214 GANITASARASANGRAHA. 101. The difference between the (given) bijus is multiplied by the square root of the base (of the quadrilateral immediately derived with the aid of those bijas). The area of (this immedi- ately) derived (primary) quadrilateral is divided (by the product. so obtained). Then, with the aid of the resulting quotient and the divisor (in the operation utilized as bijas, a second derived qua- drilateral of reference is constructed. A third quadrilateral of 101 If a and b represent the given bigas, the measures of the sides of the immediately derived quadrilateral are - Perpendicular-side = a - b Base 2ab Diagonal a + b² Ared 2ab x (a²-be) As in the case of the construction of the quadrilateral with two equal sides (vide stanza 99 ante), this rule proceeds to construct the requngd quadrilateral with thico equal sides with the aid of two derived rectangles The bijas in rolation to the first of these rectangles are - 2ab x (a - b) e., 2ab (a + b), and /Zab × (a - b). 2ab x (a - b) Applying the rule given in stanza 90 above, we have for the first rec- tangle Perpendicular-side = (a + b) x 2ab (a - b) × 2ab or 8 a²b². Base 2 x 2ab x (a + b) x √/2ab x (a - b) or tab (a - b). 2ab or 4ab (a + b). ab and 2ab Diagonal (a+b)x 2ab + (a - b) 2 x The bigas in the case of the second rectangle are The various elements of this rectangle are Perpendicular-side = 4a² b (a²-b²) ³., Base 4ab (a - b); Diagonal 4a b² + (a²-b²) or (a² + b²) ³. With the help of these two rectangles, the measures of the sides, diagonals, etc, of the required quadrilateral are ascertamed as in the rule given in stanza 99 abovc. They are Base sum of the perpendicular sides 8a²b³ + 4u²b²-(a²-bª). Top-side greater perpendicular-side manus smaller perpendicular-side = Sa³b²- [4a²b²(a²-6²)²} = (a² +6²) ². Either of the lateral sides Lesser segment of the base -6²) ². smaller diagonal-(a + b², ². smaller perpendicular-side = 4a²b² - (a² Perpendicular base of either rectangle = 4ab (a²-6³). Diagonal the greater of the two diagonals = 4ab (a² + b²). Area area of the larger rectangle = 8a²b2 x 4ab (a²-b²). It may be noted here that the measure of either of the two lateral sides is equal to the measure of the top-side. Thus is obtained the required quadrilateral with three equal sides.
