ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

RATIONAL TRAPEZIUMS 265
AD = BC = m²+n² (the sides of the trapezium)
HC = base-upright = ½ [ 4m²n²/p - p ] (segment)
AC = BD = [ 4m²n²/p + p ] (diagonal)
AH = 2 mn (altitude)
ABCD = mn [ 4m²n²/p - p ] (area)
By chosing the values of m n and p suitably, the values
of all the dimensions of the isosceles trapezium can be made
integral. Pṛthūdaka Svāmī starts with the rectangle (5, 12, 13)
and suitably takes p as 6; then he calculates out the dimensions
of the trapezium : flank sides (AD and BC) = 13, base =14,
and base = 4, altitude (AH) = 12, segments of base (DH and
HC) = 5, and 9, diagonals (AC and BD) = 15, area ABCD =
108. All these values are integers.
In this example, the rectangle chosen is (5, 12, 13) which is
AA' DH, where AD = m² + n² = 13
and DH = m² - n² = 5
whence by adding the two we have
2m² = 18
This gives the value of m = 3, and hence n = 2. Pṛthū-
daka Svāmī has taken the value of p = 6 by choice. Putting
these values of m, n and p, the values for the dimensions of the
isosceles trapezium follow from the expressions given by Brahma-
gupta.
CD = ½ ( (4.3².2² / 6) - 6 ) + ( 3²-2² ) = 9+5 = 14 (base)
Face = 9-5=4
Sides AD = BC = 3²+2² = 13
and so on for the other dimensions.
Rational Trapeziums With
Three Equal Sides
This problem is very much the same as one of the rational
isosceles tpapezium with the only difference that in this case
one of the parallel sides is also equal to the slant sides. We