac{11}{14} = 0.785714 (measured 0.785812)$
6. Now derive the π ratio from that one fraction. It is not an independent fact about the pyramid; it is 8k, and 8 × 11/14 happens to sit 0.04% from 2π.
  
  $k = \frac{s /2}{h} \Rightarrow h = \frac{s}{2 k} \Rightarrow \frac{4 s}{h} = 8 k = 8 \cdot \frac{11}{14} = 6.28571 against 2 π = 6.28319$
Answer

Both coincidences are the same coincidence, and it is a builder's fraction: a seked of 11/14. The Egyptians did not specify a slope as an angle; they specified how many palms of horizontal run go with one cubit of rise, and a cubit is seven palms. Choose 5½ palms and the seked is 11/14 = 0.785714 — against the 0.785812 the pyramid actually measures. Everything else follows. Perimeter over height is exactly 8 times the seked, so 8 × 11/14 = 6.28571 against 2π = 6.28319, and the slant-to-half-base ratio comes out at 1.6185 against φ = 1.618034. Neither π nor φ was put in. One rational slope, chosen because it was easy to set out with a rope and a plumb line, produces both to four figures.

$k = \frac{11}{14} ⟹ \frac{4 s}{h} = 6.2857 \approx 2 π and \frac{ℓ}{s /2} = 1.6185 \approx φ$

References (2)

Insight block 1 — why the factor of 1/3 cannot be shown with scissors: M. Dehn, "Ueber den Rauminhalt." Mathematische Annalen 55(3), 465–478, 1901 — the negative answer to Hilbert's third problem.
Insight block 3 — the seked, and the 5½ palms of the Great Pyramid: A. B. Chace, The Rhind Mathematical Papyrus, problems 56–60. Mathematical Association of America, 1927–29.

By the 42Math team · Updated 2026-05-12

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Mathematics lessons →

Example problems

school example - School example: compute slant height, area, and volume.
great pyramid - Great Pyramid scale approximation with real-world dimensions.
steep face - Steeper pyramid with larger face angle.
flat face - Flatter pyramid with smaller face angle.