Saturday, August 29, 2009

The Area of a Triangle

The ancient Egyptians knew how to compute the area of a triangle. There is a simple way that they might have arrived at a method for computing the area. Consider a stack of blocks which decreases by one block for each higher row as in the figure below. As can be seen the total number of blocks in n(n+1)/2. One then has to subtract the number of blocks outside the triangle which is n/2. So the total number of blocks is n²/2. If the width of a block is w and its height is h then the width of the triangle is W = n·w and its height is H = n·h. So the area of the triangle is A = ½·H·W.

(click on the image above to enlarge it)

Is it likely that the ancient Egyptians knew of the formula for the total number of blocks, n(n+1)/2? Well, consider doubling an odd unit fraction 1/n by finding the difference between 2/n and 2/(n+1). The difference is 2/[n(n+1)] so

2/n = 2/(n+1) + 2/[n(n+1)]

and, since n is odd, n+1 is even and divisible by 2. The second term is just the inverse of n(n+1)/2. It is likely that the ancient Egyptians could easily compute the total number of blocks and therefore determine the area of a triangle.

This sequence of "triangular numbers" was known to Pythagoras who studied under the Egyptian priests in Memphis, Egypt. And so it is likely that the sum of a linear series was known much earlier in time.

1 comment:

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