Lincoln index

Egyptian geometry, as that term is used in this article, refers to geometry as it was developed and used in Ancient Egypt. Ancient Egyptian mathematics as discussed here spans a time period ranging from ca. 3000 BC to ca 300 BC.

We only have a limited number of problems from ancient Egypt that concern geometry. Geometric problems appear in both the Moscow Mathematical Papyrus (MMP) and in the Rhind Mathematical Papyrus (RMP). The examples demonstrate that the Ancient Egyptians knew how to compute areas of several geometric shapes and the volumes of cylinders and pyramids. Also the Egyptians used many sacred geometric shapes such as squares and triangles on temples and obelisks.

Area

The Ancient Egyptians wrote out their problems in multiple parts. They gave the title and the data for the given problem, in some of the texts they would show how to solve the problem, and as the last step they verified that the problem was correct. The scribes did not use any variables and the problems were written in prose form. The solutions were written out in steps, outlining the process.

Areas

Object	Source	Formula (using modern notation)
triangle	Problem 51 in RMP and problems 4, 7 and 17 in MMP	$A=\frac{1}{2}bh$ b = base, h = height
rectangles	Problem 49 in RMP and problems 6 in MMP and Lahun LV.4. problem 1	$A=bh$ b = base, h = height
circle	Problems 51 in RMP and problems 4, 7 and 17 in MMP	$A=\frac{1}{4}(\frac{256}{81})d^2$ d= diameter. This uses the value 256/81 = 3.16049... for $\pi =3.1459...$
hemisphere	Problem 10 in MMP

Triangles:
The Ancient Egyptians knew that the area of a triangle is $A=\frac{1}{2}bh$ where b = base and h = height. Calculations of the area of a triangle appear in both the RMP and the MMP.

Rectangles:
Problem 49 from the RMP finds the area of a rectangular plot of land Problem 6 of MMP finds the lengths of the sides of a rectangular area given the ratio of the lengths of the sides. This problem seems to be identical to one of the Lahun Mathematical Papyri in London. The problem is also interesting because it is clear that the Egyptians were familiar with square roots. They even had a special hieroglyph for finding a square root. It looks like a corner and appears in the fifth line of the problem. We suspect that they had tables giving the square roots of some often used numbers. No such tables have been found however. Problem 18 of the MMP computes the area of a length of garment-cloth.

The Lahun PapyrusProblem 1 in LV.4 is given as: An area of 40 "mH" by 3 "mH" shall be divided in 10 areas, each of which shall have a width that is 1/2 1/4 of their length. A translation of the problem and its solution as it appears on the fragment is given on the website maintained by University College London.

Circles:
Problem 48 of the RMP compares the area of a circle (approximated by an octagon) and its circumscribing square. This problem's result is used in problem 50.

Trisect each side. Remove the corner triangles. The resulting octagonal figure approximates the circle. The area of the octagonal figure is:

$9^2-4\frac{1}{2}(3)(3)=63$ Next we approximate 63 to be 64 and note that $64=8^2$

Thus the number $4(\frac{8}{9}{)}^2=3.16049...$ plays the role of π = 3.14159....

That this octagonal figure, whose area is easily calculated, so accurately approximates the area of the circle is just plain good luck. Obtaining a better approximation to the area using finer divisions of a square and a similar argument is not simple.

Problem 50 of the RMP finds the area of a round field of diameter 9 khet. This is solved by using the approximation that circular field of diameter 9 has the same area as a square of side 8. Problem 52 finds the area of a trapezium with (apparently) equally slanting sides. The lengths of the parallel sides and the distance between them being the given numbers.

Hemisphere:
Problem 10 of the MMP computes the area of a hemisphere.

Volumes

Image of Problem 14 from the Moscow Mathematical Papyrus. The problem includes a diagram indicating the dimensions of the truncated pyramid.

Several problems compute the volume of cylindrical granaries (41, 42, and 43 of the RMP), while problem 60 RMP seems to concern a pillar or a cone instead of a pyramid. It is rather small and steep, with a seked (slope) of four palms (per cubit).

A problem appearing in section IV.3 of the Lahun Mathematical Papyri computes the volume of a granary with a circular base. A similar problem and procedure can be found in the Rhind papyrus (problem 43). Several problems in the Moscow Mathematical Papyrus (problem 14) and in the Rhind Mathematical Papyrus (numbers 44, 45, 46) compute the volume of a rectangular granary.

Problem 14 of the Moscow Mathematical Papyrus computes the volume of a truncated pyramid, also known as a frustum.

Volumes

Object	Source	Formula (using modern notation)
Cylindrical granaries	RMP 41	$V=\frac{256}{81}{r}^{2}h$ measured in cubic-cubits
Cylindrical granaries	RMP 42, Lahun IV.3	$V=\frac{32}{27}{d}^{2}h=\frac{128}{27}{r}^{2}h$ (measured in khar).
Rectangular granaries	RMP 44-46 and MMP 14	$V=wlh$ w = width, l = length, h = height
Truncated pyramid (frustum)	MMP 14	$V=\frac{1}{3}(a^2+ab+b^2)h$

Seqed

Problem 56 of the RMP indicates an understanding of the idea of geometric similarity. This problem discusses the ratio run/rise, also known as the seqed. Such a formula would be need for building pyramids. In the next problem (Problem 57), the height of a pyramid is calculated from the base length and the seked (Egyptian for slope), while problem 58 gives the length of the base and the height and uses these measurements to compute the seqed.

In Problem 59 part 1 computes the seqed, while the second part may be a computation to check the answer: If you construct a pyramid with base side 12 [cubits] and with a seqed of 5 palms 1 finger; what is its altitude? 50 year old Petroleum Engineer Kull from Dawson Creek, spends time with interests such as house brewing, property developers in singapore condo launch and camping. Discovers the beauty in planing a trip to places around the entire world, recently only coming back from .

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