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The Science of the Shulba: A Study in Early Hindu Geometry c13.s1: Figures with a given area

The Science of the Śulba: A Study in Early Hindu Geometry c13.s1

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Figures with a given area

c13.s1.1

For the construction of an altar of proper size and shape, as prescribed by the Holy Scriptures, it is sometimes necessary to describe a rectilinear figure having a given area. For instance, it is prescribed that the *vedi* of the *Pitṛyajña* must be a square of an area equal to one-ninth of that of the *Mahāvedi*.¹ Now the area of the *Mahāvedi* is given to be 972 square padas. So the problem becomes to construct a square having an area of 108 square padas. Again it is said that the *Sautrāmaṇikī-vedi* is of the form of an isosceles trapezium having an area equal to the third part of the *Mahāvedi*, that is, to 324 square padas. According to Āpastamba and Kātyāyana, this isosceles trapezium must be similar to the shape of the *Mahāvedi*. This case will be treated separately under "Similar figures." But Baudhāyana seems to have left the shape of the trapezium unrestricted. Hence there arises the problem: to find an isosceles trapezium having an area of 324 square padas. According to the tradition of some schools, it is necessary to construct square-shaped altars of area varying from 1½ to 6½ square puruṣas.²

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¹ *BŚl*, i. 81.

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*To construct a square having an area of 108 square padas.*

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Baudhāyana gives the following method:

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"One-third producer (*i.e.*, the side of a square whose area is one-third the area) of the square made with the third part of the *Mahāvedi* is that (the side of the *Paitṛkī-vedi*). Its area is the ninth part of the area (of the *Mahāvedi*)."³

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³ *BŚl*, i. 82; compare *Paṇḍit*, X (O. S.). p. 46f

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108 = 324/3 = 18²/3. So the required square will be one-third of a square on a side 18 padas long. The method for this latter construction has been taught before. Let AB be a straight line 18 padas long. Divide it into three equal parts. Let AC be one such part. Describe the square ACDE. Join AD. Describe a circle with

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² *BŚl*, ii. 14; *ĀpŚl*, viii. 3.

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Fig. 57

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centre A and radius AD cutting AB at F. Join EF. Then EF is a side of the square having an area of 108 square padas. For

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EF² & = EA² + AF², / & = EA² + AD², / & = 3AC², / & = 1/3 AB². / \therefore EF² & = 108 square padas.

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This construction is the same as to find a square three times a square of 36 sqaure padas. For 108 = 3 × 6². If 108 would have been a square number or expressible as the sum of the two square numbers, the solution would have of course been very easy.

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*To construct an isosceles trapezium having an area of 324 square padas.*

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Baudhāyana says:

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"If a square be formed with the third part of the *Mahāvedi*, its sides will be each 18 padas long. Then by making it longer on one side and shorter on the other, the sides should be determined optionally as necessary."¹

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¹ *BŚl*, i. 86-7.

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The rule does not explain clearly how the sides of the square formed with the third part of the *Mahāvedi*, are to be varied, so that the modified figure may assume the shape of an isosceles trapezium, but still retaining the same area. It seems, however, that the altitude is left unchanged, only the face and base being varied.

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Fig. 58

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Let *ABCD* be the square whose side *AB* is 18 padas long. Suppose *EFGH* to be the modified form. Also suppose EF=18x and HG=18y. Since the area must remain the same, we must have $18((18x+18y)/2)=324,$ or $x+y=2.$

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Thus we can easily obtain any number of isosceles trapeziums having the same altitude and area.

c13.s1.22

We have explained before Baudhāyana's method for the transformation of a square (or a rectangle) into an isosceles trapezium which shall have a given face. That is different from the one supposed here. But this method has one advantage over that; it does not disturb the east-west line and so its adoption, we think, to have been more probable. If the altitude also vary, we get an indeterminate equation of the form $z(x+y)=2$ where 18z is the altitude of the resulting isosceles trapezium.

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