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The Science of the Shulba: A Study in Early Hindu Geometry c11: Chapter VI. Combination of Areas

The Science of the Śulba: A Study in Early Hindu Geometry c11

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Chapter VI. Combination of Areas

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*To draw a square equivalent to* n *times a given square.*

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The solution of this proposition practically depends upon the construction of squares and rectangles having given sides. By the so-called Pythagorean Theorem, the square described on the diagonal of the given square will have an area twice as much. To draw a square thrice as much as the given square, the rule is:

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" (Construct a rectangle whose) breadth will be the measure (of a side of the given square) and length its double-producer (*i.e.*, diagonal). The diagonal of that rectangle is the treble-producer."∗1

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∗1 *BŚl*, i. 46; *ĀpŚl*, ii. 2; *KŚl*, ii. 14.

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By repeating the operation and constructing each time a rectangle whose one side will be equal to a side of the given square and another side equal to the diagonal of the rectangle constructed at the preceding stage, we shall finally arrive at one, the square on whose diagonal will be equivalent to *n* times the given square. This method is taught by almost all the writers.

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Oftentimes the process can be much shortened by a skilful device. For instance, if *n* happen to be a square number, equal to p², say, then the desired result will be obtained by drawing the square on a straight line *p* times a side of the given square. Thus Kātyāyana observes:

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" Twice the measure (of a side of a given square) is (its) fourfold-producer; thrice the measure is the ninefold-producer; four times the measure is the sixteenfold-producer. As many units of a measure as are in a cord, so many rows (or series) of squares (of that measure) there will be in a square on that cord as a side. Combine them."¹

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¹ *KŚl*, iii. 6-9; compare also *ĀpŚl*, iii. 6-7.

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If n is not a square number, simplification can be made by expressing it, when possible, as the sum of two square numbers. Thus if $n=p²+q²$ where p, q are rational quantities, then if x denote a side of the required square and c a side of the given square, we shall have $x²=(pc)²+(qc)²=(p²+q²)c²=nc².$ So that with one construction only, we shall get a square n times the given square. Kātyāyana has given some instances of this kind. He says:

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"> (If a rectangle be drawn with) one pada (a unit of linear measure) as the breadth and three padas as the length, its diagonal will be tenfold-producer. (If a rectangle be drawn with) two padas as the breadth and six padas as the length, its diagonal will be fortyfold-producer."²

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² *KŚl*, ii. 8-9.

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The first of these instances occurs also in the *Mānava Śulba*.³ Similar instances are also found in other works.

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³ *MāŚl*, v. 4-5.

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A simplification is also possible even if n is expressible as a multiple of a square number, though not as the sum of two square numbers.

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Kātyāyana gives another very elegant and simple method of finding a square equal to the sum of a number of other squares of the same size. He says:

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"> As many squares (of equal size) as you wish to combine into one, the transverse line will be (equal to) one less than that; twice a side will be (equal to) one more than that; (thus) form a triangle. Its arrow (i.e., altitude) will do that.'¹

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¹ "यावत् प्रमाणानि समचतुरस्राण्येकीकर्तुं चिकीर्षेदेकोनानि तानि भवन्ति तिर्य्यक् द्विगुणाण्येकत एकाधिकानि पार्श्विर्भवति, तस्येषुस्तत्करोति ।"—*KŚl*, vi. 5. Compare also the *Pariśiṣṭa*, verses 40-1.

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Fig. 36

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That is, if *n* be the number of equal squares to be combined together into one, form the triangle *ABC* whose base *BC* is of length (*n*—1) times a side of a square and twice the sides *AB* and *AC* of which are severally equal to (*n*+1) times a side of a square. The method of drawing the triangle, which will be consistent with the geometrical methods of the *Śulba* is this: Draw the line *BC* of length (*n*—1) times a side of a square. Fix two poles at *B* and *C*. Take a cord of length (*n*+1) times a side of a square. Fasten its two ends at the two poles and stretch the cord sidewise having taken it by the middle point. Let *A* be the point reached. Bisect *BC* at *D* and join *AD*. Then the square on *AD* will be equivalent to the sum of n given squares. For $AD² = AC² - DC²,$ $= ((n+1)/2)² a² - ((n-1)/2)² a²,$ $= na².$

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*To draw a square equivalent to the nth part of a given square.*

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After describing the method of drawing a square equivalent to 3 times a given square, Baudhāyana observes:

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"Thereby is explained the generator of the third part (*tṛtīyakaraṇī* of the square). It is the ninth part of the area."¹

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¹ *BŚl*, i. 47.

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Similar remarks occur also in the works of Āpastamba and Kātyāyana. The former says,

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"Thereby is explained the generator of the third part. Division into nine (parts)."²

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² *ĀpŚl*, ii. 3.

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and the latter,

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"Thereby is explained the generator of the third part. Division of the given measure into nine parts."³

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³ *KŚl*, ii. 15-6.

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The commentators disagree about the method actually implied in the above rules. According to some,⁴ it is this: Find a square equivalent to three times the given square. Then divide a side of this square into three equal parts. A square drawn on any of these parts will be equivalent to one-third of the given square. For, on drawing parallel lines through the points of division, the second square will be divided into nine (square) parts. Hence each part (square) is equal to one-ninth of it and so to one-third of the given square. According to others,¹ the method is this: Divide the given square into nine equal squares. Combine three of these squares into one. And it will be equivalent to the third part of the given area. Some² are of opinion that both the methods are implied by the text.

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⁴ For instance, Kapardisvāmī, Sundararāja, and Rāma.

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¹ *E.g.*, Dvārakānātha Yajvā.

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Both the interpretations are valid inasmuch as they produce the correct result. Thibaut prefers the first interpretation as it preserves in a better way the connexion of the above rule with those just preceding. But the Śulbakāras appear to have implied both the methods. Kātyāyana's rules following those noted above are:

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"The third part of the side (of the given square) produces the ninth part of it. Three of these ninth parts (on combination) will give the generator of the third part (of the given square)."³

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³ *KŚl*, ii. 17-8.

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Again this is the method adopted by Baudhāyana in constructing the *Paitṛki-vedi* which is square in shape and is equivalent to one-third the square on a side 18 padas long.∗ Further he has expressly adopted both the methods in measuring the *Sautrāmaṇīki-vedi*.⁵

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∗ *Vide infra*, p. 98f.

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⁵ *BŚr*, xix. 1.

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Proceeding in the same way we can find in general a square equivalent to the *n*th part of a given square. *Either*: First find a square equivalent to *n* times the given square. The square on the *n*th part of a side of this square will be equivalent to the *n*th part of the given square. *Or* Divide a side of the given square into *n* equal parts. On drawing parallel lines through the points of division, the given square will be divided into n² equal square parts. The square combining n of these elementary squares will be equivalent to the nth part of the given square.

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² Karavindasvāmī.

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The whole process will, of course, be much simplified if n be a square number, equal to p², say. For in that case we shall have simply to divide a side of the given square into p equal parts. The square having one of these parts as a side will be equal to nth part of the given square. Kātyāyana furnishes us with some instances of this kind:

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"> By means of half the measure (of the side of a given square) is obtained a square equivalent to the fourth part (of the given square); by one-third the measure is obtained (a square equivalent to) the ninth part; by one-fourth the measure is obtained (a square equivalent to) the sixteenth part.”¹

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¹ *KŚl*, iii. 10-11.

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Similar instances are also given by Āpastamba.²

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² *ĀpŚl*, iii. 10.

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*To draw a square equivalent to the sum of two different squares.*

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Baudhāyana gives the following method of solution of this proposition:

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Fig. 37

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"To combine two different squares, cut off from the larger a (rectangular) portion with a side of the smaller one. The diagonal of this segment will be a side of the sum."\∗1

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\∗1 *BŚl*, i. 52.

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The same method is also taught by Āpastamba² and Kātyāyana.³

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³ *KŚl*, ii. 22.

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Let ABCD be the larger square and P a side of the smaller. Cut off AF and DE making each equal to P, and complete the rectangle AFED. Join AE. Then AE² = AD² + DE² = AD² + P².

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The proof of this proposition will be evident from the Fig. 38 which, in fact, simply represents the complete constructions taught in the *Śulba*.

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Fig. 38

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\square ABCD + \square CGHI & = \triangle ADE + \triangle AEF + \triangle EGH + \triangle EHJ / & + \square BIJF, / & = \triangle ABK + \triangle AEF + \triangle HIK + \triangle EHJ / & + \square BIJF, / & = \square AEHK.

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$Or AD² + P² = AE², since CG = P$

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² *ĀpŚl*, ii. 4.

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*To draw a square equivalent to the difference of two different squares.*

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For the solution of this proposition, Baudhāyana¹ and Āpastamba² give the following rule:

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¹ *BŚl*, i. 51.

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"> To deduct a square from a square, cut off from the larger a (rectangular) segment with a side of the square which is to be deducted. Then draw a longer side of this segment diagonally across to the other longer side; and where it falls (on the other side), cut off that portion. By this cut-off portion the deduction is finished."

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Fig. 39

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Let *ABCD* be the larger square and *P* a side of the smaller square to be deducted from it. Cut off *AF* and *DE* making each equal to *P*. Join *FE*. Draw *FE* by the extremity *F* so that it falls on *AD* at *G*. Join *GE*. Then

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$GD²=GE²-DE²=AD²-P².$

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² *ĀpŚl*, ii. 5.

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This method is also taught by Kātyāyana.³

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³ *KŚl*, iii. 1.

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The proof of this proposition will appear on completing the constructions indicated in the *Śulba*. The square

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Fig. 40

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*GEPQ*, which is equal to the square *ABCD*, is comprised of four rightangled triangles each equal to *GMQ* and of the square *HKLM*. Therefore GEPQ & = GDEH + DRLN + HKLM, / & = GDRM + DEKN, / \therefore GDRM & = ABCD - DEKN, / or GD² & = AB² - DE²

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Āpastamba has given a *demonstration* of the above rule together with an illustrative example:

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"The side drawn cross becomes the diagonal (of a certain rectangle). Let it be the generator of the fourfold (of a certain square). It produces (a square equal to) both the squares which are produced separately by the cut-off portion (of the opposite side) and the other side (*i.e.*, the shorter side of the rectangular segment of the larger square or the side of the other square).

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If the other side be one puruṣa, the remainder will be three square puruṣas ; it has been stated (before).”¹

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¹ *ĀpŚl*, ii. 6.

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For the construction of a Fire-altar of proper size and shape, it is also necessary to know how to combine into a square figures of other kinds; *e.g.*, a square and a rectangle, two rectangles.² No specific rules for this purpose are found in the *Sulba-sūtra*. Hence it follows that such combinations shall have to be made with the help of the methods taught. Thus figures of every other kind are first transformed into squares and they are then combined into a square by the methods just described. This method has indeed been taught by Kātyāyana for the combination of triangles and pentagons.

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² Cf. *ĀpŚl*, xxi. 8.

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*To draw a square equivalent to two given triangles.*

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After describing a method for the transformation of an isosceles triangle or a rhombus into a square, Kātyāyana observes: “By this is explained the combination of triangles.” That is, the triangles and rhombuses to be combined should first be transformed into squares severally and the sum or difference of these squares is then found by the methods already explained. The final result can, of course, be put in the shape of a square, rectangle or triangle as required.

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*To draw a square equivalent to two given pentagons.*

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Kātyāyana has also indicated a method for the combination of pentalaterals (*pañcakarṇa*):

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“By this is also explained a method for the combination of pentalaterals too. Break up a pentalateral of equal angles into isosceles triangles; and break up a penta-lateral of unequal angles into squares.”

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Fig. 41

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Fig. 42

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Fig. 43

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The latter part of this rule appears rather to be obscure. How to break up an irregular pentagon into squares? Mahīdhara has failed to grasp the matter accurately and hence made a worse confusion.² I think Kātyāyana has in view a pentagon of the shape *ABCDE* in which *CP*=*PD*=*AM* and *BC*=*MP*=*ED*=*CP*/2 (Fig. 42); *or* in which *CP*, *PD*, *AM*, *BC*, *MP*, *ED* are all equal (Fig. 43). We find description of bricks of these shapes in the *Śulba* of Baudhāyana¹ and they are ordinarily called *haṁsamukhī* ("of the shape of the mouth of a goose"). Obviously a pentagon of that kind can be easily transformed into two or three squares of equal size. So that by combining the transformed squares we can find a square equal to the sum or difference of two or more pentagons.

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¹ *KŚl*, iv. 8.

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² Mahīdhara has once explained *ekakarṇa* = “one triangle,” *dvikarṇa* = “two triangles,” *trikarṇa* = “three triangles” and *pañcakarṇa* = “five triangles.” Later on he says *ekakarṇa* means “equal angles” and *dvikarṇa* “unequal angles.” His former interpretations are certainly wrong. If Kātyāyana had indeed meant “five triangles” by the term *pañcakarṇa*, what did he mean by the rule, “Pañcakarṇānāñca praüge apacchidyaikakarṇānāṃ dvikarṇānāṃ samacaturasre apacchidya.” [“Break up *pañcakarṇas* of *ekakarṇa* (variety) into isosceles triangles and break up *pañcakarṇas* of *dvikarṇa* (variety) into squares ?”]

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¹ *BŚl*, iii. 68, 288; compare also iii. 291-2.

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