Mathematics in Ancient and Medieval India c6.s10: Varieties of Plane Figures

Mathematics in Ancient and Medieval India c6.s10

Language: English

Varieties of Plane Figures

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Mahāvīra enlists three varieties of triangles (equi-lateral, isosceles, and scalene), five varieties of quadrilaterals (equilateral, equidichastic, equibilateral, equitrilateral and inequilateral) and eight varieties of curvilinear figures (circle, semi-circle, ellipse, conchi-form area, concave circle, convex circle, out-lying annulus, and in-lying annulus), while Śrīdhara takes the primary plane figures to be ten in number. They are rectangular (*āyata*), equilateral (*sama bhuja*), equi-bilateral (*dvi-sama bhuja*), equi-trilateral (*tri-sama bhuja*), the inequilateral quadrilateral (*visamabhuja-caturaśra*), the equilateral triangle, the scalene triangle, the isosceles triangle, the circle (*vṛtta*) and the segment of a circle (*vṛtta cāpa*).² Nārāyaṇa’s list of plane figures are also ten in numbers though he has replaced ‘the segment of a circle’ from the list of Śrīdhara by the “conch-figure”. According to Śrīdhara the area of these figures are determined by their own rules, and the areas of other figures are determined by considering their shapes in terms of the plane figures.

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² *Pāṭīgaṇita*, sū. 110–11 (Ed. by K. S. Shukla)

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*Some Geometrical Constructions from Pythagoras Theorem*

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( a ) *A right-angled triangle* can be easily constructed from Pythagoras’ theorem, when the value of any two sides are known.

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This has been expressed by algebraical formula by Brahmagupta, Mahāvīra and Bhāskara II.¹ The deduction is obvious from Pythagoras’ Theorem.

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¹ Brahmagupta gives the general solution ( ch. 12, v. 33 ) for the rightangled triangle a² + b² = c² as a = 2pq, b = p² - q², c = p² + q². His other solution ( ch. 12, v. 35 ) for the given side a is : a, ½((a²)/p - p), ½((a²)/p + p) where p is any rational number. Mahāvīra’s solution: a, 2qa/(q² - 1), ((q² + 1)/(q² - 1))a. Putting p = ((q - 1)/(q + 1))a in Brahmagupta’s second result we get Mahāvīra’s right-angled triangle. The same solution is also given by Bhāskara II ( *Līlāvatī*, rule 139 ). Mahāvīra ( *Gaṇitasārasaṃgraha*, ch. 7, v. 222½ ) constructs another rightangled triangle for the known hypotenuse c as : c, 2pqc/(p² + q²) ((p² - q²)/(p² + q²))c, Bhāskara’s solution ( *Līlāvatī*, v. 142 ) is : c, 2nc/(n² + 1), ((n² - 1)/(n² + 1))c. This readily follows from Mahāvīra’s solution by putting p/q = n. Dickson is evidently unaware of the solution given by Brahmagupta and Mahāvīra, for he ascribes the Brahmagupta and Mahāvīra results to Leonardo Fibonacci ( c. 1202 A.D. ) and Vieta ( 1580 A.D. ) Dickson — *History of Theory of Numbers*, 2, p. 192–3 ).

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(b) *Construction of scalene triangle* – In the Vedic period we have clearly pointed out, how a brick having the shape of a scalene triangle can be constructed by fusing two right angular bricks joining any two equal sides. The same principle has been adopted by Brahmagupta, Mahāvīra and Bhāskara II by the application of Pythagoras' theorem in algebraical form.¹

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¹ Brahmagupta (*Brāhmasphuṭasiddhānta*, ch. 12, v. 34) divides the scalene triangle ABC (Fig. 15) by drawing a perpendicular AD from the vertex A. The perpendicular AD evidently divides the triangle ABC into two right-triangles ADB and ADC. If AD=p, then from the previous knowledge, AB = ½((p²)/m+m), BD=½((p²)/m-m) and AC=½((p²)/n+n) and DC=½((p²)/n-n) where m, n are rational numbers. Hence BD = ½((p²)/m+(p²)/n-m-n). Mahāvīra (850 A.D.) says (Datta, B. and Singh, A.N. *Ibid.*, 2, p. 225-227) that if m²-n², 2mn, m²+n² and p²-q², 2pq, p²+q² be the sides of the two rightangled triangle and AD is so chosen that AD = 2mn=2pq, then the two triangles may be taken as ABD, ACD or as AED, ACD where ABD=AED. Then the triangle is ABC or AEC. The problem and its solutions are given in Europe by Buchet (1621 A.D.) and Cunliffe (Dickson, *History of Theory of Numbers*, 2, p. 192-3). the opposite sides of another convex quadrilateral (Vide *Brāhmasphuṭasiddhānta*, ch. 12, v. 38). Brahmagupta previously mentioned the term *caturbhūjakoṇaspṛsagvṛtta* which means that he probably considered "an inscribed quadrilateral whose vertices touch the circle" (*Ibid.*, ch. 12, v. 27).

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(c) *Construction of a quadrilateral* – A quadrilateral can be constructed, i) by joining two scalene triangles of equal bases along their base, ii) by joining four such right-triangles in a way that each pair forms a scalene triangle with equal bases. Such pairs can be constructed by suitable selection of right-angled triangle whenever necessary and iii ) by joining four scalene triangles. The number ( iii ) is not dealt by the Hindu mathematicians, but the algebraical formulae for determining the values of the quadrilateral of number ( i ) and ( ii ) were reduced from the Pythagoras' theorem by Brahmagupta.¹ Brahmagupta's method of constructing a quadri-

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Fig. 15

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1. Let a₁, b₁, c₁ and a₂, b₂, c₂ be the sides of the two right-angled triangles such that a₁²+b₁²=c₁² and a₂²+b₂²=c₂².

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Fig. 16

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Now two triangles viz. COD ( Fig. 16 ) with sides a₁( a₂, b₂, c₂ ) and AOD with sides a₂ (a₁, b₁, c₁) are made and joined with their common side OD (=a₁ a₂). On the other side of AC two triangles viz. AOB with sides b₁ (a₂, b₂, c₂) and BOC with sides b₂ (a₁, b₁, c₁) are made and joined with their common side OB = b₁ b₂. This gives a quadrilateral ABCD with sides b₁ c₂, b₂ c₁, a₁ c₂, a₂ c₁, in order. The diagonals of a quadrilateral are at right angles. From (e), p. 53, the circumradius of the triangle ABC = (b₂ c₁ × b₁ c₂)/(2b₂ b₁) = (c₁ c₂)/2.

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Again circumradius of the triangle ABD = (b₁ c₂ × a₂ c₁)/(2a₂ b₁) = (c₁ c₂)/2

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This shows that the quadrilateral ABCD is cyclic and the radius is (c₁ c₂)/2. The result is obvious from Brahmagupta's conception of forming a scalene triangle by joining two right-triangles. Brahmagupta gives another set of solution, i.e. c₁ (a₂, b₂), c₂ (a₁, b₁) or c₁ a₂, c₁ b₂, c₂ a₁, c₂ b₁ are lateral was indeed a notable achievement in this field. An instance of constructing a quadrilateral by Brahmagupta's method is given by Pṛthudakasvāmīn¹ (c. 850 A. D.) and Śrīdhara² (c. 900 A. D.). Bhāskara II³ (1150 A. D.) reproduced the method of Brahmagupta.

Linked texts (5)

Brāhmasphuṭasiddhānta — Brahmagupta, Sanskrit
Gaṇitasārasaṅgraha — Mahāvīra (mathematician), Sanskrit
Līlāvaī — Kouhala, Māhārāṣṭrī Prakrit
Triśatikā (Pāṭīgaṇitasāra) — Śrīdhara (mathematician), Sanskrit
Pāṭīgaṇita — Śrīdhara (mathematician), Sanskrit