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History of Hindu Mathematics: A Source Book v2.c17.s19: 21. Rational Quadrilaterals

History of Hindu Mathematics: A Source Book v2.c17.s19

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21. Rational Quadrilaterals

v2.c17.s19.1

**Rational Isosceles Trapeziums.** Brahmagupta has shown how to obtain an isosceles trapezium whose sides, diagonals, altitude, segments and area are all rational numbers. He says :

v2.c17.s19.2

"The diagonals of the rectangle (generated) are the flank sides of an isosceles trapezium; the square of its side is divided by an optional number and then lessened by that optional number and divided by two; (the result) increased by the upright is the base and lessened by it is the face."⁴

v2.c17.s19.3

⁴ *BrSpSi*, xii. 36.

v2.c17.s19.4

That is to say, we shall have (Fig. 5) $CD = ½((4m²n²)/p - p) + (m² - n²),$

v2.c17.s19.5

² *GSS*, vii. 158½.

v2.c17.s19.6

AB &= 1/2((4m²n²)/p-p) - (m²-n²), \\ AD &= BC = m²+n²; \end{aligned} also DH &= m²-n², / HC &= 1/2((4m²n²)/p-p), / AC &= BD = 1/2((4m²n²)/p+p), / AH &= 2mn, area ABCD = mn((4m²n²)/p-p).

v2.c17.s19.7

By choosing the values of m, n and p suitably, the values of all the dimensions of the isosceles trapezium can be made integral. Thus, starting with the rectangle (5, 12, 13) and taking p=6, Pṛthûdakasvâmî finds, by way of illustration, the isosceles trapezium whose flank sides = 13, base = 14, and face = 4. Its altitude (12), segments of base (5, 9), diagonals (15) and area (108) are also integers.

v2.c17.s19.8

Mahâvîra writes :

v2.c17.s19.9

“For an isosceles trapezium the sum of the perpendicular of the first generated rectangle and the perpendicular of the second rectangle which is generated from any (rational) divisor of half the base of the first and the quotient, will be the base; their difference will be the face; the smaller of the diagonals (of the generated rectangles) will be the flank side; the smaller perpendicular will be the segment; the greater diagonal will be the diagonal (of the isosceles trapezium); the greater area will be the area and the base (of either rectangle) will be the altitude.”¹

v2.c17.s19.10

¹ *GSS*, vii. 99½.

v2.c17.s19.11

The first rectangle (AA'DH) generated from m, n is $m² - n², 2mn, m² + n².$ If p, q be any two rational factors of half the base of this rectangle, that is, if pq = mn, the second rectangle (AB'CH) from these factors will be $p² - q², 2pq, p² + q².$ By judiciously juxtaposing these two rectangles, we shall obtain an isosceles trapezium of the type required (ABCD):

v2.c17.s19.13

Fig. 5

v2.c17.s19.14

$CD = (p² - q²) + (m² - n²),$ $AB = (p² - q²) - (m² - n²),$ $AD = BC = m² + n², if m² + n² < p² + q²,$ $DH = m² - n², if m² - n² < p² - q²,$ $AC = BD = p² + q², if p² + q² > m² + n²,$ $AH = 2mn = 2pq,$ $area ABCD = 2pq (p² - q²),$ $if 2pq (p² - q²) > 2mn(m² - n²).$

v2.c17.s19.15

The necessity of the conditions m² + n² < p² + q², m² - n² < p² - q², etc., will be at once realised from a glance at Figs. 5 and 6. The above specifications of the dimensions of a rational isosceles trapezium will give Fig. 5. But when the conditions are reversed so that m²+n²>p²+q², m²-n²>p²-q², 2pq(p²-q²) <2mn(m²-n²), the dimensions of the isosceles trapezium (Fig. 6) are :

v2.c17.s19.16

Fig. 6

v2.c17.s19.17

CD=(m²-n²)+(p²-q²), AB=(m²-n²)-(p²-q²), AC=BD=p²+q², DH=m²-n², AD=BC=m²+n², AH=2mn=2pq, area ABCD=2mn(m²-n²).

v2.c17.s19.18

**Pairs of Isosceles Trapeziums.** Mahâvîra gives the following rule for finding the face, base and equal sides of an isosceles trapezium having an area and altitude exactly equal to those of another isosceles trapezium whose dimensions are known :

v2.c17.s19.19

“On performing the *viṣama-saṅkramaṇa* between the square of the perpendicular (of the known isosceles trapezium) and an optional number, the greater result will be the equal sides of the (required) isosceles trapezium; half the sum and difference of the smaller result and the moieties of the face and base (of the known figure) will be the base and face (respectively of the required figure).”¹

v2.c17.s19.20

¹ *GSS*, vii. 173½.

v2.c17.s19.21

Let *a*, *b*, *c*, *h*, denote respectively the face, base, equal sides and altitude of the known isosceles trapezium and let *a'*, *b'*, *c'*, *h'*, denote the corresponding quantities of the required isosceles trapezium. Then, since the two trapeziums are equal in area and altitude, we must have $h'=h,$ $b'+a'=b+a, 1$ and $c'²-((b'-a')/2)²=h²,$ or $c'+(b'-a')/2c'-(b'-a')/2=h²,$ whence $c'-(b'-a')/2=r,$ and $c'+(b'-a')/2=h²/r,$ *r* being any rational number. Then $c'=½(h²/r+r), 2$ $b'-a'=(h²/r-r). 3$

v2.c17.s19.22

From (1) and (2), we get $b'=(b+a)/2+(h²/r-r)/2, 4$ $a'=(b+a)/2-(h²/r-r)/2. 5$

v2.c17.s19.23

If a=4, b=14, c=13, h=12, taking r=10, we shall have¹ a'=34/5, b'=56/5, c'=61/5.

v2.c17.s19.24

¹ *GSS*, vii. 174½.

v2.c17.s19.25

It has been stated above that, if *m*, *n*, *p*, *q* are rational numbers such that m²± n²<p²± q², we must have $a=(p²-q²)-(m²-n²),$ $b=(p²-q²)+(m²-n²),$ $c=m²+n²,$ $h=2mn=2pq.$

v2.c17.s19.26

Substituting these values in (2), (4), (5) we get the dimensions of the equivalent isosceles trapezium as $a' &= (p² - q²) - (4p²q²/r - r)/2, / b' &= (p² - q²) + (4p²q²/r - r)/2, / c' &= (4p²q²/r + r)/2.$

v2.c17.s19.27

If m² ± n² > p² ± q², the sides of the pair of isosceles trapeziums equal in area and altitude will be $a &= (m² - n²) - (p² - q²), / b &= (m² - n²) + (p² - q²), / c &= p² + q²; / a' &= (m² - n²) - (4m²n²/r - r)/2, / b' &= (m² - n²) + (4m²n²/r - r)/2, / c' &= (4m²n²/r + r)/2.$

v2.c17.s19.28

These two isosceles trapeziums will also have equal diagonals.

Rational Trapeziums with Three Equal Sides

v2.c17.s19.t1.1

This problem is nearly the same as that of the rational isosceles trapezium with this difference that in this case one of the parallel sides also is equal to the slant sides. Brahmagupta states the following solution of the problem :

v2.c17.s19.t1.2

“The square of the diagonal (of a *generated* rectangle) gives three equal sides; the fourth (is obtained) by subtracting the square of the upright from thrice the square of the side (of that rectangle). If greater, it is the base; if less, it is the face.”¹

v2.c17.s19.t1.3

¹ *BrSpSi*, xii. 37.

v2.c17.s19.t1.4

The rectangle generated from m, n is given by $m² - n², 2mn, m² + n².$

v2.c17.s19.t1.5

If ABCD be a rational trapezium whose sides AB, BC, AD are equal, then $AB = BC = AD = (m² + n²)²,$ $CD = 3(2mn)² - (m² - n²)² = 14m²n² - m⁴ - n⁴,$ or CD = 3(m² - n²)² - (2mn)² = 3m⁴ + 3n⁴ - 10m²n².

v2.c17.s19.t1.6

In particular, putting m = 2, n = 1, Pṛthûdakasvâmî deduces two rational trapeziums with three equal sides, *viz.*, (25, 25, 25, 39) and (25, 25, 25, 11).

v2.c17.s19.t1.7

The first solution is also given by Mahâvîra who indicates the method for obtaining it. He says :

v2.c17.s19.t1.8

"For a trapezium with three equal sides (proceed) as in the case of the isosceles trapezium with (the rectangle formed from) the quotient of the area of a generated rectangle divided by the square-root of its side multiplied by the difference of its elements and divisor; and that (formed) from the side and upright."

v2.c17.s19.t1.9

That is to say, from any rectangle (m² - n², 2mn, m² + n²), calculate $(2mn(m² - n²))/(√(2mn(m - n))) = √(2mn(m + n)).$

v2.c17.s19.t1.10

Then from √(2mn(m - n)), √(2mn(m + n)) form the rectangle $8m²n², 4mn(m² - n²), 4mn(m² + n²). 1$

v2.c17.s19.t1.11

Again from 2mn, m² - n² form another rectangle $6m²n² - m⁴ - n⁴, 4mn(m² - n²), (m² + n²)² 2$

v2.c17.s19.t1.12

By the juxtaposition of the rectangles (1) and (2) we get Brahmagupta's rational trapezium with three equal sides : CD = 8m²n² + (6m²n² - m⁴ - n⁴) = 14m²n² - m⁴ - n⁴, AB = 8m²n² - (6m²n² - m⁴ - n⁴) = (m² + n²)² = AD = BC, if m² + n² < 4mn.

v2.c17.s19.t1.13

¹ *GSS*, vii. 101½.

v2.c17.s19.t1.14

The segment (CH), altitude (AH), diagonals (AC, BD) and area of this trapezium are also rational, for CH = 6m²n² - m⁴ - n⁴, AH = 4mn(m² - n²), AC = BD = 4mn(m² + n²), area ABCD = 32m³n³(m² - n²).

v2.c17.s19.t1.15

**Rational Inscribed Quadrilaterals.** Brahmagupta formulated a remarkable proposition: To find all quadrilaterals which will be inscribable within circles and whose sides, diagonals, perpendiculars, segments (of sides and diagonals by perpendiculars from vertices as also of diagonals by their intersection), areas, and also the diameters of the circumscribed circles will be expressible in integers. Such quadrilaterals are called Brahmagupta quadrilaterals.

v2.c17.s19.t1.16

The solution given by Brahmagupta is as follows:

v2.c17.s19.t1.17

"The uprights and bases of two right-angled triangles being reciprocally multiplied by the diagonals of the other will give the sides of a quadrilateral of unequal sides: (of these) the greatest is the base, the least is the face, and the other two sides are the two flanks."¹

v2.c17.s19.t1.18

¹ *BrSpSi*, xii. 38.

v2.c17.s19.t1.19

Taking Brahmagupta's integral solution, the sides of the two right triangles of reference are given by m² - n², 2mn, m² + n² ; p² - q², 2pq, p² + q² ;

v2.c17.s19.t1.20

where *m*, *n*, *p*, *q* are integers. Then the sides of a Brahmagupta quadrilateral are $. (m² - n²)(p² + q²), & (p² - q²)(m² + n²), / 2mn(p² + q²), & 2pq(m² + n²). (A)$

v2.c17.s19.t1.21

Mahâvîra says :

v2.c17.s19.t1.22

“The base and the perpendicular (of the smaller and the larger derived rectangles of reference) multiplied reciprocally by the longer and the shorter diagonals and (each again) by the shorter diagonal will be the sides, the face and the base (of the required quadrilateral). The uprights and bases are reciprocally multiplied and then added together; again the product of the uprights is added to the product of the bases; these two sums multiplied by the shorter diagonal will be the diagonals. (These sums) when multiplied respectively by the base and perpendicular of the smaller figure of reference will be the altitudes; and they when multiplied respectively by the perpendicular and the base will be the segments of the base. Other segments will be the difference of these and the base. Half the product of the diagonals (of the required figure) will be the area.”¹

v2.c17.s19.t1.23

¹ *GSS*, vii. 103½.

v2.c17.s19.t1.24

If the rectangle generated from *m*, *n* be smaller than that from *p*, *q*, then, according to Mahâvîra, we obtain the rational inscribed quadrilateral of which the sides are $(m² - n²)(p² + q²)(m² + n²), (p² - q²)(m² + n²)²,$ $2mn(p² + q²)(m² + n²), 2pq(m² + n²)² ;$ whose diagonals are $2pq(m² - n²) + 2mn(p² - q²)(m² + n²),$ $(m² - n²)(p² - q²) + 4mnpq(m² + n²);$ whose altitudes are $2pq(m² - n²) + 2mn(p² - q²) 2mn,$ $(m² - n²)(p² - q²) + 4mnpq (m² - n²);$ whose segments are $2pq(m² - n²) + 2mn(p² - q²) (m² - n²),$ $(m² - n²)(p² - q²) + 4mnpq 2mn;$ and whose area is $½ 2pq(m² - n²) + 2mn(p² - q²) (m² - n²)(p² - q²)$ $+ 4mnpq (m² + n²)².$

v2.c17.s19.t1.25

Śrîpati writes :

v2.c17.s19.t1.26

“Of the two right triangles the sides and uprights are reciprocally multiplied by the hypotenuses; of the products the greatest is the base, the smallest is the face and the rest are the two flank sides of a quadrilateral with unequal sides.”∗1

v2.c17.s19.t1.27

∗1 *SiŚe*, xiii. 42.

v2.c17.s19.t1.28

Bhâskara II gives the rule :

v2.c17.s19.t1.29

“The sides and uprights of two optional right triangles being multiplied by their reciprocal hypotenuses become the sides : in this way has been derived a quadrilateral of unequal sides. There the two diagonals can be obtained from those two triangles. The product of the uprights, added with the product of the sides, gives one diagonal; the sum of the reciprocal products of the uprights and sides is the other.”∗2

v2.c17.s19.t1.30

∗2 *L*, p. 51.

v2.c17.s19.t1.31

Bhâskara II³ illustrates by taking the right triangles (3, 4, 5) and (5, 12, 13) so that the resulting cyclic quadrilateral is (25, 39, 60, 52). The same example was given before by Mahâvîra¹ and Pṛthûdakasvâmî.² This cyclic quadrilateral also appears in the *Triśatikâ* of Śrîdhara³ and in the commentary of the *Āryabhaṭîya* by Bhâskara I (522). The diagonals of this quadrilateral are, states Bhâskara II, 56 (= 3.12 + 4.5) and 63 (= 4.12 + 3.5) (Fig. 7). He then observes:

v2.c17.s19.t1.32

³ *L*, p. 52.

v2.c17.s19.t1.33

¹ *GSS*, vii. 104½.

v2.c17.s19.t1.34

² *BrSpSi*, xii. 38 (*Com.*).

v2.c17.s19.t1.35

³ *Triś*, Ex. 80.

v2.c17.s19.t1.36

"If the figure be formed by changing the arrangement of the face and flank then the second diagonal will be equal to the product of the hypotenuses of the two right triangles (of reference), *i.e.*, 65." (Fig. 8).

v2.c17.s19.t1.37

Fig. 7

v2.c17.s19.t1.38

Fig. 8

v2.c17.s19.t1.39

By taking the right triangles (3, 4, 5) and (15, 8, 17) Bhâskara II gets another cyclic quadrilateral (68, 51, 40, 75), whose diagonals are (77, 85), altitude is 308/5, segments are 144/5 and 231/5, and area is 3234.⁴ (Fig. 9). With the sequence of the sides (68, 40, 51, 75) the diagonals are (77, 84) (Fig. 10), and with (68, 40, 75, 51) they are (84, 85). (Fig. 11).

v2.c17.s19.t1.40

⁴ *L*, pp. 46ff.

v2.c17.s19.t1.41

Fig. 9

v2.c17.s19.t1.42

Fig. 10

v2.c17.s19.t1.43

Fig. 11

v2.c17.s19.t1.44

The deep significance of Brahmagupta's results has been demonstrated by Chasles¹ and Kummer.²

v2.c17.s19.t1.45

¹ M. Chasles, *Aperçu historique sur l'origine et development des méthodes en géométrie*, Paris, 1875, pp. 436ff.

v2.c17.s19.t1.46

² E. E. Kummer, "Über die Vierecke, deren Seiten und Diogonalen rational sind," *Journ. für Math.*, XXXVII, 1848, pp. 1-20.

v2.c17.s19.t1.47

In fact, according to the sequence in which the quantities (A) are taken, there will be two varieties of Brahmagupta quadrilaterals having them as their sides : (1) one in which the two diagonals intersect at right angles and (2) the other in which the diagonals intersect obliquely. The arrangement (A) gives a quadrilateral of the first variety. For the oblique variety, the sides are in the following order : $.(p² - q²)(m² + n²), & (m² - n²)(p² + q²), / 2mn(p² + q²), & 2pq(m² + n²) ; (B)$ $or .(p² - q²)(m² + n²), & 2mn(p² + q²), / (m² - n²)(p² + q²), & 2pq(m² + n²). (C)$

v2.c17.s19.t1.48

Bhâskara II points out that the diagonals of the Brahmagupta quadrilateral are in the (A) variety, $2pq(m² - n²) + 2mn(p² - q²), 4mnpq + (p² - q²)(m² - n²) ;$ in (B), $2pq(m² - n²) + 2mn(p² - q²), (p² + q²)(m² + n²) ;$ and in (C), $4mnpq + (p² - q²)(m² - n²), (p² + q²)(m² + n²).$ The diameter of the circumscribed circle in every case is (p² + q²)(m² + n²).

v2.c17.s19.t1.49

Gaṇeśa (1545) shows that the quadrilateral is formed by the juxtaposition of four right triangles obtained by multiplying the sides of each of two rational right triangles by the upright and base of the other. He writes :

v2.c17.s19.t1.50

“A quadrilateral is divided into four triangles by its intersecting diagonals. So by the juxtaposition of four triangles a quadrilateral will be formed. For that purpose the four triangles are assumed in this manner : Take two right triangles formed in the way indicated

v2.c17.s19.t1.51

Compare also L. E. Dickson, “Rational Triangles and Quadrilaterals,” *Amer. Math. Mon.*, XXVIII, 1921, pp. 244-250.

v2.c17.s19.t1.52

before. If the upright, base and hypotenuse of a rational right triangle be multiplied by any arbitrary rational number, there will be produced another right triangle with rational sides. Hence on multiplying the sides of each of the two right triangles by an optional number equal to the base of the other and again by an optional nnmber equal to the upright of the other, four right triangles will be obtained by the judicious juxtaposition of which the quadrilateral will be formed."

v2.c17.s19.t1.53

He then shows with the help of specific examples (see Figs. 12, 13 & 14) that we can obtain in this way from the same set of two rational right triangles two varieties of rational convex quadrilaterals: One in which the diagonals intersect each other perpendicularly; and the other in which they do so obliquely.

v2.c17.s19.t1.54

Fig. 12

v2.c17.s19.t1.55

Fig. 13

v2.c17.s19.t1.56

Fig. 14

Inscribed Quadrilaterals having a Given Area

v2.c17.s19.t2.1

Mahâvîra proposes to find all rational inscribed rectangles having the same given area (A, say). He says:

v2.c17.s19.t2.3

$altitude = 2mn/s,$ $side = (m² + n²)/s;$ where s is an arbitrary rational number chosen such that s²A > 2mn(m²-n²).

v2.c17.s19.t2.4

For the construction of an inscribed trapezium of three equal sides Mahâvîra gives the following rule :

v2.c17.s19.t2.5

"The square of the given area is divided by the cube of an optional number and then increased by that optional number ; half the result gives the (equal) sides of a generated trapezium of three equal sides (having the given area) ; twice the optional number diminished by the side is the base ; and the given area divided by the optional number is the altitude."¹

v2.c17.s19.t2.6

¹ *GSS*, vii. 150.

v2.c17.s19.t2.7

In other words, the dimensions of an inscribed trapezium of three equal sides having a given area A will be

v2.c17.s19.t2.8

$side = ½((A²)/(s³) + s),$ $base = 2s - ½((A²)/(s³) + s),$ $altitude = A/s.$

v2.c17.s19.t2.9

To find inscribed quadrilaterals having a given area Mahâvîra gives the following rule :

v2.c17.s19.t2.10

"Break up the square of the given area into any four arbitrary factors. Half the sum of these factors is diminished by them (severally). The remainders are the sides of an (inscribed) quadrilateral with unequal sides."²

v2.c17.s19.t2.11

² *GSS*, vii. 152. This result follows from the fact that the area of a cyclic quadrilateral is √((s - a)(s - b)(s - c)(s - d)).

v2.c17.s19.t2.12

**Triangles and Quadrilaterals having a Given Circum-Diameter.** Mahâvîra proposes to find all rational triangles and quadrilaterals inscribable in a circle of given diameter. His solution is :

v2.c17.s19.t2.13

"Divide the given diameter of the circle by the calculated diameter (of the circle circumscribing any generated figure of the required kind). The sides of that generated figure multiplied by the quotient will be the sides of the required figure."¹

v2.c17.s19.t2.14

¹ *GSS*, vii. 221½.

v2.c17.s19.t2.15

In other words, we shall have to find first a rational triangle or cyclic quadrilateral ; then calculate the diameter of its circum-circle and divide the given diameter by it. Dimensions of the optional figure multiplied by this quotient will give the dimensions of the required figure of the type.

v2.c17.s19.t2.16

It has been found before (p. 227) that the sides of a rational triangle are proportional to $m² + n², p² + q², (p² - q²) ± (m² - n²)$ and its altitude is proportional to 2mn (or 2pq), m, n, p, q being any rational numbers such that mn = pq. The diameter of the circle circumscribed about this triangle is proportional to $((m² + n²)(p² + q²))/2mn.$

v2.c17.s19.t2.17

Then the sides of a rational triangle inscribed in a circle of diameter D will be $2mnD/(p² + q²), 2mnD/(m² + n²), 2mnD((p² - q²) ± (m² - n²))/((m² + n²) (p² + q²));$ and its altitude will be $(D(2mn)²)/((m² + n²)(p² + q²)).$

v2.c17.s19.t2.18

The dimensions of a rational inscribed quadrilateral, as stated by Mahâvîra, have been noted before (p. 236). The diameter of its circum-circle is $(p² + q²)(m² + n²)².$ Then, according to Mahâvîra, the sides of a rational quadrilateral inscribed in a circle of diameter D, are $D(2mn/(m² + n²)), D((m² - n²)/(m² + n²)), D(2pq/(p² + q²)), D((p² - q²)/(p² + q²));$ its diagonals are $2pq (m² - n²) + 2mn (p² - q²) D/((p² + q²)(m² + n²)),$ $(m² - n²)(p² - q²) + 4mnpq D/((p² + q²)(m² + n²));$ and its area is $(D²)/(2 (p² + q²)(m² + n²)) 2pq(m² - n²) + 2mn(p² - q²)$ $× (m² - n²)(p² - q²) + 4mnpq;$ so that the sides, diagonals and area are all rational.

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