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214
GANITASIRASANGBAHA.
1014. The difference between the (given) bijzs is multiplied by the square root of the base (of the quadrilateral immediately derived with the aid of those bijas). The area of (this immediately) derived (primary) quadrilateral is divided (by the product 80 obtained). Then, with the aid of the resulting quotient and the divisor (in the operation utilized as bijas, a second derived quadrilateral of reference is constructed. A third quadrilateral of
101. If a and b represent the given bijas, the meusures of the sides of the immediately dorived quadrilateral are :
Perpondicular-ride =a- b? Buse = 2ab Diagonal= 2% + 1
Aros = 2ab x (a - b) As in the case of the construction of the quadrilateral with two equal sides (vide stanza 99 ante), this rulo proceeds to construct the roquired quadrilateral with threo equal sides with the aid of two derived roctangles. The bijas in relation to the first of theso rectangles are :2ab a– bo):
1.A., V 2ab (a + b), and V2ab (a - b). V Zab * (a - b) Applying the rulo given in stanza 90 above, we have for the first roc.
tangle: Porpendicular-side = (a + b) x 2ab - (a - b)? 2ab or 8 a?bo. Bnso = 2 x V 2ab (a + b) x V 2ab (a - b) or tab (a - b).
Diagonal = (a + b)2ab + (a - b) * 2ab or 4að (a+ b*). The dijas in tho case of the second rectangle are: 4-and 2ab. The various elements of this rectangle are :
Porpendicular-side = 4a6% - (a - b) ; Bage=4ab (az - 6');
Diagonal = 4a b2 + (a - bay or (a + b*) With the holp of theso two rectangles, tho moasuros of the siden, diugonals, eto., of the roquirod quadrilateral aro ascertained as iu tho rule given in stansa 991 above. Thoy aro:
Base = sum of the perpendioular sides = 8ab' + 4a98-a'-6")". Top-side = greater perpendioular-side nuinua smaller perpendioular-side
= Sab* - 4a?b!-(a'-6")* = (a'+)'. Either of the lateral sidus = smaller diagonal=(a+b). Losgor segment of the base = smaller perpendionlar-sido = 4a"6" - (a'
--0')', Perpondioular = base of either rectangle = fab (a'-'). Diagonal tho grenter of the two diagonale =4ab (a+b).
Area = area of the larger reotangle 80'' x 4ab (a'-6"). It may be noted here that the measure of either o! the two lateral sides is equal to the moosure of the top-side. Thus is obtained the required quadrilateral with threo equal sides.
