Mathematics in Ancient and Medieval India c6.s9: Technical Terms in the Jaina and other Works

Mathematics in Ancient and Medieval India c6.s9

Language: English

Technical Terms in the Jaina and other Works

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¹ For technical terms under (a), (b) and (c), vide Kapadia’s note in the Preface of *Gaṇitatilaka* of Śrīpati.

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(a) In the *Sūryaprajñapati* (c. 300 B. C.) the following terms are traced :—

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² Indische Studien, **10**, p. 274.

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(b) In the *Bhagavatī-sūtra* (c. 300 B. C.), we find the geometrical terms as follows :—

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(b) Technical terms in the *Āryabhaṭīya* (ch. on *Gaṇita*).

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( c ) In the *Tattvārthādhigamasūtra* of Umāsvati ( c. 150 B. C ), are found the following terms :

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*Basis of Nomenclature of the Geometrical Figures.*

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There are two fundamental bases usually adopted for naming geometrical figures depending on the relation between i) sides constituting the figure and ii) angles contained in the figure. Shortly they may be called side basis and angle basis. In Indian treatises both these types are found to occur. As already stated in the list for technical terms of the *Śulbasūtras*, the following terms are found to occur : *same-caturasra* ( square ), *dīrgha-caturasra* ( rectangle ) which are based on the relation of the sides and the terms *karṇa* ( angle ),

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( e ) Technical terms in the *Brāhmasphuṭasiddhānta* ( ch. 12 ) of Brahmagupta ( 628 A. D. ).

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*eka-karṇa* (a geometrical figure of equal angles), *dvi-karṇa* (a geometrical figure of two equal angles), *tri-karṇa* (triangle) etc. based on relations between the angles. In the *Sūryaprajñapati* (300 B. C.) both these types are represented respectively by *sama-caturasra* (square), *viṣama-caturasra* (quadrilateral other than square) having side basis, and *sama-catuṣkoṇa* (square or rectangle), *viṣama-catuṣkoṇa* (quadriangle other than square or rectangle) having angle basis. Both Āryabhaṭa I (c. 476 A.D.) and Brahmagupta (628 A. D.) however used the side basis depending on the terms *tribhuja*, *caturbhūja* and also *caturarasra* (square), *viṣama-caturasra* (quadrilateral having all unequal sides).

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Euclid (c. 300 B. C.) in his *Elements* divides the rectilinear figures according to the number of their sides¹ but later on, he introduced the angle nomenclature also.² The Romans simply followed the Greek usuage³. The early Egyptians together with the Babylonians, Hebrews and Arabs are said to have followed only the side nomenclature. Since the Indian School made use of both the angle and the side nomenclature from the time of the *Śulbasūtras* (c. 600 B. C.) i.e. earlier than Euclid, there is no reason to suggest that the Indians borrowed their ideas from those of the Greeks, on the other hand it might be otherwise.

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¹ *Tri-pleuron*, *tetra-pleuron*, *poly-pleuron*.

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² *Tri-gonon*, *tetra-gonon*.

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³ Tropfke, J. *Geschichte der Elementer Mathematik*, Bd. IV, p. 60–61, 1923.

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*Propositions, Areas and Miscellaneous Problems*

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a) *Similar triangle* : In the *Śulbasūtras*, many similar figures were drawn, though there is no mention of the actual proportional relationship of the corresponding parts of the two similar figures. In the *Sūryasiddhānta* (c. 400 A. D.) and other astronomical treatises we meet with such proportional relationship between two similar figures.¹ The result like, “*Ksitijyā* (earth sine) multiplied by the *Vyāsārdha* (radius) and then divided by the *dina-vyāsadala* (day-radius) gives the *caradala* (ascensional difference)” is the direct application of this proportional relations of the similar triangles. Āryabhaṭa I (c. 476 A. D.) used this proportional relationship of the corresponding sides in case of similar triangles to solve shadow problems.² Similar application of many other astronomical and geometrical problems was found in the works of Brahmagupta³ and other later scholars.⁴

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¹ Vide Burgess’s edition of the *Sūryasiddhānta*, Calcutta, p. 61, 1935.

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² *Āryabhaṭīya* (gaṇita) v. 15 & 16.

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³ *Brāhmasphuṭasiddhānta*, ch. 12, v. 54.

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⁴ *Gaṇitasārasaṃgraha*, ch. 9; *Līlāvatī*, rule 238–45.

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(b) *The area of a triangle:* The theorems relating to the areas are found in Euclid (c. 300 B. C.) and apparently were common property long before his time. Egyptian surveyors, even before the time of Euclid were in the habit of finding the areas of the triangular and rectangular fields. Indians too, were able to calculate the area of a triangle (=½), rectangle (=length) and isosceles trapezium [=½ (sum of the parallel sides) × distance between the parallel sides] in connection with the construction of altars on the ground.⁵ The same formulae are also found in the works of Āryabhaṭa I¹. (c. 499 A. D.), and others². For area of the scalene triangle³ Bhāskara I (c. 600 A. D.) perhaps knew the area of a triangle with sides a, b and c to be √(s(s-a)(s-b)(s-c)) where 2s = a + b + c. This formula may be obtained in terms of the sides as follows :

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⁵ Vide, p. 121–22.

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¹ *Āryabhaṭīya* (gaṇita), v. 6.

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³ *Āryabhaṭīya bhaṣya*, ii, 6½; vide also Shukla, K. S. *Gaṇita*, 22, No. 2, p. 63, 1971.

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Fig. 12

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AC² = AD² + CD² = (AB-BD)² + CD² = AB² + (BD² + CD²)
- 2AB.BD = AB² + BC² - 2 AB.BD or b² = c² + a² - 2.c.BD ∴ BD = (c² + a² - b²)/2c = ½ [ c + (a² - b²)/c ] = x, say and AD = c - x = ½ [ c - (a² - b²)/c ] According to Bhāskara I CD = √(a² - x²) or √(b² - (c - x)²) Now CD = √(a² - x²) = √(a² - ((c² + a² - b²)/2c)²) = √(((2ac)² - (c² + a² - b²))/(4c²))

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² *Līlāvatī* of Bhāskara II, rule 164.

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$= √(((2ac + c² + a² - b²)(2ac - c² - a² + b²))/(4c²))$ $= √(((c + a + b)(c + a - b)(b + c - a)(b - c + a))/(4c²))$ $= √((2s(2s - 2b)(2s - 2a)(2s - 2c))/(4c²))$ $= 4 √((s(s - a)(s - b)(s - c))/(4c²))$

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Again, $△ ABC = ½ AB.CD$ $= ½. c. 4 √((s(s - a)(s - b)(s - c))/(4c²))$ $= √(s(s - a)(s - b)(s - c))$

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This result perhaps was also known to Brahmagupta¹ and Mahāvīra.² The same formula is also found in the work of Bhāskara II³ ( c 1150 A. D. ). A similar formula is given by the Greek author Heron⁴ in the first century A. D.

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¹ *bhujakṛtyantarabhūrhṛtahīnayutābhurdvi bhājitā bādhe svābādhāvargonādbhujavargānmūlamvalamba* । (*Brāhmasphuṭasiddhānta*; ch. 12, v. 22 ).

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² *Gaṇitasārasaṃgraha*, ch. 7, v. 50.

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³ *Līlāvatī*, rule 167.

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⁴ Smith, D. E. *History of Mathematics*, 2, p. 287, Dover Publication.

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( c ) *The area of a cyclic quadrilateral* with sides a,b,c and d, is √((s-a)(s-b)(s-c)(s-d)) where 2s = a + b + c + d. This formula is given by Brahmagupta⁵ ( c. 628 A. D. ), Mahāvīra¹ ( c. 850 A. D. ), Śrīdhara² ( c. 850–950 A. D. ) and Bhāskara II,³ though these Indian scholars did not mention that the quadrilateral is cyclic. The proof may be obtained on the following lines.

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⁵ *Brāhmasphuṭasiddhānta*, ch. 12, v. 21.

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¹ *Gaṇitaśārasaṃgraha*, ch. 7, v. 50.

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³ *Līlāvatī*, rule 167.

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From Fig. 12, △ ABC = ½. AB.CD = ½. c. R Sin A = ½ c.b Sin A = ½bc. (R SinA)/R [ Vide chapter on Trigonometry. ] = ½.a.R Sin C = ½ab Sin C = ½ (ab R SinC)/R .........(1)

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Again, BC² = CD² + BD² = CD² + (AB - AD)² = (CD² + AD²) + AB²
- 2AB.AD = AC² + AB² - 2AB.AD or a² = b² + c² - 2c.R Cos A = b² + c² - 2c.b. Cos A ......(2) = b² + c² - 2bc. (R Cos A)/R .

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Fig. 13

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Now, in Fig. 13, By (2) in △ ABC, AC² = a² + b² - 2ab (R Cos B)/R ......(3)

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² *Pāṭīgaṇita*, sū., 117; ( Ed. by Shukla ).

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In △ ADC, $AC² = c² + d² - 2cd (R Cos D)/R$ $= c² + d² - 2cd (R Cos(180⁰--B))/R$ $= c² + d² + 2cd (R Cos B)/R$ $∴ a² + b² - (2ab R Cos B)/R = c² + d² + 2cd (R Cos B)/R$ $or (R Cos B)/R = (a² + b² - c² - d²)/(2(ab + cd)) ··· ··· ··· (4)$ Now (R Sin B)² = R² - R² Cos²B = R²1 - ((R Cos B)/R)² $= R²[1 - (a² + b² - c² - d²)/(2(ab + cd))²]$ $= R²(2(ab + cd)² - (a² + b² - c² - d²)²)/(4(ab + cd)²)$ $= R²((a + b)² - (c - d)²(c + d)² - (a - b)²)/(4(ab + cd)²)$ $= R²((a + b + c--d)(a + b--c + d)(c + d + a--b)(c + d--a + b))/(4(ab + cd)²)$ $= R² (2(s - d). 2(s - c). 2(s - b). 2(s - a))/(4(ab + cd)²) ( when$ $2s = a + b + c + d.)$ $= R² (16.(s - a)(s - b) (s - c) (s - d))/(4(ab + cd)²)$ $∴ (R Sin B)/R = 2 √(((s - a)(s - b)(s - c) (s - d))/((ab + cd))) ··· ···(5)$ Now quad. ABCD $= +$

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$= ½ ab(R Sin B)/R + ½cd (R Sin D)/R [ From (1) ]$ $= ½ ab(R Sin B)/R + ½ cd(R Sin (180⁰-B))/R$ $= ½ ab(R Sin B)/R + ½ cd(R Sin B)/R$ $= ½(ab+cd)(R Sin B)/R$ $= √((s-a)(s-b)(s-c)(s-d)) [ From (5) ]$

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Āryabhaṭa II (c. 950 A. D.) has however pointed out that this rule does not give the accurate area of any quadrilateral. (*Mahāsiddhānta*, xv. 69).

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(d) *Diagonals of a quadrilateral* are given by Brahmagupta¹ (628 A. D.), Mahāvīra² (c. 850 A. D.) and Bhāskara II³ (c. 1150 A. D.). They gave the following lengths of the diagonals of a quadrilateral in terms of the given sides, which is valid only for a cyclic quadrilateral. If a, b, c, d be the sides in order, and d₁ and d₂ be the two diagonals, then:

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¹ *Brāhmasphuṭasiddhānta*, ch. 12, v. 28.

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³ *Līlāvatī*, rule 190.

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$d₁ = √(((ac+bd)(ad+bc))/(ab+cd)) and$ $d₂ = √(((ac+bd)(ab+cd))/(ad+bc))$

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The proof is simple and follows from result (2) under Fig. 13.

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$We have, AC² = a² + b² - 2ab(R Cos B)/R$

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² *Gaṇitasārasaṃgraha*, ch. 7, v. 54.

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= a² + b² - 2ab.(a² + b² - c² - d²)/(2(ab + cd)) [ From 4 ] = (2(ab + cd)(a² + b²) - 2ab(a² + b² - c² - d²))/(2(ab + cd)) = (2cd(a² + b²) + 2ab (c² + d²))/(2(ab + cd)) = ((ac + bd)(ad + bc))/(ab + cd) ∴ AC = d₁ = √(((ac + bd) (ad + bc))/(ab + cd)) Similarly it may be shown, BD = d₂ = √(((ac + bd)(ab + cd))/(ad + bc)) Again, d₁².d₂² = (ac + bd)² ∴ d₁.d₂ = ac + bd.

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The formulae were obtained by Ptolemy ( c. 150 A. D. ), though the exact method followed by him is not known. The first of these formulae i.e. the expression for d₁ was rediscovered by W. Snell¹ in 1619 A. D. Brahmagupta's method for the determination of the diagonals of a cyclic quadrilateral constitutes a remarkable contribution in geometry of the time.

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¹ Smith, D. E. *History of Mathematics*, 2, p. 287, Dover Publication.

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( *e* ) *The circum radius of a triangle* as given by Brahmagupta² is bc/2p where b and c are two sides of the triangle and p is the triangle drawn through their point of intersection. No proof is given. The proof is obvious In any triangle, AD = R Sin B (vide Fig. 12). = c Sin B = (c.R Sin B)/R

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² *Brāhmasphuṭasiddhānta*, ch. 12, v. 27.

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= (c . b)/(R. 2) [ a/(R Sin A) = b/(R Sin B) = c/(R Sin C) = 2 in a circumscribed circle, ]

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∴ p = AD = bc/2R

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or R = bc/2p

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(f) *Area of a circle* is given by Āryabhaṭa¹ I and and Bhāskara II² as ½ c.r. = ² where c = circumference and r = radius. Gaṇeśa gives a very simple proof in which he has divided a circle into even number of equal triangles by drawing radius to the circumference. Half of the total number of triangles are so fitted with remaining triangles that they form a rectangular area having breadth = radius, and length = half the circumference i.e. area = r.½c = ².

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¹ *Āryabhaṭīya* (Gaṇita), v. 7.

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(g) *The surface area*³ of a sphere was given Bhāskara II as 4². According to him the area of a circle multiplied by four is the net covering the surface of the ball having the same radius.

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³ *Līlāvatī*, rule 203.

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(h) *The value of the chord of a circle* is given by Āryabhaṭa I⁴ by the relation as CD² = AD.DB, when the chord CDE intersects the diameter ADB (Fig. 14a.) of a circle. CD has been defined as R Sin BC or R Sin AC by Bhāskara I (vide his *Āryabhaṭīyabhāṣya*). This rule is given by Brahmagupta⁵ (c. 628 A.D.) and Bhāskara II.¹ In justification of this rule Bhāskara I has selected several interesting problems such as the hawk and rat problems, the lotus problems and the

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² *Līlāvatī*, rule 203.

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⁴ *Āryabhaṭīya* (Gaṇita), v. 17½.

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⁵ *Brāhmasphuṭasiddhānta*, ch. 12, v. 41.

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¹ *Līlāvatī*, rule 206-7.

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(a) (b) (c) Fig. 14

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crane and fish problems etc. Some of these problems also occur in later works. The problems are as follows :

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*Problem 1* : “A hawk is sitting on the top of a pole whose height is 18 (cubits). A rat who has gone out of his dwelling (at the foot of the pole) to a distance of 81 (cubits), while returning towards his dwelling, afraid of the hawk, is killed by the cruel ( bird ) on the way. Say how far has he gone towards his hole, and also the ( horizontal ) motion of the hawk ( the speeds of the rat and the hawk being the same". ( *Āryabhaṭīya Bhāṣya*, ii, 17; reappeared also in Yallaya's commentary ).

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Bhāskara I ascribes such problems to previous writers. His method of solution runs as follows :

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Let AOB be the horizontal diameter of the circle of centre O. CDE, the vertical chord intersects AOB at D. Imagining that the hawk is sitting at C and the rat at B, they see each other. The rat then runs to his hole at D, but is killed by the hawk at O. Here CO = BO, Since speed of the hawk and rat are same. Given CD = 18 cubits, DB = 81 cubits, since CD² = AD × DB, AD = 4 cubits. \therefore DO &= DB - OB / &= DB - OA / &= DB - (DO + AD) / or DO &= 1/2(DB - AD) / &= 1/2(81 - 4) = 1/2 \ . \ 77 / &= 38 1/2 cubits / Similarly BO &= 1/2(DB + AD) / &= 1/2(81 + 4) = 1/2.85 = 42 1/2 cubits Hence the rat has crossed a distance of 42½ cubits, and the horizontal movement of the hawk along CO is DO = 38½ cubits.

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*Problem 2* : "A lotus flower of 6 *aṅgulas* just dips ( into the water ) when it advances through a distance of 2 cubits. I want to know the height of the lotus plant and the depth of the water." (Similar problems occur in the works of Bhāskara II and Narāyaṇa).

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In this problem (Fig. 14b), AOB is the vertical diameter of the circle of centre O and EDC, the horizontal chord, denotes the water level. OA is the lotus stalk, AD lotus flower, and O the root of the lotus, and E and C the points where the lotus flowers just dips into the water. Here AD=6 angulas=¼ cubits DC=2 cubits and DB=(DC²)/AD = (2²)/(¼) = 2² × 4 = 16 cubits. Like that of previous problem, OA = height of the plant &= \textstyle1/2\ (DB+AD) / &= \textstyle1/2\ (16 + \textstyle1/4) / &= 65/8 = 8\textstyle1/8 cubits OD &= depth of the water / &= OA-AD = \textstyle65/8 - 1/4 = 7 7/8 cubits.

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*Problem 3* : "There is a reservoir of water of dimensions 6 × 12. At the east-north corner there is a fish and at the west-north corner there is a crane. For fear of him, the fish, crossing the reservoir, hurriedly went towards the south in an oblique direction but was killed by the crane who came along the sides of the reservoir. Give out the distances travelled by them assuming that their speeds are the same." (*Āryabhaṭīya bhaṣya*, ii. 17; reappears in the Raghunatha Raja's commentary. See also *Gaṇitakumudī*, Kṣetravyāhara, p. 38–39, Ex. 29–31).

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Let CDEF be the reservoir (Fig. 14c) of which DE=CF=12 and CD=FE=6, suppose C=east-north corner, and F=west-north corner. At first the fish was at C and the crane at F. Then out of fear the fish swims along CO and the crane moving along FE and EO, kills the fish at O. Then CO = FE + EO = EB + EO (EB is drawn equal to FE).

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A circle is drawn with O as centre and OC as radius, which will pass through B.

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We have DB = DE + EB = DE + FE = 12 + 6 = 18 Let DB produced meet the circle at A. ∴ AD = (CD²)/DB = (6²)/18 = 36/18 = 2 ∴ AB = AD + DB = 2 + 18 = 20 ∴ OC = 10

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Therefore distance covered by the crane = 10 = distance covered by the fish.

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( i ) *The volume of solid bodies*. Āryabhaṭa I's formula for the volume of a sphere and that of a triangular pyramid are not quite correct.¹ His formula for a sphere is √(π).³ or 1.47 ³, the correct value being 4/3³ or 1.33. .³. The correct value was given by Bhāskara II.² The formula for that of the triangular pyramid is given incorrectly by Āryabhaṭa I ( *Gaṇita-pāda*, verse 8 ) as ½ × area of the triangular base × height. The correct formula for a square pyramid is given by Bhāskara II as ⅓ × radius × square base. For volume of the sphere, Bhāskar II has divided it into as many pyramids as there are unit square areas on the surface, its height being equal to the radius. Hence the volume of the sphere becomes equal to ⅓ of the sphere i.e. 4/3³.

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¹ *Āryabhaṭīya* ( *Gaṇita* ), v. 7.

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² *Līlāvatī*, rule 203.

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The expression for the volume of a cone occurs first in India in the work of Brahmagupta. He used the term *sūcī*, meaning thereby generally a pyramid with a base of any form. The base may be a circle and hence the term includes a cone as well. According to Brahmagupta¹ (c. 628 A. D.) *volume of a cone (or pyramid* = ⅓ of the base. This formula reappears in the works of Mahāvīra² (c. 850 A. D.), Āryabhaṭa II³ (c. 950 A. D.), Nemichandra⁴ (c. 975 A. D.), Śrīpati⁵ (1039 A. D.) and Bhāskara II⁶ (1150 A. D.). The formula refered to by Kaye⁷ is (Circumference/6)². This formula no doubt occurs in the works of the above writers except Mahāvīra. But in connection with this formula it has been specifically mentioned by all of them that the formula is to be employed only in case of "the measurements of the maunds grain" (*rāśi-vyāvahāra*), not in other cases. It has been further remarked by Brahmagupta, Śrīpati and Bhāskara II that in that case the height of the maund must be assumed to be equal to the circumference of the base divided by 9, 10 or 11 according to calculates two approximate values called *vyavahārika gaṇita* or ‘approximate value’ ( A ) and *antra gaṇita* or ‘gross volume’ ( G ). Mahāvīra calls them respectively *karmāntika-phala* and *auṇḍra-phala*. It is stated : A = ( area from half the sum of the linear dimensions of face and base ) × height, and G = ½ ( area of face + area of base ) × height. The accurate volume ( V ) of the frustum is then given to be V = ⅓ ( G – A ) + A. Now for the frustum of a right circular cone noted before, $A = ((r + R)²)/2 π × h, G = ½ (π r² + π R²) × h$ ∴ V = (π)/3 (r² + R² + rR) × h as stated before.

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¹ *Brāhmasphuṭasiddhānta*, ch. 12, v. 44.

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² Compare *Gaṇitasārasaṃgraha*, ch. 8, v. 17½ and 20½.

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³ *Mahāsiddhānta*, ch. 15, v. 105.

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⁴ *Trilokasāra* of Nemichandra, Gāthā 19.

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⁵ *Siddhāntaśekhara*, ch. 13, v. 44.

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⁶ *Līlāvatī*, vide Sudhākara Dvivedi's edition, p. 65; Colebrooke, *Hindu Algebra*, p. 98.

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⁷ Kaye, G. R. *Indian Mathematics*, p. 39.

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For a *volume of a frustum of a pyramid* the sides of whose upper face are a, b and the corresponding sides of whose lower face are a', b', the approximate value will be $A = ((a + a')/2) ((b + b')/2) × h, G = ½ (ab + a'b') h$ Then the accurate volume of the frustum of the wedge will be $V = ⅓ [½ (ab + a'b') h - ((a + a')/2) ((b + b')/2) h]$ $+ ((a + a')/2) ((b + b')/2) h . . . . . (1)$ On reduction we easily obtain, $V = ⅙ [ab + a'b' + (a + a')(b + b')] × h ... ... (2)$ The formula reappears in this reduced form in the works of Āryabhaṭa II¹, Śrīpati² and Bhāskara II³ the kinds of grain. Nemicandra has considered the case (for finer grains) in which the height is one-eleventh of the circumference of the base.¹ So there is absolutely no doubt that the formula was intended only for a rough calculation.

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¹ *Mahāsiddhānta*, ch. 15, v. 106.

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² *Siddhāntaśekhara*, ch. 13, v. 45.

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³ *Līlāvatī*, Ibid., p. 45.

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¹ Gāthā 23, in which Nemicandra has specified the kinds of seeds for which this assumption is to be made.

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The formula of the *volume of a frustum* appears explicitly in India first in the works of Śrīdhara.² The frustum is the part of a solid figure such as a cone or a pyramid, cut off by a plane parallel to the base. If d and D denote the diameters of the upper and lower face of the frustum of a right circular cone and h its height, then its volume V will be given by,

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$V = h/24√(10[ d² + D² + ( d + D )² ])$ $= (π)/3 ( r² + R² + rR )h$

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² *Triśatikā*, rule 58. This clearly shows that Śrīdhara knew how to find the volume of a complete right circular cone though he had not explicitly recorded it.

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where r, R denote the radii of the upper and lower face and π = √10, the value adopted by Śrīdhara. It also reappears in the works of Āryabhaṭa II³ and Mahāvīra.⁴ Brahmagupta⁵ followed by Mahāvīra⁶ gave a general method of calculating the volume of frustum of a solid, such as a pyramid, a cone and a wedge whose upper and lower faces are parallel. He first The formula for the volume of a truncated wedge (frustum of a pyramid) is given in the *Chiu-Chang Suan-Shu*¹ and *Mong-hsia Pi-t'an*² of Ch'en Huo (died in 1075 A. D.) as

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³ *Mahāsiddhānta*, ch. 15, v. 106.

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⁴ *Gaṇitasārasaṃgraha*, ch. 8, v. 17½.

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⁵ *Brāhmasphuṭasiddhānta*, ch. 12, v. 45-6.

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⁶ *Gaṇitasārasaṃgraha*, ch. 8, v. 9-11½.

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¹ Mikami, Y. *The Development of Mathematics in China and Japan*, p. 15, 1913.

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$⅙ [ ( 2a + a' ) b + ( 2a' + a ) b' ] × h ... ... ... (3)$

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This expression is of course easily reducible to the second Hindu form. But the first Hindu form i. e. the Brahmagupta-Mahāvīra form is considerably different from the Chinese form. Putting a = b, a' = b' in (3), we get the formula for the volume of a frustum of a pyramid with a square base as ⅓ ( a² + a'² + aa' ) h. This particular formula occurs in the *Chiu Chang Suan-Shu*³, *Chang Ch'iu-Chien Suan-Ching*⁴ and Heron's *Stereometry*.⁵ It was also known to the ancient Egyptians.⁶ Cantor⁷ followed by Tropfke⁸ and Smith⁹ thinks that Brahmagupta's rule was meant for such a particular case only. But we do not think so.¹⁰ For there is nothing in Brahmagupta's definition of his rule to suggest such a limitation. These writers were probably misled by an example of a well with a square face (a case of a truncated square pyramid turned up-side down) given by the commentator Pṛthudakasvāmī in illustration of Brahmagupta’s rule. Mahāvīra’s rule is absolutely free from any kind of limitation. For an illustration of it, he has given examples relating to a truncated pyramid with a triangular, square and rectangular base, a frustum of a right circular cone and also a truncated wedge.¹

Linked texts (15)

Brāhmasphuṭasiddhānta — Brahmagupta, Sanskrit
Gaṇitasārasaṅgraha — Mahāvīra (mathematician), Sanskrit
Līlāvaī — Kouhala, Māhārāṣṭrī Prakrit
Sūryasiddhānta — Maya, Sanskrit
Āryabhaṭīya — Āryabhaṭa, Sanskrit
Dhātusutta — Pāli
Sūriyapannatti — Ardhamāgadhī Prakrit
Viyāhapannatti — Sudharman, Ardhamāgadhī Prakrit
Tiloyasāra — Nemicandra Siddhāntacakravartin, Śaurasenī Prakrit
Siddhāntaśekhara — Śrīpati, Sanskrit
Mahāsiddhānta (Āryasiddhānta) — Āryabhaṭa II, Sanskrit
Gaṇitatilaka — Śrīpati, Sanskrit
Tattvārthasūtra — Umāsvāti, Sanskrit
Triśatikā (Pāṭīgaṇitasāra) — Śrīdhara (mathematician), Sanskrit
Pāṭīgaṇita — Śrīdhara (mathematician), Sanskrit