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Revision difference : Egyptian geometry areas calculated in, cubits, khets and setats
Version 21 Version 20
A scribal area of triangles and other shapes area calculation method is reported by three \PMlinkexternal{Rhind Mathematical Papyrus}{http://ahmespapyrus.blogspot.com/2009/01/ahmes-papyrus-new-and-old.html} problems \PMlinkexternal{RMP 53-55}{http://rmp50-60.blogspot.com/}. The scribal geometry utilized quotients and remainders in an arithmetic context that was looked like scribal weights and measures, algebra, and/or 2/n tables calculations. A scribal area of triangles and other shapes area calculation method is reported by three \PMlinkexternal{Rhind Mathematical Papyrus}{http://ahmespapyrus.blogspot.com/2009/01/ahmes-papyrus-new-and-old.html} problems \PMlinkexternal{RMP 53-55}{http://rmp50-60.blogspot.com/}. The scribal geometry utilized quotients and remainders in an arithmetic context that was looked like scribal weights and measures, algebra, and/or 2/n tables calculations.
RMP 53 calculated the area of two triangles, of 45/8 setat and 63/8 setat, and a third area of an undefined shape by the note, 1/10 of 1 3/8 mh added to 10 cubits of land (COL) is the desired area. A setat was 100 cubit by 100 cubit, or 10,000 square cubits. A cubit of land (COL), or mh, was one cubit wide by 100 cubits long, or 1/100 setat. RMP 53 calculated the area of two triangles, of 45/8 setat and 63/8 setat, and a third area of an undefined shape by the note, 1/10 of 1 3/8 mh added to 10 cubits of land (COL) is the desired area. A setat was 100 cubit by 100 cubit, or 10,000 square cubits. A cubit of land (COL), or mh, was one cubit wide by 100 cubits long, or 1/100 setat.
The first triangle had an altitude of 5 khet and a base of 9/4 khet. Using the area of a triangle formula, 1/2 the base times the altitude, 5*(9/4)*(1/2)= (45/8) = 5 5/8 setat. The first triangle had an altitude of 5 khet and a base of 9/4 khet. Using the area of a triangle formula, 1/2 the base times the altitude, 5*(9/4)*(1/2)= (45/8) = 5 5/8 setat.
The second triangle had an altitude of 7 khet and a base of 9/4 khet. Using the area of a triangle formula, 1/2 the base times the altitude, Ahmes calculated 7*(9/4)*(1/2) = 63/8 = 7 7/8 setat The second triangle had an altitude of 7 khet and a base of 9/4 khet. Using the area of a triangle formula, 1/2 the base times the altitude, Ahmes calculated 7*(9/4)*(1/2) = 63/8 = 7 7/8 setat
The third calculation found the area of undefined shape discussed by: The third calculation found the area of undefined shape discussed by:
11/8 mh = 110/8 mh + 10 mh = 23 3/4 mh = 1/8 setat + 11 1/4 mh 11/8 mh = 110/8 mh + 10 mh = 23 3/4 mh = 1/8 setat + 11 1/4 mh
since 12 1/2 mh = 1/8 setat. since 12 1/2 mh = 1/8 setat.
Scholars have suggested that a truncated pyramid or a triangle defined the third shape. Scholars have suggested that a truncated pyramid or a triangle defined the third shape.
To assist the decoding of the third RMP 53 area RMP 54, and RMP 55 setat and mh data have been consulted. To assist the decoding of the third RMP 53 area, RMP 54, and RMP 55 details have been consulted.
RMP 54 partitioned 7/10 setat by 10, 5, 2 1/2 and 1 1/4 segments. Proof was provided by multiplying one setat by 7/10, 14/10, 28/10 and 56/10 within a quotient and remainder context. A quotient setat and a scaled remainder mh were scaled as the 2/n table and a ro unit in hekat (volume unit) were scaled, by writing: RMP 54 segmented 7/10 setat by 10, 5, 2 1/2 and 1 1/4 segments. Proof was provided by finding setat + COL areas of 7/10, 14/10, 28/10 and 56/10 setats segments within a quotient and remainder, that looked like a 2/n table context:
a. (7/10)*(4/4) = 28/40 = (24 + 3)/40 = 3/8 setat + 300/40 mh = 5/8 setat + 7 1/2 mh a. (7/10)*(4/4) = 28/40 = (24 + 3)/40 = 3/8 setat + 300/40 mh = 5/8 setat + 7 1/2 mh
b. (14/10)*(4/4) = 56/10 = (55 + 1)/40 = 11/8 setat + 100/4 mh = 1 3/8 setat + 2 1/2 mh b. (14/10)*(4/4) = 56/10 = (55 + 1)/40 = 11/8 setat + 100/4 mh = 1 3/8 setat + 2 1/2 mh
c. (28/10)*(2/2) = 56/20 = (55 + 1)/20 = 11/4 setat + 100/20 mh = 2 3/4 setat + 5 mh c. (28/10)*(2/2) = 56/20 = (55 + 1)/20 = 11/4 setat + 100/20 mh = 2 3/4 setat + 5 mh
d. (56/10) = (55 + 1)/10 = 11/2 setat + 100/10 COL = 5 1/2 setat + 10 mh d. (56/10) = (55 + 1)/10 = 11/2 setat + 100/10 COL = 5 1/2 setat + 10 mh
Ahmes may have also made calculations thinking in mh unuts. For example, Ahmes may have made calculations in mh unuts, or by following the 2/n table scaling method.
Ahmes shorthand partition of 7/10 setat, (1/2 + 1/5) setat, may have focused upon 1/5 setat written as 20 mh. Knowing 12 1/2 mh was 1/8 setat, an answer may have been recorded by: For example:
(1/2 + 1/5)setat = (1/2 + 1/8 + (20 - 12 1/2 mh) = 5/8 setat + 7 1/2 mh. 7/10 setat, written as (1/2 + 1/5) setat, may have written 1/5 setat as 20 mh. Knowning that 12 1/2 mh = 1/8 setat
(1/2 + 1/5)setat = (1/2 + 1/8) = 5/8 setat + (20 - 12 1/2 mh) = 7 1/2 mh = 5/8 setat + 7 1/2 mh.
RMP 55 takes 3/5 of 5 setat to obtain 3 setat by three steps: RMP 55 takes 3/5 of 5 setat to obtain 3 setat by three steps:
a. 1/2 setat + 10 mh a. 1/2 setat + 10 mh
b. 1 1/8 setat + 7 1/2 mh b. 1 1/8 setat + 7 1/2 mh
c. 1 3/8 setat + 2 1/2 setat c. 1 3/8 setat + 2 1/2 setat
d. adding steps a. and c, knowing that 12 1/2 mh = 1/8 setat d. adding steps a. and c, knowing that 12 1/2 mh = 1/8 setat
(1/2 setat + 10 mb) + (1 3/8 setat + 2 1/2 mh) = 2 7/8 setat + 12 1/2 mh = 3 setat (1/2 setat + 10 mb) + (1 3/8 setat + 2 1/2 mh) = 2 7/8 setat + 12 1/2 mh = 3 setat
\begin{thebibliography}{9} \begin{thebibliography}{9}
\bibitem{1} Milo Gardner, \emph{The Egyptian Mathematical Leather Roll Attested Short Term and Long Term, History of Mathematical Sciences}, Hindustan Book Company, 2002. \bibitem{1} Milo Gardner, \emph{The Egyptian Mathematical Leather Roll Attested Short Term and Long Term, History of Mathematical Sciences}, Hindustan Book Company, 2002.
\bibitem{2} Milo Gardner, \emph{An Ancient Egyptian Problem and its Innovative Solution, Ganita Bharati}, MD Publications Pvt Ltd, 2006. \bibitem{2} Milo Gardner, \emph{An Ancient Egyptian Problem and its Innovative Solution, Ganita Bharati}, MD Publications Pvt Ltd, 2006.
\bibitem{3}Richard Gillings, \emph{Mathematics in the Time of the Pharaohs}, Dover Books, 1992. \bibitem{3}Richard Gillings, \emph{Mathematics in the Time of the Pharaohs}, Dover Books, 1992.
\bibitem{4} Oystein Ore, \emph{Number Theory and its History}, McGraw-Hill Books, 1948, Dover reprints available. \bibitem{4} Oystein Ore, \emph{Number Theory and its History}, McGraw-Hill Books, 1948, Dover reprints available.
\bibitem{5} T.E. Peet, \emph{Arithmetic in the Middle Kingdom}, Journal Egyptian Archeology, 1923. \bibitem{5} T.E. Peet, \emph{Arithmetic in the Middle Kingdom}, Journal Egyptian Archeology, 1923.
\bibitem{6} Tanja Pommerening, \emph{"Altagyptische Holmasse Metrologish neu Interpretiert" and relevant phramaceutical and medical knowledge, an abstract, Phillips-Universtat, Marburg, 8-11-2004, taken from "Die Altagyptschen Hohlmass}, Buske-Verlag, 2005. \bibitem{6} Tanja Pommerening, \emph{"Altagyptische Holmasse Metrologish neu Interpretiert" and relevant phramaceutical and medical knowledge, an abstract, Phillips-Universtat, Marburg, 8-11-2004, taken from "Die Altagyptschen Hohlmass}, Buske-Verlag, 2005.
\bibitem{7} Gay Robins, and Charles Shute \emph{Rhind Mathematical Papyrus}, British Museum Press, Dover reprint, 1987. \bibitem{7} Gay Robins, and Charles Shute \emph{Rhind Mathematical Papyrus}, British Museum Press, Dover reprint, 1987.
\bibitem{8} L.E. Sigler, \emph{Fibonacci's Liber Abaci: Leonardo Pisano's Book of Calculation}, Springer, 2002. \bibitem{8} L.E. Sigler, \emph{Fibonacci's Liber Abaci: Leonardo Pisano's Book of Calculation}, Springer, 2002.
\bibitem{9} Hana Vymazalova, \emph{The Wooden Tablets from Cairo:The Use of the Grain Unit HK3T in Ancient Egypt, Archiv Orientalai}, Charles U Prague, 2002. \bibitem{9} Hana Vymazalova, \emph{The Wooden Tablets from Cairo:The Use of the Grain Unit HK3T in Ancient Egypt, Archiv Orientalai}, Charles U Prague, 2002.
\end{thebibliography} \end{thebibliography}