Editing History of geometry (section)

===Egyptian geometry===
{{main|Egyptian geometry}}

The ancient Egyptians knew that they could approximate the area of a circle as follows:<ref name="School Mathematics">Ray C. Jurgensen, Alfred J. Donnelly, and Mary P. Dolciani. Editorial Advisors Andrew M. Gleason, Albert E. Meder, Jr. ''Modern School Mathematics: Geometry'' (Student's Edition). Houghton Mifflin Company, Boston, 1972, p.&nbsp;52. {{ISBN|0-395-13102-2}}. Teachers Edition {{ISBN|0-395-13103-0}}.</ref>

::::Area of Circle ≈ [ (Diameter) x 8/9 ]<sup>2</sup>.

Problem 50 of the [[Ahmes]] papyrus uses these methods to calculate the area of a circle, according to a rule that the area is equal to the square of 8/9 of the circle's diameter. This assumes that [[pi|{{pi}}]] is 4&times;(8/9)<sup>2</sup> (or 3.160493...), with an error of slightly over 0.63 percent. This value was slightly less accurate than the calculations of the [[Babylonia]]ns (25/8 = 3.125, within 0.53 percent), but was not otherwise surpassed until [[Archimedes]]' approximation of 211875/67441 = 3.14163, which had an error of just over 1 in 10,000.

Ahmes knew of the modern 22/7 as an approximation for {{pi}}, and used it to split a hekat, hekat x 22/x x 7/22 = hekat;{{cn|date=July 2018}} however, Ahmes continued to use the traditional 256/81 value for {{pi}} for computing his hekat volume found in a cylinder.

Problem 48 involved using a square with side 9 units. This square was cut into a 3x3 grid. The diagonal of the corner squares were used to make an irregular octagon with an area of 63 units. This gave a second value for {{pi}} of 3.111...

The two problems together indicate a range of values for {{pi}} between 3.11 and 3.16.

Problem 14 in the [[Moscow Mathematical Papyrus]] gives the only ancient example finding the volume of a [[frustum]] of a pyramid, describing the correct formula:
:<math>V = \frac{1}{3} h(a^2 + ab + b^2)</math>
where ''a'' and ''b'' are the base and top side lengths of the truncated pyramid and ''h'' is the height.