We will consider now the fundamental area problem in geometry: Given a plane figure, find its area. While solutions to this problem were well known for many common shapes, it is the systematic way in which Greek geometers addressed the problem that was new. Let us follow some of this development.
First of all, what is area? The simple answer is that it is a measure of the size of a plane figure. But as soon as one attempts to make such a measure, one needs an appropriate scale to measure against. For the same figure might measure 5 square feet on one scale and 0.4645 square meters on another. Which measure is right? Well, both are, of course. This just illustrates that area, like length and volume, are measured relative to some unit. The measure is dependent on the unit used. Our example also illustrates the importance of the square in area measure. The square foot and square meter represent areas determined by a square that measures one foot, or one meter, to a side. In fact, the calculation of the area of a figure was called its quadrature, a word that comes from the Latin for square, quadratus (four-sided). The Greek geometers made this same inference: to measure the area of a figure meant for them either to determine how many times bigger it was than a given square, or--what was equivalent for them--to construct a square with the same area as the given figure. Notice that this definition is independent of which unit (square foot or square meter) is used. It matters only that some square be given as unit.
How do we use this idea to measure the area of a rectangle? In the simple case in which the dimensions of the rectangle are multiples of the dimensions of the given square, the answer is straightforward: one can tile the rectangle with a number of copies of the square that is precisely the product of the relative dimensions of its sides: we find that area = base¥height.
And since the square is itself a special kind of rectangle, it follows that the area of a square satisfies
What if the dimensions of the rectangle are not simple multiples of the dimensions of the square? Then we look for a smaller unit for which both figures are multiples. For instance, a rectangle measuring three-and-a-half units by one third of a unit and a unit square can both be measured by a square that is one-sixth of a unit to a side. In terms of the smaller unit, the original unit has an area of 6 ¥ 6 = 36 and the rectangle an area of (3.5 ¥ 6) ¥ (1/3 ¥ 6) = 21 ¥ 2 = 42, so in terms of the original unit, the rectangle has an area of 42/36 = 7/6 square units.
For the parallelogram, we can dissect the figure into two pieces (by dropping a perpendicular from one corner to the opposite side) that can be recombined into a rectangle with the same base and height.
So the area of the parallelogram also satisfies area = base ¥height.
A duplication of a triangle produces a parallelogram with the same base and height:
So the area of the triangle is area = (1/2)(base ¥height).
In theory, this now allows for the quadrature of any polygon, because it is always possible to triangulate a polygon, as indicated in this diagram:
As the area of the polygon is the sum of the areas of the triangles it composes, the quadrature is effected for any figure with straight sides.
The real problem for area calculations comes when we begin to ask the question of figures with curved sides, the quintessential example being the circle. Ancient pre Greek mathematicians dealt with this problem in various ways, but nearly all presented methods for the quadrature of the circle that were good approximations at best. While we know today that the area of a circle with radius r or diameter d has the form
this only says that the area is some fixed multiple of the square of the radius; it does not divulge what we know about the constant p. The fact that we use a special symbol for this number belies the difficulty behind the problem. Egyptian papyri, for example, record finding the area of the circle by means of the calculation
If you wish to turn a circle into a square, divide the diameter into 8 parts, and again one of these eight parts into 29 parts; of these 29 parts, remove 28, and moreover the sixth part (of the one part left) less the eighth part (of the sixth part).
The Brahmin author means to say that the area of the circle is equal to the area of a square whose side is a certain fraction of the diameter of the circle, that fraction being
We will return
to a discussion of the quadrature of the circle, since work on this problem
helped to fuel the development of ideas that are central to calculus. For
now, we will look at an area problem that was thought might provide a satisfactory
resolution to the problem of squaring the circle. (It is worth remarking
that today, the phrase "squaring the circle" is used to describe any really
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last modified 8/28/02
Copyright (c) 2000. Daniel E. Otero