History of Hindu Mathematics: A Source Book v2.c17.s19.t2: Inscribed Quadrilaterals having a Given Area

History of Hindu Mathematics: A Source Book v2.c17.s19.t2

Language: English

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 :

"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.

Linked texts (1)

Gaṇitasārasaṅgraha — Mahāvīra (mathematician), Sanskrit