# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४१०

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214
GAṆITASĀRASAṄGRAHA.

101${\displaystyle {\tfrac {1}{2}}}$. The difference between the (given) bījas is multiplied by the square root of the base (of the quadrilateral immediately derived with the aid of those bījas). 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 bījas, a second derived quadrilateral of reference is constructed. A third quadrilateral of

101${\displaystyle {\tfrac {1}{2}}}$. If a and b represent the given bījas, the measures of the sides of the immediately derived quadrilateral are :-

Perpendicular-side ${\displaystyle =a^{2}-b^{2}}$
Base ${\displaystyle =2ab}$
Diagonal ${\displaystyle =a^{2}+b^{2}}$
Area ${\displaystyle =2ab\times (a^{2}-b^{2})}$

As in the case of the construction of the quadrilateral with two equal sides (vide stanza 99${\displaystyle {\tfrac {1}{2}}}$ ante), this rule proceeds to construct the required quadrilateral with three equal sides with the aid of two derived rectangles. The bījas in relation to the first of these rectangles are :-

${\displaystyle {\tfrac {2ab\times (a^{2}-b^{2}):}{{\sqrt {2ab}}\times (a-b)}}{\text{i.e.,}}{\sqrt {2ab}}\times (a+b){\text{, and }}{\sqrt {2ab}}\times (a-b)}$

Applying the rule given in stanza 90${\displaystyle {\tfrac {1}{2}}}$ above, we have for the first rectangle :

Perpendicular-side${\displaystyle =(a+b)^{2}\times 2ab-(a-b)^{2}\times 2ab{\text{ or }}8a^{2}b^{2}.}$
Base${\displaystyle =2\times {\sqrt {2ab}}\times (a+b)\times {\sqrt {2ab}}\times (a-b){\text{ or }}4ab(a^{2}-b^{2}).}$
Diagonal${\displaystyle =(a+b)^{2}\times 2ab+(a-b)^{2}\times 2ab{\text{ or }}4ab(a^{2}+b^{2}).}$

The bījas in the case of the second rectangle are: ${\displaystyle a^{2}-b^{2}-(a^{2}-b^{2}){\text{ and }}2ab.}$

The various elements of this rectangle are :

Perpendicular-side${\displaystyle =4a^{2}b^{2}-(a^{2}-b^{2})^{2};}$
Base${\displaystyle =4ab(a^{2}-b^{2});}$
Diagonal${\displaystyle =4a^{2}b^{2}+(a^{2}-b^{2}){\text{ or }}(a^{2}+b^{2})^{2}.}$

With the help of these two rectangles, the measures of the sides, diagonals, etc., of the required quadrilateral are ascertained as in the rule given in stanza 99${\displaystyle {\tfrac {1}{2}}}$ above. They are :

Base =sum of the perpendicular-sides ${\displaystyle =8a^{2}b^{2}+4a^{2}b^{2}-(a^{2}-b^{2})^{2}.}$
Top-side = greater perpendicular-side minus smaller perpendicular-side ${\displaystyle =8a^{2}b^{2}-\left\{4a^{2}b^{2}-(a^{2}-b^{2})^{2}\right\}=(a^{2}+b^{2})^{2}.}$
Either of the lateral sides = smaller diagonal ${\displaystyle (a^{2}+b^{2})^{2}}$
Lesser segment of the base = smaller perpendicular-side ${\displaystyle 4a^{2}b^{2}-(a^{2}-b^{2})^{2}.}$
Perpendicular = base of either rectangle ${\displaystyle 4ab(a^{2}-b^{2}).}$
Diagonal = the greater of the two diagonals ${\displaystyle 4ab(a^{2}+b^{2})}$
Area = area of the larger rectangle ${\displaystyle 8a^{2}b^{2}\times 4ab(a^{2}-b^{2}).}$

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.