Math 409/409G History of Mathematics Egyptian Geometry

Egyptians and the 3 – 4 – 5 triangle The Egyptians of 2000 B.C. didn’t know about the Pythagorean Theorem; after all, Pythagoras wasn’t born until 572 B.C. But they did know about the 3 – 4 – 5 triangle. In fact, Egyptian architects made great use of this triangle to create right angles. Here’s how they used it.

They knotted a loop of rope into 12 equally long segments.

Then they shaped the rope into a 3 – 4 – 5 triangle to produce a right angle.

By 1850 B.C., the Egyptians had developed formulas for the area of some planar figures and the volume of some 3-dimensional solids. Their formulas have been handed down to us in the 25 problems of the Moscow Papyrus. Not all of their formulas were correct.

Moscow Papyrus Problem 14 This problem deals with finding the volume of a truncated square pyramid and is one of the correct Egyptian formulas. Before looking at this problem, let’s review some terms.

What’s a pyramid? A pyramid is named after its base. At the right we have a square pyramid. The segment through the center of the base and the vertex of the pyramid is perpendicular to the base and is the height of the pyramid.

What’s a truncated pyramid? It’s simply a pyramid with its top lopped off by a plane parallel to its base. Such a solid is also called a frustum of a pyramid. At the right is a frustum of a right circular cone.

Statement of Problem 14 “A truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top.”

Egyptian solution to Problem 14 “You are to square this 4, result 16. You are to double the 4, result 8. You are to square the 2, result 4. You are to add the 16, the 8, and the 4, result, 28. You are to take a third of 6, result 2. You are to take 28 twice, result 56. See, it is 56. You will find it right.”

How does this solution give us a formula for a truncated pyramid? R 1 “You are to square this 4, result 16. R 2 You are to double the 4, result 8. R 3 You are to square the 2, result 4. R 4 You are to add the 16, the 8, and the 4, result, 28. R 5 You are to take a third of 6, result 2. R 6 You are to take 28 twice, result 56. R 7 See, it is 56. You will find it right.”

Here’s the Egyptian solution to the formula for a truncated, square pyramid R 1 “You are to square this 4, result 16.” R 1 = b 2

R 2 “You are to double the 4, result 8.” R 2 = ab

R 3 “You are to square the 2, result 4. R 3 = a 2

R 4 “You are to add the 16, the 8, and the 4, result, 28.” R 1 = 16 R 1 = b 2 R 2 = 8 R 2 = ab R 3 = 4 R 3 = a 2 R 4 = a 2 + ab + b 2

R 5 “You are to take a third of 6, result 2. R 5 =

R 6 “You are to take 28 twice, result 56.” R 4 = 28R 4 = a 2 + ab + b 2 R 5 = 2R 5 = R 6 = (a 2 + ab + b 2 )

R 7 “See, it is 56. You will find it right.” R 6 = 56 R 6 = (a 2 + ab + b 2 ) The volume of a truncated pyramid is: V = (a 2 + ab + b 2 )

This ends the lesson on Egyptian Geometry