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Scribal areas of triangles and areas of other shapes were reported in three Rhind Mathematical Papyrus problems RMP 53-55. The scribal geometry utilized quotients and remainders in an arithmetic context that followed scribal weights and measures that partitioned one hekat by substituting 64/64, 10 hin, 64 dja or 320 ro. Answers to each problem were multiplied by its divisor to return 64/64, 10 hin, 64 dja, 320 ro), algebra, and/or 2/n tables (converting 2/n by LCM multipliers, recalling 2/97 times 56/56) 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.
The area first triangle, with an altitude of 5 khet, and a base of 9/4 khet, was found by triangle formula:
1/2 the base times the altitude, 5*(9/4)*(1/2)= (45/8) = 5 5/8 setat.
The second triangle area, with an altitude of 7 khet, and a base of 9/4 khet, was found by the 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:
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.
Alternative views suggest that the third shape may have defined a truncated pyramid (base 6, top 3, height 95/18), or a triangle (base 6, and altitude 95/12).
To assist the decoding of the third RMP 53 area RMP 54, and RMP 55 scribal guidelines 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:
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
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
Ahmes may have also made calculations thinking in mh unuts. For example,
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:
(1/2 + 1/5)setat = (1/2 + 1/8 + (20 - 12 1/2 mh) = 5/8 setat + 7 1/2 mh.
RMP 55 takes 3/5 of 5 setat to obtain 3 setat by four steps (a to d):
a. 1/2 setat + 10 mh
b. (1 + 1/8) setat + (7 + 1/2) mh
c. (1 + 3/8) setat + (2 + 1/2) setat
d. adding steps a and c, and knowing (12 + 1/2) mh = 1/8 setat
(1/2 setat + 10 mh) + [(1 + 3/8) setat + (2+ 1/2)mh] = (2 + 7/8)setat + (1 + 2 + 1/2) mh = 3 setat
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