One of the problems of geometry that famously can't be solved by compass and
ruler methods is called
squaring the circle which, oddly enough, doesn't
mean multiplying a circle by itself: it means constructing a square whose area
is equal to that of the circle. In contrast, however, there are several ways to
construct a square whose area is equal to that of a given rectangle (i.e. to
square a rectangle). I'll give three here; the second, relying on Pythagoras's
nicely for long thin rectangles; the other two are better suited to those
whose sides are tolerably similar in length.
These ways of squaring a rectangle, in turn, let us construct a square whose area is equal to that of any given parallelogram. Drop a perpendicular from an abtuse corner onto the side opposite the longer of the two into that corner; this cuts off a triangle that we can slide to the other end of the sides perpendicular to the cut; reattaching it results in a rectangle (that we could equally obtain by shearing the parallelogram parallel to its longer sides until its other sides are perpendicular), whose square root we can take. We can likewise turn a triangle into a parallelogram of equal area, and thus also a square of equal area. Cut the triangle with a line through the centres of two of its sides; this line is parallel to the third side and half as long. Half-turn the triangular piece cut off by this about one end of the cut-line. This maps half of the edge the chosen end was on onto that edge's other half; half of the other bisected edge onto an edge parallel to the other half of that edge; and the cut to an extension of itself, forming with it a line parallel to the triangle's third side and of the same length. Our triangle thus becomes a parallelogram; which we can transform via a rectangle to a square, preserving area all the way.
Now, any polygon can be cut up into triangles, for each of which we can construct a square of equal area. We can add the area of any two squares, by using their sides as perpendiculars of a right-angle triangle, whose hypotenuse square is the sum of the first two; so we can add up the squares from the triangles into which we cut up our polygon. We're thus able to construct a square whose area is equal to that of any polygon (i.e. to square an arbitrary polygon). Since comparing the areas of squares is just a matter of comparing the lengths of their sides, this lets us compare the areas of arbitrary polygons.
The simplest way to take a square root of a rectangle's area is to use the intersecting chords theorem – this says that, where two chords of a circle cross, multiplying together the lengths of the parts into which either cuts the other, we get the same product (i.e. area of a rectangle with the two parts as its side) from either chord. We can apply this to a diameter, whose two parts are a side of our rectangle and its extension by a length equal to the other side, crossing a perpendicular chord that extends that other side. The diameter necessarily bisects this chord, so a square on either half of the chord has the same area as the rectangle.
If you think carefully about this, you may notice that this is actually using the half-sum (arithmetic mean) of the rectangle's edges as hypotenuse of a right-angle triangle with the half-difference as one of the other sides; the third side of the triangle is then our half-chord (geometric mean) perpendicular to the diameter. The next method takes the same approach, but uses a different construction to do it, that works better for long thin rectangles.
Pythagoras's theorem is usually stated in terms of adding two squares to make a third; but it can equally be expressed in terms of subtracting the square on one perpendicular side from the square on the hypotenuse, to get a figure with the same area as the square on the other perpendicular side. That's a bit more of a mouthful, but it leads to one interesting result. Embed the subtracted square on a perpendicular side in the square on the hypotenuse, with a corner in common: we're left with a strip of the square on the hypotenuse, whose width is the difference between the lengths of the hypotenuse and the chosen perpendicular side; this strip's area is equal to that of the square on the other perpendicular side.
Now, in general, if we take one square out of another, we'll be left with such a strip: and we can cut it in two pieces that we can reassemble into a rectangle. One side of this is the thickness of the strip, which was the difference between the sides of the two original squares. The diagram below shows that the other side of the rectangle is in fact the sum of the two squares' sides: cut along the red line, as indicated, and move one limb of the strip to abut the end of the other.
So, back in our right-angle triangle, we have the square on a perpendicular side equal to a rectangle whose sides are the sum of and difference between the other two sides. The nice thing about this is that we can run it backwards: given a rectangle, we can construct a square with the same area. When we halve the sum and difference of the rectangle's sides, we get back the sides of the two squares whose difference was the rectangle.
If we cut off, from one end of the long side of our rectangle, a length equal to the short side, then bisect the remainder, the mid-point of this cuts the long side into the hypotenuse and other side of the right-angle triangle we need. So construct two semicircles: the diameter of one is the remainder we bisected (so its radius is the non-hypotenuse part of the whole) and the other's diameter is the hypotenuse that's the longer part into which the first's mid-point splits the long side of the rectangle. Connect the ends of the hypotenuse to where these two semicircles meet; the connecting lines meet in a right-angle and one of them is the non-hypotenuse side we used as radius for one of our semicircles. The square on the third side of the resulting triangle thus has the same areas as the rectangle we started with.
Both Euclid's proof and the proof by similar triangles, of Pythagoras's theorem, show how to cut the square on the hypotenuse into two rectangles, each of which has area equal to one of the squares on the other two sides. Interpret one of a rectangle's longer sides as one of the sides, perpendicular to the hypotenuse, of the square on the hypotenuse and recreate the right-angle triangle in whose proof our rectangle would appear by taking the opposite (equally long) side to be the line that cuts that square into two rectangles, each of which is equal to the square on a perpendicular side of the triangle.
First, we construct the putative hypotenuse by extending a shorter side of our rectangle to the length of a longer side. We then use this line as diameter for a semi-circle on its side opposite the rectangle; the long side of our rectangle, that has an end part-way along this putative hypotenuse, when extended, must cut the semi-circle in the right-angle corner of our triangle. We only actually need one of the perpendicular sides of that triangle, the one that shares a corner with our rectangle; this side's length is the square root of the rectangle's area. Still, the other side helps us, in the usual way, to construct a square on this side, that we can be sure has the same area as the rectangle we started with:
Euclid's proof, furthermore, gives us a way to transform the rectangle into the square using a shear, a rotation and a second shear; these operations don't change area, so the rectangle and square necessarily have the same area.Valid CSS ? Valid HTML ? Written by Eddy.