Ganitasara Sangraha

SR No.: 020800
Original Sutra Author: N/A
Author: Mahaviracharya, M Rangacharya
Publisher: Government of Madras
Publication Year: 1912
Total Pages: 523
Language: Hindi, Sanskrit
Classification: Book_Devnagari
File Size: 22 MB

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GANITASĀRASANGRAHA.
1013. The difference between the (given) bijas 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 so 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 bījas, the measures of the sides of the immediately derived quadrilateral are:
Perpendicular-side=a? - 62 Base = 2ab Diagonal= a2 + 12
Area = 2ab x (a - b) As in the case of the construction of the quadrilateral with two equal sides (vide stanza 99$ ante), this rule proceeds to construct the required quadrilateral with three equal sides with the aid of two derived rectangles. The būjas in relation to the first of these rectangles are :2ab x (a - 82):
i.e., 2ab x (a + b), and v2ab x (a - b). V 2ab + (a - b) Applying the role given in stanza 90) above, we havo for the first rec
tangle : Perpendicular-side = (a + b) ? x 2ab - (a - b) ? x 2ab or 8 abo. Base = 2 x V 2ab x (a + b) x V 2ab x (a - b) or 4ab (a- b).
Diagonal = (a + b)2 x 2ab + (a - b) ? x 2ab or 4a) (a + b). The bējas in the case of t.be second rectangle are: a - b and 2ab. The various elements of this rectangle are :
Perpendicular-side = 4a 1o - (a? - 6%) ; Base = 4ab (a2 - 12);
Diagonal = 4a2 b2 + (a? - baya or (u? + b) With the help of tbese two rectangles, the measures of the sides, diagonals, etc., of the required quadrilateral are ascertained as in the rule given in stanza 99% above. They are :
Base = sum of the perpendicular sides=82b2 + 4a2b? -(a* - 12). Top-side greater perpendicular-side ninus smaller perpendicular-side
= Saab% - 4a2j2 – (a? – 62)2 = (a’ +62)2. Either of the lateral sides=smaller diagonala? + b2, 2. Lesser segment of the base = smaller perpendicular-sido = 4a2b2 - (a*
-6?). Perpendionlar = base of either rectangle == 4ab (a? -69). Diagonal = the greater of the two diagonals =4ab (a? +62).
Area area of the larger reotangle =80262 x 4ab (as - 62). 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.