Notice: This material will be included in a forthcoming (summer 2000) book with the tentative title Experiencing Geometry in Euclidean, Spherical, and Hyperbolic Spaces. This new book will be an expanded and updated version of Experiencing Geometry on Plane and Sphere. This material is in draft form and may not be duplicated or quoted without the author's written permission, except for purposes of review or trying out the material with students. As always comments are welcome and will affect the final draft. Send comments to dwh2@cornell.edu.
Chapter 14
The diagonal of an oblong produces by itself both the areas which the two sides of the oblong produce separately.
— Baudhayana, Sulbasutram, Sutra 48 [A: Baudhayana]
In the last chapter, we showed that two polygons on a sphere with the same area are equivalent by dissection because both are equivalent to the same biangle. In this chapter, we will prove an analogous result on the plane by showing that every planar polygon is equivalent by dissection to a square.
In the process of exploring this dissection theory, we will follow a path through a corner of the forest of mathematics — a path that has delighted and surprised the author many times. We will bring with us the question: What are square roots? Along the way we will confront relationships between geometry and algebra of real numbers, in addition to similar triangles, the Pythagorean Theorem (the quote above is a statement of this theorem written before Pythagoras), and possibly the oldest written proof in geometry (at least 2,600 years old). This path will lead to the solutions of quadratic and cubic equations in Chapter 15. We will let the author's personal experience lead us on this path.
Square Roots
When I was in eighth grade, I asked my teacher, "What is a square root?" I knew that the square root of N was a number whose square was equal to N, but where would I find it? (Hidden in that question is "How do I know it always exists?") I knew what the square roots of 4 and 9 were — no problem there. I even knew that the square root of 2 was the length of the diagonal of a unit square, but what of the square root of 2.5 or of p?
At first, the teacher showed me a square root table (a table of numerical square roots), but I soon discovered that if I took the number listed in the table as the square root of 2 and squared it, I got 1.999396, not 2. (Modern-day pocket calculators give rise to the same problem.) So I persisted in asking my question — What is the square root? Then the teacher answered by giving me THE ANSWER — the Square Root Algorithm. Do you remember the Square Root Algorithm — that procedure, similar to long division, by which it is possible to calculate the square root? Or perhaps more recently you were taught the "divide and average" method which goes like this:
If A1 is an approximation of the square root of N then the average of A1 and N/A1 is an even better approximation which we could call A2 . And then the next approximation A3 is the average of A2 and N/A2 . In equation form this becomes
An+1 = (1/2)(An+(N/An)).
For example, if A1 = 1.5 is an approximation of the square root of 2, then
A2 = 1.417··· , A3 = 1.414216···
and so forth are better and better approximations.
But wait! Most of the time these algorithms do not calculate the square root — they only calculate approximations to the square root. The algorithms have an advantage over the tables because I could, at least in theory, calculate approximations as close as I wished. However, they are still only approximations and my question still remained — What is the square root which these algorithms approximate?
My eighth-grade teacher then gave up, but later in college I found out that some modern mathematicians answer my question in the following way: "We make an assumption (the Completeness Axiom) which implies that the sequence of approximations from the Square Root Algorithm must converge to some real number." And, when I continued to ask my question, I found that in modern mathematics the square root is a certain equivalence class of Cauchy sequences of rational numbers, or a certain Dedekind cut. Finally, I let go of my question and forgot it in the turmoil of graduate school, writing my thesis and beginning my mathematical career.
Later, I started teaching the geometry course that is the basis for this book. One of the problems in the course is the following problem.
Problem 14.1. A Rectangle Dissects into a Square
Show that, on the plane, every rectangle is equivalent by dissection to a square.
Suggestions
Problems 14.1 and 14.2 can be done in any order. So if you get stuck on one problem, you can still go on to the other. In Problems 14.1 and 14.2, it is especially important to make accurate models and constructions — rough drawings will not show the necessary length and angle relationships.
Problem 14.1 is one of the oldest problems in geometry, so you may have guessed (correctly!) that it is one of the more complex ones. This problem is interesting for more than just historical reasons. You are asked to prove that you can cut up any rectangle (into a finite number of pieces) and rearrange the pieces to form a square, like so:
Figure 14.1. Dissecting a rectangle into a square.
Since you are neither adding anything to the rectangle nor removing anything, the area must remain the same, so ab = x2, or x = 
. What you are really finding is a geometric interpretation of a square root.
Let us look at a proof which is similar to the proofs in many standard geometry textbooks:
Let s = HK = (AE)(HD)/AD = s(a - s)/a = s - s2/a = s - b.
Therefore, we have D EFK @ D RCD, D EBR @ D KHD.
Figure 14.2. Textbook proof.
I was satisfied with the proof until in the second year of the course I started sensing student uneasiness with it. As I listened to their comments, I noticed questions being asked by the students: "What is The students and I also noticed that the facts used about similar triangles in the proof above are usually proved using the theory of areas of triangles. Thus, this proof could not be used as part of a concrete theory of areas of polygons, which was our purpose in studying dissection theory in the first place. Notice that the above proof also assumes the existence of the square which in analysis is based on the Completeness Axiom. The conclusion here seems to be that it would be desirable instead to construct the square root x. That started me on an exploration which continued on and off for many years.
Now let us solve a few problems in dissection theory:
Here are three methods for constructing x. For all three constructions we will use a rectangle like the one shown in Figure 14.1, with the longer side b as the base and the shorter side a as the height.
For the first construction, Figure 14.3, take the rectangle and lay a out to the left of b. Use this base line as the diameter of a circle. The length x that you're looking for is the perpendicular line from the left side of the triangle to where it intersects the circle:
Figure 14.3. First construction of x.
The second construction, Figure 14.4, is similar but this time put a on the inside of the base of the rectangle. Now the side x you are looking for is the segment from the lower left corner of the rectangle to the point at which a perpendicular rising from the place you put a intersects the circle:
Figure 14.4. Second construction of x.
The third construction is a bit more algebraic than the others, and doesn't directly involve a circle:
Figure 14.5. Third construction of x.
This construction can be used together with the result of Problem 14.2 in order to obtain a proof by subtraction.
For all of these constructions, it is imperative that you use accurate models. Whatever method you choose, make the rectangle and the square overlap as much as possible, and see how to fit the other pieces in. Then, you have to prove that all of the sides and angles line up properly. Note that it is much better to solve this problem geometrically rather than by only trying to work out the algebra — actually do the construction and proceed from there. Finally, you don't have to use one of the constructions shown here. If these don't make sense to you, then find one of your own that does.
And remember not to use results about similar triangles since we normally need results about areas in order to prove these results as we will do in Problem 14.3.
Baudhayana's Sulbasutram
While reading an unrelated article, I ran across an item that said that the problem of changing a rectangle into a square appeared in the Sulbasutram. "Sulbasutram" means "rules of the cord" and the several Sanskrit texts collectively called the Sulbasutra were written by the Vedic Hindus starting before 600 BC and are thought to be compilations of oral wisdom which may go back to 2000 BC. (See for example, A. Seidenberg article The Ritual Origin of Geometry [Hi: Seidenberg (1961)].) These texts have prescriptions for building fire altars, or Agni. However, contained in the Sulbasutra are sections which constitute a geometry textbook detailing the geometry necessary for designing and constructing the altars. As far as I have been able to determine these are the oldest geometry (or even mathematics) textbooks in existence. There are at least four versions of the Sulbasutram by Baudhayana, Apastamba, Katyayana, and Manava. The geometric descriptions are very similar in these four books and I will only use Baudhayana's version here, see [AT: Baudhayana].
The first chapter of Baudhayana's Sulbasutram contains geometric statements called "Sutra." Sutra 54 is what we asked you to prove in Problem 14.1. It states:
If you wish to turn an oblong2 into a square, take the shorter side of the oblong for the side of square. Divide the remainder into two parts and inverting join those two parts to two sides of the square. Fill the empty place by adding a piece. It has been taught how to deduct it.
Here is a diagram for Sutra 54:
Figure 14.6. Sutra 54.
So our rectangle has been changed into a figure with an "empty place" which can be filled "by adding a piece" (a small square). The result is a large square from which a small square has to be removed (or "deducted").
Now Sutra 51:
If you wish to deduct one square from another square, cut off a piece from the larger square by making a mark on the ground with the side of the smaller square which you wish to deduct; draw one of the sides across the oblong so that it touches the other side; by this line which has been cut off the small square is deducted from the large one.
Figure 14.7. Sutra 51: Construction of side of square.
We wish to deduct the small square (a2) from the large square. Sutra 51 tells us to "scratch up" with the side of the smaller square — this produces the line AB and the oblong ABCD. Now, if we "draw" the side CD of the large square to produce an arc, then this arc intersects the other side at the point E. The sutra then claims that BE is side of the desired square whose area equals the area of the large square minus the area of the small square. This last assertion follows from Sutra 50, which we ask you to prove in Problem 14.2.
See Figure 14.8 for the drawing that goes with Sutra 50:
If you wish to combine two squares of different size into one, scratch up with the side of the smaller square a piece cut off from the larger one. The diagonal of this cutoff piece is the side of the combined squares.
Figure 14.8. Sutra 50.
Be sure you see why Sutra 50 is a statement of what we call the Pythagorean Theorem.
A. Seidenberg, in an article entitled The Ritual Origin of Geometry [Hi: Seidenberg], gives a detailed discussion of the significance of Baudhayana's Sulbasutram. He argues that it were written before 600 BC (Pythagoras lived about 500 BC and Euclid about 300 BC). He gives evidence to support his claim that it contains codification of knowledge going "far back of 1700 BC" and that knowledge of this kind was the common source of Indian, Egyptian, Babylonian, and Greek mathematics. Together Sutras 50, 51, and 54 describe a construction of a square with the same area as a given rectangle (oblong) and a proof (based on the Pythagorean Theorem) that this construction is correct. You can find stated in many books and articles that the ancient Hindus, in general, and the Sulbasutram, in particular, did not have proofs or demonstrations, or they are dismissed as being "rare." However, there are several sutras in [AT: Baudhayana] similar to the ones discussed above. I suggest you decide for yourself to what extent they constitute proofs or demonstrations.
Baudhayana avoids the Completeness Axiom by giving an explicit construction of the side of the square. The construction can be summarized in Figure 14.9:
Figure 14.9. Baudhayana's construction of square root.
This is the same as Euclid's construction in Proposition II-14 (see [A: Euclid, page 409]). But Euclid's proof is much more complicated. Note that neither Baudhayana nor Euclid gives a proof of Problem 14.1 because the use of the Pythagorean Theorem obscures the dissection. However, they do give a concrete construction and a proof that the construction works. Both Baudhayana and Euclid prove the following theorem which uses equivalence by subtraction.
Theorem: For every rectangle R there are squares S1 and S2 such that R + S2 is equivalent by dissection to S1 + S2 and thus R and S1 have the same area.
Notice that both Baudhayana's and Euclid's proofs of this theorem and your proof of Problem 14.1 avoid assuming that the square root exists (and thus avoid the Completeness Axiom). They also avoid using any facts about similar triangles. These proofs explicitly construct the square and show in an elementary way that its area is the same as the area of the rectangle. There is no need for the area or the sides of the rectangle to be expressed in numbers. Also given a real number, b, the square root of b can be constructed by using a rectangle with sides b and 1. In Problem 14.3 we will use these techniques to prove basic properties about similar triangles.
So, finally, we have an answer to our question — What is a square root? It is "an answer" because many other solutions given from different points of view may be on the horizon. In particular, the Addendum to this chapter contains a geometric method based on Baudhayana for finding numerical approximations to many square roots.
Problem 14.2. Equivalence of Squares
Prove the following: On the plane, the union of two squares is equivalent by dissection to another square.
Suggestions
This is closely related to the Pythagorean Theorem. There are two general ways to approach this problem: You can use Problem 14.1 or you can prove it on its own, which will result in a proof of the Pythagorean Theorem — hence, you can't use the Pythagorean Theorem to solve this problem because you will be proving it!
To see how this problem relates to the Pythagorean Theorem, think about the following statement of the Pythagorean Theorem:
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
This is not just an algebraic equation — the squares referred to are actual geometric squares.
Figure 14.10. Pythagorean Theorem.
The term "completing the square" likewise came from geometry, and has some application in this problem as well as in Problem 14.1. As in the other dissection problems, make the three squares coincide as much as possible, and see how you might get the remaining pieces to overlap. You might start by reflecting the square with side c over the side c on the triangle. Then prove that the construction works as you say it does.
Any Polygon Can Be Dissected into a Square
If you put together Problems 13.1, 14.1, and 14.2, a surprising result is created:
Theorem. On the plane, every polygon is equivalent by dissection to a square.
Figure 14.11. Every polygon dissects into a square.
Problem 14.3. Similar Triangles
Near the beginning of this chapter we gave a textbook proof of Problem 14.1 which used properties of similar triangles. Later you found a proof that did not need to use similar triangles. Now you are ready to give a dissection proof of:
If two triangles have corresponding angles congruent, then the corresponding sides of the triangles are in the same proportion to one another.
Suggestions
Look at your proof of Problem 14.1. It probably shows implicitly that Problem 14.3 holds for a pair of similar triangles in your construction. For more generality, let q be one of the angles of the triangles and place the two q's in VAT position in such a way that they form two parallelograms as Figure 14.12.
Figure 14.12. Similar triangles.
Show that the two parallelograms are equivalent by dissection and use that result to show that ac = bd or in equivalent form a/d = c/b. You may find it clearer if you start by looking at the special case of q = p/2.
Three-Dimensional Dissections and Hilbert's Third
In 1900, David Hilbert delivered a lecture before the International Congress of Mathematicians in which he listed 23 problems "from the discussion of which an advancement of science may be expected." These problems are now called Hilbert's Problems.
Hilbert's Third Problem3. Is it possible to specify two tetrahedra of equal bases and equal altitudes which can in no way be split up into congruent tetrahedra, and which cannot be combined with congruent tetrahedra to form two polyhedra which themselves could not be split up into congruent tetrahedra?
Shortly after Hilbert's lecture, Max Dehn found such tetrahedra and also proved that a regular tetrahedron is not equivalent by dissection to a cube. Thus there is no possibility of dissecting polyhedra into cubes. To show these results, Dehn proved the following:
Theorem. If P and Q are two polyhedra in 3-space that are equivalent by dissection (or by subtraction), then the dihedral angles (see Chapter 19) of P are, mod p, a rational linear combination of the dihedral angles of Q. That is, if ai are the dihedral angles of P and bj , are the dihedral angles of Q, then there are integers ni, mj, a, b such that
åi niai + ap = åj mjbj + bp.
Addendum: Numerical Approximations of Square Roots
We earlier described the "divide and average" (D&A) method for approximating the numerical value of a square root. This method is also sometimes called "Newton's method" and seems to be the most widely used and most efficient modern numerical method. On a January, 1990, visit to the Sankaracharya Mutt in Konchipuram, Tamilnadu, India, I was surprised to discover that there is a method based on Baudhayana's Sulbasutra which produces for many square roots the same approximations as the D&A method but with significantly less computations.
Baudhayana's Sutra 52 states that the diagonal of a square is the side of another square with two times the area of the first square. Thus, if we consider the side of the original square to be one unit, then the diagonal is the side (or root) of a square of area two, or simply the square root of 2, that is Sutras 61 and 62 contain the following prescription for finding the length of the diagonal of a square:
Increase the length [of the side] by its third and this third by its own fourth less the thirty-fourth part of that fourth. The increased length is a small amount in excess (savi´e¸a)4.
Thus the above passage from the Sulbasutram gives the approximation:
I use » instead of = indicating that the Vedic Hindus were aware that the length they prescribed is a little too long (savi´e¸a). In fact my calculator gives:
and the Sulbasutram's value expressed in decimals is
So the question arises — how did the Vedic Hindus obtain such an accurate numerical value? Unfortunately, there is nothing that survives which records how they arrived at this savi´e¸a.
There have been several speculations5 as to how this value was obtained, but no one as far as I can determine has noticed that there is a step-by-step method (based on geometric techniques in the Sulbasutram) that will not only obtain the approximation:
but can also be continued indefinitely to obtain as accurate an approximation as one wishes.
This method will in one more step obtain:
where the only numerical computation needed is 1154 = 2[(34)(17) - 1] and, moreover, the method shows that the square of this approximation is less than 2 by exactly
The interested reader can check that this approximation is accurate to eleven decimal places.
In the Sulbasutram the agni are described as being constructed of bricks of various sizes. Mentioned in made in Sutra 3-7 of square bricks of side 1 pradesa (span of a hand, about 9 inches) on a side. Each pradesa was equal to 12 angula (finger width, about 3/4 inch) and one angula was equal to 34 sesame seeds laid together with their broadest faces touching. Thus the diagonal of a pradesa brick had length:
1 pradesa + 4 angula + 1 angula - 1 sesame thickness.
I do not believe it is purely by chance that these units come out this nicely. Notice that this length is too large by roughly one-thousandth of the thickness of a sesame seed. Presumably there was no need for more accuracy in the building of altars!
Baudhayana's Sulbasutram does not how he found the savi´e¸a. But let us see if his geometric methods discussed above hint at a method for finding numerical approximations of square roots.
Construction of the Savi´e¸a for the Square Root of Two
If we apply Sutra 54 to the union of two squares each with sides of 1 pradesa we get a square with side 1½ pradesa from which a square of side ½ pradesa had been removed. See the Figure 14.6.
Now we can attempt to take a strip from the left and bottom of the large square — the strips are to be just thin enough that they will fill in the little removed square. The pieces filling in the little square will have length 1/2 and six of these lengths will fit along the bottom and left of the large square. The reader can then see that strips of thickness (1/6)(1/2) pradesa (= 1 angula) will (almost) work:
Figure 14.13. Refining the construction.
There is still a little square left out of the upper right corner because the thin strips overlapped in the lower left corner. Notice that
We can get directly to Figure 14.14. Alternative construction of first approximation.
We now have that two square pradesas are equal to a large square minus a small square. The large square has side equal to 1 pradesa plus 1/3 of a pradesa plus 1/4 of 1/3 of a pradesa, or 1 pradesa and 5 angulas and the small square has side of 1 angula. To make this into a single square we may attempt to remove a thin strip from the left side and the bottom just thin enough that the strips will fill in the little square. Since these two thin strips will have length 1 pradesa and 5 angulas or 17 angulas we may cut each into 17 rectangular pieces each 1 angula long. If these are stacked up they will fill the little square if the thickness of the strips is 1/34 of an angula (or Figure 14.15. Further approximation of the square root of two.
Thus with error expressed by
I write "2·1" instead of "2" to remind us that for Baudhayana (and, in fact, for most mathematicians up until near the end of the 19th Century) that If we again follow the same procedure of removing a very thin strip from the left and bottom edges and cutting them into 2[1154(1154/2)-1] = (1154)2 - 2 = 1,331,714
and thus that the next approximation (savi´e¸a) is
The difference between 2·1 and the square of this savi´e¸a is
This method will work for any number N which you can first express as the area of the difference of two squares, N·1 = A2 - B2, where the side A is an integral multiple of the side B. For example,
I find that the easiest way for me to see that these expressions are valid is to represent them geometrically in a way that would also have been natural for Baudhayana. To illustrate:
Figure 14.16. Finding savi´e¸as for 5 and 2½.
Figures 14.13 and 14.14 give other examples. The reader should try out this method to see how easy it is to find savi´e¸as for the square roots of other numbers, for example, 3, 11, 2¾.
Fractions in the Sulbasutram
You have probably noticed that all the fractions above are expressed as unit fractions, but this is not always the case in the Baudhayana's Sulbasutram. For example, in Sutra 69 he discusses how to find a length which is an approximation to the diagonal of a square whose side is the "third part of" 8 prakramas (which equals 240 angulas). He describes the construction:
... increase the measure [the 8 prakramas] by its fifth, divide the whole into five parts and make a mark at the end of two parts.
In more modern notation if we let D equal 8 prakramas, then this gives the approximation of the diagonal of a square with side (1/3)D as
This is equivalent to If you attempt to find the savi´e¸as for other square roots you will find it convenient to use non-unit fractions. For example, by starting with this picture:
Figure 14.17. Finding the square root of thirteen.
you can make slight modifications in the above method to find:
Comparing with the Divide-and-Average (D&A) Method
Today the most efficient method usually taught to find square roots is called "divide-and-average". It is also sometimes called Newton's method. If you wish to find the square root of N then you start with an initial approximation a0 and then take as the next approximation the average of a0 and N/a0. In general, if an is the nth approximation of the square root of N, then an+1 = ½(an + (N/an)). The interested reader can check that if you start with [1+(1/3)+(1/12)] = [17/12] = 1.416666666667 as your first approximation of However, Baudhayana's method uses significantly less computations (in addition, of course, to the drawings either on paper or in one's mind). For example, look at the following table which compares the methods for the first four approximations. For Baudhayana's method at the n-th stage let kn denote the number of thin pieces added into the missing square and let cn denote the correction term that is added .
1 + (1/3) + (1/4)(1/3) 1.414215686 (577/408) c2 = - (1/34)(1/4)(1/3) 1.414213562 (665857/470832) c3 = -(1/1154)(1/34)(1/4)(1/3) 1.414213562 = (886731088897/627013566048) c4 = -(1/1331714) c3 Notice that the (10-digit) calculator reaches its maximum accuracy at the third stage. At this stage the Baudhayana method obtained more accuracy (it can be checked that it is accurate to 12-digits) and the only computation required was (34)2-2 = 1154 which can easily be accomplished by hand. Baudhayana's approximations are numerically identical to those attained in the D&A method using fractions, but again with significantly less computations. Of course, Baudhayana's method has this efficiency only if you do not change Baudhayana's representation of the approximation into decimals or into standard fractions. At the fourth stage the Baudhayana method is accurate to less than
2[(13317142-2)(1331714)(1154)(34)(4)(3)]-1
or roughly 24-digit accuracy with the only calculation needed being
(1154)2-2 = 1331714.
Notice that in Baudhayana's fourth representation of the savi´e¸a for the square root of 2:
the unit is first divided into 3 parts and then each of these parts into 4 parts and then each of these parts into 1154 parts and each of these parts into 133174 parts. Notice the similarity of this to standard USA linear measure where a mile is divided into 8 furlongs and a furlong into 220 yards and a yard into 3 feet and a foot into 12 inches. Other traditional systems of units work similarly except for the metric systems where the division is always by 10. Also, some carpenters I know when they have a measurement of Conclusions
Baudhayana's method can not come even close to the D&A method in terms of ease of use with a computer and its applicability to finding the square root of any number. However, the Sulbasutra contains many powerful techniques, which, in specific situations have a power and efficiency that is missing in more general techniques. Numerical computations with the decimal system in either fixed point or floating point form has many well-known problems. (See, for example, [RN: Turner (1991)].) Perhaps we will be able to learn something from the (apparently) first geometry text in the world and devise computational procedures that combine geometry and numerical techniques. 2
Here "oblong" means "rectangle." 3
See [Z: Sah] and [Z: Boltyanski] for a discussion of this problem and its history. 4
This last sentence is translated by some authors as "The increased length is called savi´e¸a". I follow the translation of "savi´e¸a" given by B. Datta on pp. 196-202 in The Science of the Sulba, University of Calcutta, 1932; see also G. Joseph (The Crest of the Peacock, I.B. Taurus, London, 1991) who translates the word as "a special quantity in excess".
be the side of the square equivalent by dissection to the rectangle with sides a and b. Place the square, AEFH, on the rectangle, ABCD, as shown in Figure 14.2. Draw ED to intersect BC in R and HF in K. Let BC intersect HF in G.1 From the similar triangles D KDH and D EDA we have HK/AE = HD/AD, or
?" "How do you find it?" Those used to be my questions!
Pause, explore, and write out your ideas before going to the next page.
. The Sanskrit word for this length is dvi-karani or, literally, "that which produces 2".
.
See Datta Op.cit. for a discussion of several of these, some of which are also discussed in G. Joseph, Op. cit.
,
,
.
.
by considering the following dissection:
pradesa). Without a microscope we will now see the two square pradesas as being equal in area to the square with side 
pradesa. But with a microscope we see that the strips overlap in the lower left corner and thus that there is a tiny square of side 
still left out.
is still a little in excess. We can now perform the same procedure again by removing a very very thin strip from the left and bottom edges and then cutting them into 
pradesa lengths in order to fill in the left out square. If w is twice the number of
lengths in 
pradesa, then the strips we remove must have width
pradesa. We can calculate w easily because we already noted that there were 17 segments of length
and each of these segments was divided into 34 pieces and then one of these pieces was removed. Thus w = 2[34(17)-1] = 1154 and
.
denoted the side (a length) of a square with area 2.
length pieces, then the reader can check that the number of such pieces must be
.
.
.
.
being approximated by 1.44.
.
, then the succeeding approximations are numerically the same as those given by Baudhayana's geometric method.
D&A - calculator
D&A - Fractions
Baudhayana's Method
a1 = 1.416666667
17/12
a2 = ½(a1 + (2/a1)) =
½[(17/12)+ 2(12/17)] =
k2 = 2[(3·4)+4+1] = 34
a3 = ½(a2 + (2/a2)) =
½[(577/408)+2(408/577)] =
k3 = (34)2-2 = 1154
a4 = ½(a3 + (2/a3)) =
½[(665857/470832)+2(470832/665857)]
k4 = (1154)2-2 = 1331714 
,
inches are likely to work with it as 
, or 2 inches plus a half inch minus an eighth of that half — this is a clearer image to hold onto and work with. From Baudhayana's approximation it is easier to have an image of the length of 
than it is from the D&A's (886731088897/627013566048).
1
In the case that ABCD is so long and skinny that K ends up between G and F, we can, by cutting ABCD in half and stacking the halves, reduce the proof to the case above.