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