Geometry

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Deep Secrets Geometry

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Unfortunately, a great many school children are mislead into believing is 3+1/7 = 3.142857 - accurate to < 1/100. It is a common fallacy is that the only computed as 3+1/8 using the observation below that the area of a circle of radius is "close to" the area of a square 8 units on a side. Until recently, Archimedes of Syracuse (250 BC) was generally consider the first person to calculate pi to some accuracy; however, as we shall see below the Egyptians already knew Archimedes (250B.C.) value of = 256/81 = 3 + 1/9 + 1/27 + 1/81, (the suggestion that the egyptians used 3 + 1/13 + 1/17 + 1/160 = 3.1415 for is at best implicit) exhibited in the problem 50 below. The astronomer Ptolemy, of Alexandria AD 150, knew 3+10/71 < <3+1/7 while in China in the fifth century, Tsu Chung-Chih calculate pi correctly to seven digits. Today, we "only" know to 50 billion decimal places.

For More info on Pi click here for Dons Pi Page.

Note 1 khet is 100 cubits, and 1 meter is about 2 cubits. A setat is a measurement of area equal to what we would call a square khet.

	An alternate conjecture exhibiting the value of is that the egyptians easily observed that the area of a square 8 units on a side can be reformed to nearly yield a circle of diameter 9.

Rhind papyrus Problem 50. A circular field has diameter 9 khet. What is its area.

The written solution says, subtract 1/9 of of the diameter which leaves 8 khet. The area is 8 multiplied by 8, or 64 setat. Now it would seem something is missing unless we make use of modern data: The area of a circle of diameter d is (d/2)² =d²/4. Now assume 64 = 9²/4 = 81/4, then = 3 + 1/9 + 1/27 + 1/81 ~ 3.1605. But 3 + 1/9 + 1/27 + 1/81 is a number, presumably, intrinsically more pleasing to the egyptians than
3 + 1/13 + 1/17 + 1/160.

THE PYTHAGOREAN THEOREM WORKING!!!

Moscow Papyrus Problem 10. line-by-line translation

Example of calculating [the surface area of] a basket [hemisphere]. geometry/
You are given a hemisphere with a mouth [magnitude]

of 4 + 1/2 [in diameter].

What is its surface?

Take 1/9 of of 9 [since]

the basket is half an egg [hemisphere]. You get 1.

Calculate the remainder [when subtracted from 9] which is 8.

Calculate 1/9 of 8.

You get 2/3 + 1/6 + 1/18.

Find the remainder of this 8

After subtracting 2/3 + 1/6 + 1/18. You get 7 + 1/9.

Multiply 7 + 1/9 by 4 + 1/2.

You get 32. Behold this is its surface [area]!

You have found it correctly.

In our notation and method here is what occurred.
Let d be the diameter and S be the surface area.
S = 2d(8/9)(8/9)d =

The problem and its solution can be interpreted as follows: Find the area of a hemisphere (a basket of half an egg] of diameter 4 + 1/2. The surface area of a hemisphere is
= 256/81 = 3 + 1/9 + 1/27 + 1/81

Egyptian Mathematicians

Abu Kamil Shuja
Ahmes
Diophantus
Euclid

Heron
Hypatia
Hypsicles
Menelaus

Pappus
Ptolemy
Serenus
Theon

Yunus

http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egypt_geometry.html