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. Triangles: The Ancient Egyptians knew that the area of a triangle is 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 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: Next we approximate 63 to be 64 and note that Thus the number 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 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
Several problems compute the volume of cylindrical granaries, 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 of four palms. 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. Several problems in the Moscow Mathematical Papyrus and in the Rhind Mathematical Papyrus 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.
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 needed for building pyramids. In the next problem, the height of a pyramid is calculated from the base length and the seqed, 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 and with a seqed of 5 palms 1 finger; what is its altitude?
