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The Shulbasutras of Baudhayana, Apastamba, Katyayana and Manava: With Text, English Translation and Commentary c8: Part III: Commentary

The Śulbasūtras of Baudhāyana, Āpastamba, Kātyāyana and Mānava: With Text, English Translation and Commentary c8

Language
English

Contents

Part III: Commentary

Chapter 1: Units of Measurements, Construction of Squares and Rectangles, Knowledge of Surd, Theorem of Square on the Diagonal and Related Problems

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UNITS OF MEASUREMENTS, CONSTRUCTION OF SQUARES AND RECTANGLES, KNOWLEDGE OF SURD, THEOREM OF SQUARE ON THE DIAGONAL AND RELATED PROBLEMS.

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In the first two chapters Baudhāyana has given a summary of geometrical knowledge and some results of mathematical interest required for the construction of sacrificial altars. How the knowledge was used in connection with the measuring of grounds and placing of different layers of bricks has been discussed in detail in subsequent chapters. This chapter deals specifically with the units of measurements of altars, methods of construction of squares and rectangles, application of surd numbers, and the theorem of square on the diagonal of a rectangle.

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UNITS OF MEASUREMENTS

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**1.3.** Baudhāyana's table of units of measurements runs as follows:

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| 1 *aṅgula* | = | 14 *aṇus* | = | 34 *tilas*; |
| 1 small *pada* | = | 10 *aṅgulas*; | | |
| 1 *prādeśa* | = | 12 *aṅgulas*; | | |
| 1 *pada* | = | 15 *aṅgulas*; | | |
| 1 *īṣā* | = | 188 *aṅgulas*; | | |
| 1 *akṣa* | = | 104 *aṅgulas*; | | |
| 1 *yuga* | = | 86 *aṅgulas*; | | |
| 1 *jānu* | = | 32 *aṅgulas*; | | |
| 1 *śamyā* | = | 36 *aṅgulas*; | | |
| 1 *bāhu* | = | 36 *aṅgulas*; | | |
| 1 *prakrama* | = | 2 *padas*; | | |
| 1 *aratni* | = | 2 *prādeśas* | = | 24 *aṅgulas*; |
| 1 *puruṣa* | = | 5 *aratnis* | = | 120 *aṅgulas*; |
| 1 *vyāma* | = | 5 *aratnis*; | | |
| 1 *vyāyāma* | = | 4 *aratnis*; | | |
| 1 *aṅgula* | = | ¾ incha (approx.). | | |

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Āpastamba (*Aśl.* 6.5, 15.4) has prescribed the same values of Baudhāyana for *īṣā*, *akṣa*, *yuga*, *puruṣa*, *vyāyāma*, *aratni* and *prādeśa*. These units of Āpastamba have been used by both Kātyāyana (*Kśl.* 2.1, 5.9) and Mānava (*Mśl.* 2.1, 4.4). The unit *pada* has been made equal to 12 *aṅgulas* by Kātyāyana (*Kśl.* 5.9). The term *vitasti* has been used in place of *prādeśa* by these two latter *śulbakāras* as well as by Kauṭilya in his *Arthaśāstra*b, while its value remains the same. Mānava (*Mśl.* 4.2-4.4) has supplied some more units, e. g. 6 *tuṇḍa*=1 *bāla* of 3 years old calf; 3 *bālas* =½ mustard seed; 2 mustard seeds=1 *yava*a; 1 *aṅgula*=6 *yavas*; and 1 *prādeśa*=10 *vitastis*.

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a Fleet, 233.

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b Shamasastry (2), 117.

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The units like *aṅgula*, *pada*, *prakrama*, *prādeśa*, *bāhu*, *aratni* carry a long tradition and have been used earlier in the *Saṃhitās* and Brāhmaṇic literature in the same sense as these have been used in the *Śulbasūtras*.b

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CONSTRUCTION OF SQUARES AND RECTANGLES

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**1.4-1.5.** *Square.* Baudhāyana has described here two methods of construction of squares.

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*First Method.* Let *XY* be the given cord and *U* a mark at its middle (Fig. 1 (a)); *EW*, the *prācī* of the figure = *XY*; *O* the middle point of *EW* obtained corresponding to *U* of *XY*, where a pole is fixed.

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Fig.1. (a) Cord. (b) Square on a given length.

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A circle with *O* as centre and *OE* as radius is drawn (Fig. 1(b)). Then *EW* is a diameter of the circle along east-west line. Two other circles with *E* and *W* as centres and *EW* as radius are separately drawn. The points of intersection of these two circles are denoted by *N'* and *S'* The line *N'S'* fixes the second diameter *NS* of the circle whose centre is *O*. Again fastening the two ties once at *E*, *W*, *N* and *S* and drawing arcs, the points *A*, *B*, *C*, *D* are fixed. Then *ABCD* gives the required square.

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*Second Method.* Let *XS*, the given measure (*pramāṇa*) be *a* and *XY*, the increased cord, 2*a* (Fig. 2(a)). *S* is the mark at the middle of *XY*; then *XS* measures the length for *prācī*. *T* is the *nyañcana* mark, so that *ST*=½*a*—¼*a*=¼*a*. *U* is another mark at the middle of *SY*. *XT*, the diagonal (*akṣṇayā*) = a+¼a=5/4a *TY*, the breadth (*tiryaṅmānī*) = 2a-5/4ι=¾a Clearly, a²+(¾a)²=(5/4ι)² In other words, XY²+YT²=XT².

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a Shamasastry (1), 153, 55.

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b Macdonell and Keith, II, 577, 584.

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∴ XYT is a right-angled triangle (Fig.2(b)).

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Fig. 2. (a) Cord. (b) Right-angled triangle. (c) Square on a given length.

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For the construction of any geometrical figure intended in the *śulbasūtra* XY is always stretched along east-west line, known as *pārśvamāni*, YT along north-south line, known as *tiryaṅmāni*, and XT along the diagonal known as *akṣṇayārajju*. Now it is easy to see how the right-angled triangle XYT has been used for the construction of the square.

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The corners L and K (*śroṇi* points), M and N (*aṃsa* points) are fixed with the help of the point T of the triangle XYT [Fig.2(c)]. Then by using the half-cord UY, the points C, B, D and A are marked such that WC=WB=ED=EA. The figure ABCD gives the required square.

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1.6-1.7. *Rectangle and isosceles trapezium.* Baudhāyana's method of construction of rectangle with the help of a cord runs as follows :

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Let XY be a piece of cord taken equal to the desired breadth of the rectangle [Fig. 3 (a)]; S, a mark at the middle of XY;

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Fig. 3. (a) Cord. (b) Rectangle having desired length and breadth.

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*E*, *W*, the *prācī* poles; *P* and *Q*, *G* and *H*, poles at equal distances apart on both sides of each of the *prācī* poles. *O*, the mark assigned by the middle mark *S* when ties at *X* and *Y* are fixed at *P* and *Q* and stretched by *S*; and *A*, the point designated by the middle mark *S*, when both ties at *X* and *Y* are fixed at *E* and stretched.

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In a similar way, the other corner points, *B*, *C*, *D*, are traced [Fig. 3(b)]. Hence *ABCD* is the required rectangle.

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In the rule (*Bśl.* 1.7), Baudhāyana hints at the method of construction of an isosceles trapezium shorter in one side. For this purpose, a mark on the cord according to desired length is to be given and the rest is similar to that of *Bśl.* 1.6.

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**1.8.** *Square.* Baudhāyana describes here another method of construction of a square as follows:

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Let *XS* be the cord of given measure *a* [Fig. 4(a)] ;
XY=1½a=3/2a ;
SY=½ a;
*T* the *nyañcana* mark.
∴ XT=a+(⅓·½ a-⅙·½ a)=a+1/12 a=13/12 a
and TY=½ a-1/12a=5/12 a
The relation a²+(5/12 a)² = (13/12 a)² holds,
i.e., XY²+YT²=XT².

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Fig. 4. (a) Cord. (b) Right-angled triangle. (c) Square having given side.

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The right-angled triangle *XYT* [Fig. 4(b)] has been used to fix the corner points *L*, *M*, *N*, and *K* of the required construction. The points *C*, *D*, *A* and *B* are fixed such that WC=ED=EA=WB=a/2 [Fig. 4(c) ]. *ABCD* gives the required square.

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SURD NUMBERS

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**1.9-1.11.** Here Baudhāyana states that in a square *ABCD*, AC²=2 AB² [since AB=BC] or AC=√2 AB, where *AC* is known as the *dvikaraṇī* of the measure *AB*. If AB=a, AC=√2 a, where *a* is the measure. The result is sometimes considered by scholars as a particular case of the more generalized rule given by Baudhāyana in *Bśl.* 1.10. But Baudhāyana gave no such hint. On the other hand, he has tried to establish a more generalized result on the basis of this statement. According to him, when the measure of the side of a square is *a*, its diagonal is √2 a. Then again the measure of the diagonal of a rectangle having sides *a* and √2 a, is √3 a, for a²+(√2 a)² = (√3a)²; √3a is known as the *tṛkaraṇī*. This result has been extended

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Fig. 5. Square on the diagonal. Fig. 6. Producer of *tṛkaraṇī* and *tṛtīyākaraṇī*.

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to obtain the value of *tṛtīyākaraṇī* by both Āpastamba (*Āśl.* 2.2 and 2.3) and Kātyāyana (*Kśl.* 2.10 and 2.11). The commentators Kapardisvāmī, Sundararāja and Rāma have expressed in identical terms the value as well as meaning of this term. According to them, a square on the producer *AB* (=√3a) when divided into nine equal parts by means of three parallel lines drawing from both sides, produces the square, *EBGF* which is one-ninth of the square *ABCD* (Fig. 6). Then EB²=1/9 AB² or EB =⅓ AB =⅓ √3 a. =1/(√3) a.

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The producer *EB* is known as *tṛtīyākaraṇī* =1/(√3) a, where *a* is the side of the original square.

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THEOREM OF SQUARE ON THE DIAGONAL

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**1.12-1.13.** The theorem states that in a rectangle *ABCD*, AC²=AB²+BC² (Fig. 7). This is a most general statement and is enunciated first by Baudhāyana. The proposition is stated almost in identical language by Āpastamba (*Āśl.* 1.4), Kātyāyana (*Kśl.* 2.7) and Mānava (*Mśl.* 10.10). Baudhāyana further says that the theorem is easily verified from the following relations: $3²+4² =5²$ $12²+5² =13²$ $15²+8² =17²$ $7²+24²=25²$ $12²+35²=37²$ $15²+36²=39².$

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No proof of this theorem is given by Baudhāyana and other *śulba* writers, since it is beyond their tradition to do so. Zeuthen, Cantor, Vogt, Cajori and Heath have expressed the view that the general statement was possibly the result of an induction from a small number of cases of right-angled triangles having sides in rational numbers known to them. But this is not the actual case. Our discussions on rational rectangles and construction of geometrical figures amply justify that the general character of the theorem was rightly understood by the *śulbakāras*.

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Fig. 7

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Fig. 8

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A number of conjectures by Heath, Bürk, Müller, Thibaut, Datta and others as to the way the proof of the theorem could have been arrived at are available. A few of them are discussed in what follows.

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(i) According to Heatha, the problem of transformation of a square into a rectangle given by Baudhāyana in *Bśl.* 2.3. formed the basis of the proof. For, square *ABCD* drawn on the diagonal *AD* of the right-angled triangle *AGD* is equivalent to four equilateral triangles, while its sides *GD* and *AG* produce two each (Fig. 16). This has also established Bürk’s hypothesis.b

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(ii) The combination of two different squares as described by Baudhāyana in *Bśl.* 2.1 (Fig. 12) might have laid the foundation of the general statement of the theorem. This is according to Müllerc.

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(iii) Thibautd opined that the *śulbakāras* were observant of the fact that the square on *AD* is equivalent to four equal triangles, one of which is equal to half of the square on *OA* or *OD*, i.e. the squares on *OA* and *OD* together are equivalent to four equal triangles (Fig. 8). This pattern of arrangement of equilateral triangles are actually found in the first layer of construction with bricks in the *vakrapakṣaśyenacit* as described by Baudhāyana.

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(iv) According to Datta,e the construction of the *paitṛki vedi* established the theorem of square on the diagonal. The altar is mentioned in the *Śatapatha Brāhmaṇa* (XIII.8.1.5) as a square with its corners pointed towards the cardinal directions. It is referred to by Baudhāyana (*Bśl.* 3.11) and also by Kātyāyana (*Kśl.*2.2), where the method of its construction in detail has appeared. The square *EGWH* obtained by joining the middle points of a square *ABCD* (of area 2 sq. *puruṣas*) is the *paitṛki vedi* and is half (in area) of the original square (Fig. 29). The original square *ABCD* is a square on its east-west line *EW*. *EW* is again the diagonal of the newly formed square *EGWH*. This is undoubtedly a convincing proof (since *EW*²=2 *EG*²).

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a Heath, 352.

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b Bürk, 55, 556.

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c Müller, 194-95.

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d Datta (2), 111.

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e Datta (2), 113-115.

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(v) The knowledge of *dvikaraṇī*, *tṛkaraṇī*, discussed by Baudhāyana (*Bśl*. 1.9—1.11,) led in a way to the theorem of square on the diagonal.

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(vi) Both Āpastamba (*Āśl*. 3.7) and Kātyāyana (*Kśl*. 3.7) gave an ingenious method for calculating the area of a square or rectangle, thereby establishing the theorem of square on the diagonal. According to this method, if there are *p* units in *AB* and *q* units in *BC*, then the rectangle *ABCD* has *pq* square units, which can be obtained by drawing *p* number of parallel lines through *p* units of *AB* and *q* number of parallel lines through *q* points of *BC* (Fig. 9). This proves directly that AC² = AB² + BC².

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Fig. 9

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(vii) Kātyāyana (*Kśl*. 2.4 and 2.5) has considered a rectangle of breadth 1 *pada* and length 3 *padas*, whose diagonal is a 10 fold producer. According to Dattaa, this justifies the statement of the theorem of square on the diagonal, as may be seen from Fig. 10. In the square *ABCD*, DH = CG = CK = BF = AE = 1 pada.

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Fig. 10,

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Fig. 11

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Now *sq*. *ABCD*

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= sq on BK + sq on DH + 4 tr. AEH
= AH² + AE² + 4 tr. AEH.

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Again *sq* ABCD = EH² + 4 tr. AEH ∴ AH² + AE² = EH² (viii) Dattaa has given another proof of this theorem. Four rectangles each equal to a given rectangle having breadth a, length b and the diagonal c are so constructed that the diagonal of each rectangle forms the side of a square (Fig. 11). Then $c² = (a + b)² - 4 (½ ab)$ or, c² = a² + b².

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a Datta (2), 115-116.

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Although such specific constructions and arguments are not supplied as proofs by the *śulbakāras*, ample evidence is left by them in their details of constructions to believe that the proofs of the theorem of square on the diagonal of a rectangle were known to them.

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*Pythagorean Theorem in other Culture Areas.*

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The theorem of square on the diagonal of a rectangle is usually known as Pythagorean theorem after the name of Pythagoras (*c.* 540 B.C.). In fact, the relation 3² + 4² = 5² and some such relations have been used by Pythagoras, but evidence of any general statement regarding this is not yet available. Actual proof was first given by Euclid (*c.* 300 B.C.). Proclus (*c.* 460 A.D.), the commentator of Euclid's *Elements*b remarked: "For my part, while I admire those who first observed the truth of the theorem, I marvel more at the writer of the *Elements*, not only because he made it first by a most lucid demonstration, but because he compelled assent to the still more general theorem by the irrefragable arguments of science in the sixth book." Heath has quoted this with the remark:c "It is difficult for us to be more positive than Proclus was".

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It is fairly certain that the practical use of the theorem was current in old Babylonian times (*c.* 1800-1600 B.C.). The evidence for this is found in certain Babylonian cuneiform tablets. No general theorem was found to have been mentioned. It has been conclusively proved by Neugebauer that Pythagoras derived his "Number theorem of Universe" as well as the so-called Pythagorean theorem from cuneiform tablets.d The Chinese knew of a similar relation which appeared in *Chou Pei* (4th century B.C.), but it really became well known from the time of its first commentator Chao Chun Chhinge (3rd century A.D.). A proof of the theorem was given by Bhāskara IIf (1150 A.D.). According to Needham, Bhāskara II's treatment was derived from the *Chou Pei*.g This is not true, for the proof of Bhāskara II and that given in *Chou Pei* can readily be deduced from a number of constructions described already in the *Śulbasūtras*. .

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a Datta (2), 117

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b *Elements*, Book I, prop. 47.

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c Heath (3), 96.

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d Neugebauer, 28-42.

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e Needham, 95.

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f *Bījagaṇita*, 70.

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g Needham, 19.

Chapter 2: Transformation of Geometrical Figures

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This chapter deals with the method of combination or difference of two separate squares into a square and the transformation of a square into a rectangle, an isosceles trapezium or a circle and vice versa.

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CONSTRUCTION OF A SQUARE BEING SUM OF, DIFFERENCE BETWEEN, TWO SQUARES

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**2.1-2.2.** These two rules of Baudhāyana give the methods of construction of a square as the sum and difference of two different squares.

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Here three technical terms, *hrasīyasaḥ*, *varṣīyasaḥ* and *vṛddhram* are used. According to Kapardisvāmī,a *hrasīyasa* means the side of the smaller square, *varṣīyasa* the side of the larger square and *vṛddhram* the rectangular portion (*dīrghacaturaśram*).

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*Method of combination* (*samāsa*).

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For the combination of a smaller square *EBGF* with another square *ABCD*, this rule of Baudhāyana suggests that the rectangular portion *ABGH* is cut off by the side of the smaller square whose side is equal to *BG*. Then *AG* of this cut-off portion will be the side of the combined square (Fig. 12).

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Fig. 12.

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Fig. 13.

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Evidently, $AG² = AB² + BG² = sum of two squares.$ The same method is also given by Āpastamba (*Āśl*. 2.4) and Kātyāyana (*Kśl*. 2.13).

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PROOF : Dattab has suggested the following proof of this proposition (Fig. 13). $sq. ABCD + sq. ECGF$ $= tr. ABO + tr. AOI + tr. OFG + tr. OFJ + sq. IJED$ $= tr. ADH + tr. AOI + tr. HEF + tr. OFJ + sq. IJED$ $= sq. AOFH$ $or, AB² + CG² = AO²$

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a *Āśl*. Mysore 73, 39.

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b Datta (2), 77.

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*Method of difference (nirhāra).*

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To construct a square equal to the difference between a smaller square *EBGF* and other square *ABCD*, the rule *Bśl.* 2.2 suggests that the rectangular portion *ABGH* is cut off by the side *BG* of the smaller square. Then the side *GH* of the cut off portion is allowed to fall on *AB*, and *P* is the point where it falls. Here *GH* = *GP*. Then *BP* is the side of a square which is equal to the difference of the squares *ABCD* and *EBGF* (Fig. 14).

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Fig. 14.

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Fig. 15.

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Evidently, BP² & = GP² - BG² / & = GH² - BG² / & = AB² - BG² / & = difference of two squares ABCD and EBGF.

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The method is also given by Āpastamba (*Āśl.* 2.5) and Kātyāyana (*Kśl.* 3.1).

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Proof : The following proof based on the knowledge of the *śulbakāras* is due to Dattaa (Fig. 15).

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Now, sq. *PGFH* = 4 tr. *PGI* + sq. *IJKL* & = 2 tr. PGI + 2 tr. PGI + sq. IJKL / & = rect. PBGI + rect. PBGI + sq. IJKL / & = (rect. PBGI + sq. IJKL) + rect. PBGI / & = (rect. PBGI + sq. IJKL) + sq. MBGJ + rect. PMJI / & = (rect. PBGI + sq. IJKL + rect. PMJI) + sq. MBGJ / & = (rect. PBGI + sq. IJKL + rect. JGEK) + sq. MBGJ / & = sq. PBEL + sq. MBGJ / or, sq PBEL & = sq. PGFH - sq. MBGJ / \therefore BP² & = PG² - BG² / or BP² & = AB² - BG²

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Transformation of a Square into a Rectangle

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**2.3-2.4.** Baudhāyana has given two methods for transformation of a square into a rectangle.

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According to the first method, a square is transformed into a rectangle, such that the diagonal of the square equals the longer side of the rectangle. The method is also given by Kātyāyana (*Kśl.* 3.4).

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a Datta (2), 79.

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The square ABCD is divided by its diagonal AC (Fig. 16). The portion ADC is again divided into two equal halves by GD and each is transferred to occupy the position AEB and BFC. Then AEFC is the required rectangle. For, $sq. ABCD = tr. ABC + tr. AGD + tr. GCD$ $= tr. ABC + tr. AEB + tr. BFC$ $= rect. AEFC.$

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The method is limited in scope, for it only turns a square into a rectangle, the longer side of which is equal to the diagonal of the square.

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The second method concerns the transformation of a square into a rectangle of which one side is given. The same rule is also given by Āpastamba (*Āsl.* 3.1). Both Baudhāyana and Āpastamba gave no clear exposition of the second half of this *sūtra*. To substantiate this rule, Thibaut,a considered as an instance a square of side 5 units and changed it into a rectangle of 3 units by 8⅓ units. Likewise, Bürkb transformed a square of side 6 units into a rectangle of 4 units by 9 units as follows.

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Fig. 16.

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Fig. 17.

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The sq. ABCD is broken into a rect. ABEF making its side BE (=4 units) equal to the desired shorter side of the rectangle, and rectangle OECP (where EC=2 units), together with a square FOPD. The rectangle OECP is transferred to the other side, and GAFH is its new position. Next the smaller square FOPD (2 units × 2 units) is changed into a rectangle (of 1 unit by 4 units) and IGHJ becomes its new position (Fig. 17). Hence BI (6 + 2 + 1 = 9 units) is the length of the new rectangle. Similarly, if we change a square of 7 units into a rectangle of 5 units by 49/5 (= 7 + 2 + ⅘) units, we have to construct a rectangle of ⅘ unit by 5 units from a square of 2 units by 2 units. This is actually no solution to the problem since the transformation of square FOPD to a rectangle IGHJ is again a problem of fundamental nature.

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The commentators Dvārakānātha Yajvā and Sundararāja have described a general method as follows : *yāvadicchaṃ pārśvamānyau prācyau vardhayitvā uttarapūrvāṃ karṇarajjumāyacchet sā dīrgha caturaśramadhyasthāyāṃ samacaturaśra tiryaṅmānyāṃ yatra nipatati tata uttaraṃ hitvā dakṣiṇāṃsaṃ tiryaṅmānīṃ kuryāt taddīrghacaturaśraṃ bhavati/* This means: Having increased upto the desired length the two sides (*pārśvamānī*)

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a Thibaut (1), 246.

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b Bürk, 56, 334.

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c Thibaut (1), 247.

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towards east, the diagonal-cord is stretched towards north-east corner. The (diagonal) line cuts the breadth (*tiryaṅmānī*) of the square lying inside the rectangle; the northern portion is cut off (by drawing a line through this point parallel to *prācī*); the southern side becomes the breadth (*tiryaṅmānī*) of the (desired) rectangle.

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In Fig. 18, the sides *BA* and *CD* of the square *ABCD* are increased to *E* and *F* respectively, so that each of the sides *BE* and *CF* becomes equal to the given length *PQ*. The diagonal cord *CE* cuts the side *AD* at *O*. Then the northern portion *EBHG* is cut off by drawing a line *HG* passing through *O* parallel to the *prācī* line *BA*. Now *GHCF* is the required rectangle.

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Fig. 18. Square into a rectangle. Fig. 19. Square into a rectangle of given side.

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This is a general and perfectly satisfactory method. Both Thibaut and Bürk did not consider this method as that of Baudhāyana on the ground that Baudhāyana himself mentioned this method as *anyaśca prakāraḥ*, meaning 'another method'. Baudhāyana's method was to cut off from a given square a rectangle of side smaller than that of the square while Dvārakānātha's method started from the construction of a rectangle of side greater than that of the square. From our discussion it is clear that in the methods suggested by both Baudhāyana and Dvārakānātha, the final result of constructing a rectangle equivalent to a square is the same but their methods of attaining it are different. For this difference, Sundararāja gave the same line of argument as that of Dvārakānātha in transforming a square into a rectangle with the remark, *ayamatra prakāraḥ*a meaning, 'this is the method taught here'. To keep a symmetry with the original *sūtra* of Baudhāyana, Dattab put the method of Dvārakānātha in the following form.

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From the square *ABCD*, the portion *AGHD* is cut off, such that AG = DH = PQ, the side of the required rectangle. The diagonal *AH* is produced to meet *BC* (produced) at *E*. The rectangle *ABEF* is completed. Then *AGKF* is the equivalent rectangle (Fig. 19).

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For, tr. ABE = tr. AFE, tr. AGH = tr. ADH and tr. HCE = tr. HKE. Hence rectangle GC = rectangle DK.

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Now sq. ABCD &= rect. AH + rect. GC / &= rect. AH + rect DK / &= rect. AK.

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a *Āsl. Mysore* 49.

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b *Datta* (2), 90.

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TRANSFORMATION OF A RECTANGLE INTO A SQUARE

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**2.5.** This is a most general method of transforming a rectangle into a square given by Baudhāyana. The same method is also taught by Āpastamba (*Āśl.* 2.7) and Kātyāyana (*Kśl.* 3.2). Baudhāyana’s method runs as follows.

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Let *ABCD* be the given rectangle (Fig. 20). The portion *ABFE* is cut off such that AE = AB = the breadth of the rectangle. The remaining portion *EFCD* is cut off into two equal` halves. One half *GHCD* is placed on the other side and its new position becomes *BKLF*. A small square *FLMH* is fitted at the corner.

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Fig. 20. A rectangle into a square.

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Fig. 21. A square into an isosceles trapezium.

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Now, rect. ABCD = sq. AKMG - sq. *FLMH*, which shows that the rectangle *ABCD* is expressed as the difference of two squares. Since the method of *nirhāra* has already been taught before by Baudhāyana (*Bśl.* 2.2), a square equal to the difference of the two squares mentioned above is found by allowing the side *KM* to fall at *P* over *BH*. Then the square on *BP* will be equal to the difference of two squares, which is equal to the area of the given rectangle.

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For, BP² & = PK² - BK² / & = MK² - FL² / & = sq. ABFE + rect. EFHG + rect. FBKL / & = sq. ABFE + rect. EFHG + rect. DGHC / & = rect. ABCD.

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In the case of a rectangle of very great length, Kātyāyana’s (*Kśl.* 3.3) advice is to cut it again and again by its breadth, combine the pieces by the *samāsa* method (*Bśl.* 2.1) and finally to achieve the result by applying the *nirhāra* method (*Bśl.* 2.2). This is clearly no improvement upon the method given by Baudhāyana.

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TRANSFORMATION OF A SQUARE OR RECTANGLE INTO AN ISOSCELES TRAPEZIUM

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**2.6.** By this method a square as well as a rectangle are changed into a trapezium of given side (smaller than the side of the square).

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The square *ABCD* is required to be transformed into an isosceles trapezium *AGCE*, whose shorter side *AE* is equal to the given length *PQ* (Fig. 21). The rectangular portion *EFCD* is divided into two equal halves and the half *ECD* is shifted to its other side, such the *AGB* is its new position. Hence *AGCE* is the required isosceles trapezium.

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For, sq. *ABCD* = rect. *ABFE* + tr. *EFC* + tr. *ECD*
= rect. *ABFE* + tr. *EFC* + tr. *AGB*
= trap. *AGCE*

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This method of transformation was known earlier in the *Śatapatha Brāhmaṇa* (*Śat. Br.* 10.2.1.4).

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TRANSFORMATION OF A RECTANGLE OR SQUARE INTO A TRIANGLE OR RHOMBUS

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**2.7-2.8.** Baudhāyana has given a method of constructing a triangle or a rhombus whose area is equal to that of a given rectangle. For the construction of a triangle as described in *sūtra* (*Bśl.* 2.7), a square is to be constructed whose area will be twice that of the given rectangle. A rectangle twice the area is first constructed and then transformed into a square by the method described in *Bśl.* 2.5. Let *ABCD* be the transformed square and *E* the middle point of *AD*. Join *EB*, *EC*. Then *EBC* is the required triangle whose area is equal to that of the given rectangle (Fig. 22).

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Fig. 22. A rectangle into a triangle.

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Fig. 23. A rectangle into a rhombus.

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For, tr. *EBC* = ½ sq. *ABCD* = given rectangle.

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For the construction of a rhombus as in *sūtra* (*Bśl.* 2.8), let the rectangle *ABCD* be so constructed that its area is double that of the given rectangle. Let *E*, *F*, *G*, *H* be the middle points of *AB*, *BC*, *CD* and *DA* respectively. Join *EF*, *FG*, *GH* and *HE* to produce the required rhombus *EFGH* (Fig. 23).

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For, rhombus *EFHG*
= tr. *EFH* + tr. *GFH*
= ½ (rect. *ABFH* + rect. *CDHF*)
= ½ rect. *ABCD*

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This is given by both Āpastamba (*Āśl.* 12.8) and Kātyāyana (*Kśl.* 4.4).

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TRANSFORMATION OF A SQUARE INTO A CIRCLE.

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**2.9.** The following method of transforming a square into a circle is given by Baudhāyana. The same method has also been taught by Āpastamba (*Āśl.* 3.2), Kātyāyana (*Kśl.* 3.11) and Mānava (*Mśl.* 1.8).

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Let *ABCD* be the given square and *O* its centre. The half diagonal *OA* is drawn over the east-west line *OE*, such that OA = OE. Then a circle with radius *OF* equal to *OG* plus ⅓ of *GE* i.e. *GF*, is drawn to give the required circle (Fig. 24).

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Fig. 24.

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Here, radius [t] = OF & = OG + GF / & = OG + ⅓ GE / & = OG + ⅓ (OA - OG).

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Let 2a be the side of the square *ABCD*. $OF = a + ⅓ (a √2 - a)$ or r = a [1 + ⅓ (√2 - 1)], where OF = r or r = a/3 (2 + √2)

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As per *Bśl* 2.12 (*vide infra*), √2 is given by, $√2 & = 1 + ⅓ + 1/3.4 - 1/3.4.34 / & = 577/408 = 1.4142156$

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Baudhāyana’s more refined value of π is given by (*Bśl.* 4.15), $π & = 4 (1 - ⅛ + 1/8.29 - 1/8.29.6 + 1/8.29.6.8) / & = 3.0885.$

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Using the above value of √ 2 and π, the area of the transformed circle = π r² = 3.9989 a², which is in close agreement with the area of the given square, 4a².

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If we take π = 3 (*Bśl.* 4.15), area of the circle becomes 3.885 a², which falls far short of the area of the given square. Āpastamba made an additional remark on the method of circling a square as *sānityā maṇḍalaṁ yāvaddhīyate tāvadāgantu*, which makes also the interpretation equally difficult as to whether the method is exact or inexact one. The commentator Kapardisvāmī has broken up *sānityā* as *sā* and *anityā* concluding that the method is an inexact one. The passage has been interpreted by Karavindasvāmī as follows: ‘The circle is exactly as large as the square, for as much the circle falls short, so much comes in.’a Thibaut, Bürk and Datta have referred to the same difficulty as to the real sense in which these words were used by Āpastamba.b

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However, Dvārakānātha Yajvāc, commentator of *Baudhāyana śulba* has proposed the following correction to the formula of Baudhāyana, which gives better result: $r = [a + a/3 (√2 - 1)] (1 - 1/118)$

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a Āśl. Mysore, 50.

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b Datta (2), 142-43.

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c Thibaut (2), 10, 21.

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The problem of quadrature has also been discussed by Drenckhahn,a Chakrabarty,b and Gurjar.c

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TRANSFORMATION OF A CIRCLE TO A SQUARE

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**2.10-2.11.** Baudhāyana describes two methods of finding quadrature of a circle.

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*First Method.*

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If 2a be the side of a square and d the diameter of the circle, then

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$2a = 7d/8 + [ d/8 - 28d/8.29 + ( d/8.29.6 - d/8.29.6.8 ) ]$

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or, 2a = d - d/8 + d/8.29 - d/8.29 ( ⅙ - 1/6.8 )

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*Second Method.*

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$2a = d - 2/15 d$

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This result is also given by both Āpastamba (*Āśl.* 3.3) and Kātyāyana (*Kśl.* 3.12)

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*Rationale.*

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(A) The rationale of the result obtained from the first method is given by Thibaut, Cantor and Müller as follows :

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(i) Thibautd has suggested that the result was possibly obtained from the previous result of circling a square, r = a/3 (2 + √2) by inversion.

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$For, 2a = 3/(2 + √2) d$

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$= 1224/1393 d √2 = 577/408$

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$= d ( ⅞ + 1/8.29 - 1/8.29.6 + 1/8.29.6.8 )$

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since, 1) 1/8th of 1393 = 174⅛.

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$. 2) & 7/8th of 1393 = 1218 7/8 (less by 5 1/8 from 1224) / 3) & 1/8.29 th of 1393 = 6 (approx) / 4) & 1/8.29.6 th of 1393 = 1 / 5) & 1/8.29.6.8 of 1393 = 1/8 } & (i.e. 6 - 1 + 1/8 = 5 1/8 / & surplus by 5 1/8 from 1224)$

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c Gurjar (2), 11-16.

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More or less the same method is given by Cantor.e (ii) Müller's derivationa is as follows : $2a = 3/(2 + √2) d = (3 √2)/(2 √2 + 2) d = 3/2 · (√2)/(1 + √2) d$ $= ( 3/2 · (17 - 1/34)/(29 - 1/34) ) d = ( (51 - 3/34)/(58 - 2/34) ) d √2 = 17/12 - 1/12.34$ $= ( 1 - (7 + 1/34)/(58 - 2/34) ) d$ $= d - ⅛ · ( (56 + 8/34)/(58 - 2/34) ) d = d - ⅛ [ 1 - (2 - 10/34)/(58 - 2/34) ] d$ $= d - ⅛ [ 1 - 1/29 ( 1 - (10/34 - 2/34.29)/(2 - 2/34.29) ) ] d$ $= d - ⅛ d [ 1 - 1/29 1 - ⅙ ( 1 - (4 + 5/29)/(34 - 1/29) ) ]$ $= d - ⅛ d + 1/8.29 d [ 1 - ⅙ 1 - ⅛ ( 1 - (2 - 41/29)/(34 - 1/29) ) ]$

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a Drenckhahn, 1-13.

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b Chakrabarty (2), 23-28.

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d Thibaut (1), 254.

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e Datta (2), 145.

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Thus, $2a = d - d/8 + d/8.29 - d/8.29 ( ⅙ - 1/6.8 ) - d/8.29.6.8 · (2 - 41/29)/(34 - 1/29)$

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The last term is neglected, it being very small.

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However, Dvārakānāthab has suggested a more correct result of the above formula as follows : $2a = [ d - d/8 + d/8.29 + d/8.29 ( ⅙ - 1/6.8 ) ] × ( 1 + ½ · 3/133 )$

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(B) The rationale of the second method may be obtained as follows :

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The average of two squares, one circumscribed and the other inscribed, determines the approximate area of the circle.

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a Müller, 201.

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b Thibaut (2), 10, 21.

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∴ Area of the circle = (4r² + 2r²)/2 = 3r²

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Since the square is taken to be equal in area to the circle, 4a² = 3r² or a = (√3)/2 r

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The value of √3 may be obtained by the method of successive approximation as follows:

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(i) √A = √(a² + c) = a + c/(2a + 1),

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where 2a + 1 is the difference between the squares of c and the next positive integer. Therefore,
√3 = √(1² + 2) = 1 + ⅔ = 5/3

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(ii) For finding the next approximation e, √A is written as

c8.t2.117

√A = a + c/(2a + 1) + e
Then squaring both sides and cancelling the value of e², since it is very small, the value of e is obtained.
Here √3 = 5/3 + e

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Squaring and cancelling the value of e² we get
10/3 e + 25/9 = 3
or e = 1/15

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then √3 = 1 + ⅔ + 1/15 = 26/15

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Obviously,

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a = ½ · √3 r = ½ · 26/15 r = 13/15 r
= r - 2/15 r [ r = radius ]
or 2 a = d - 2d/15 [ d = 2r = diameter ]

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*The value of √2*

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**2.12 :** The value of √2 given by Baudhāyana is √2 = 1 + ⅓ + 1/3.4 - 1/3.4.34 (approx.)

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The same *sūtra* is also given by Āpastamba (*Āśl.* 1.6) and Kātyāyana (*Kśl.* 2.9).

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In decimal fraction, the above value of √2 = 1.4142156. According to modern calculation, √2 = 1.4142135. Thus it is clear that the ancient Indians attained a remarkable degree of accuracy in calculating an approximate value of √2. The *śulbakāras* gave methods, for constructing a square equal to the sum of two equal squares, but gave no method of calculating the value of its diagonal.

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Thibaut, Rodet, Datta, and others gave possible methods of solution for arriving at the value as follows:—

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(i) *Thibaut's proof.*a

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Now, 17² = 2.12² - 1. Thibaut argued, by how much the side 17 must be diminished in order that the square on it may be 2.12² exactly. Since 2 × 17 × 1/34 = 1, he observed, two strips each of 1/34 (approximately) are to be cut off from a square with 17 as side to obtain the square 2.12² (i.e. 12² + 12²).

c8.t2.129

Hence, (17 - 1/34)² = 2.12²

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