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
tough problem.)

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last modified 8/28/02

Copyright (c) 2000. Daniel E. Otero