पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४०७

CHAPTER VII--MEASUREMENT OF AREAS.

at the numerical measure of the perpendicular-side and of the other side, when the numerical measure of the hypotenuse is known:-

97${\tfrac {1}{2}}$.^[1] The operation of saṅkramaṇa, conducted between (an optionally chosen exact) divisor of the square of the measure of the perpendicular-side and the resulting quotient, gives rise to the measures of the hypotenuse and of the other side (respectively). Similarly (the same operation of saṅkramaṇa) in relation to the square of the measure of the other side (gives rise to the measures of the perpendicular-side and of the hypotenuse). Or, the square root of the difference between the squares of the hypotenuse and of a (suitably chosen) optional number forms, along with that chosen number, the perpendicular-side and the other side respectively.

An example in illustration thereof.

98${\tfrac {1}{2}}$. In the case of a certain (geometrical) figure, the perpendicular-side is 11 in measure; in the case of another figure, the (other) side is 60; and in the case of (still) another figure the hypotenuse is 61. Tell me in these cases the measures of the unmentioned elements.

The rule regarding the manner of arriving at a quadrilateral figure having two equal sides (with the aid of the given bījas):-

99${\tfrac {1}{2}}$.^[2] The perpendicular-side of the primary figure derived (with the aid of the given bījas), on being added to the perpendicular-side (in another figure) derived with the aid of the two optionally chosen) factors of half the base of (this original) derived

---

97${\tfrac {1}{2}}$.^  This rule depends on the following identities :-

${\begin{aligned}I.&\left\{{\tfrac {(a^{2}-b^{2})^{2}}{(a-b)^{2}}}\pm (a-b)^{2}\right\}\div 2=a^{2}+b^{2}{\text{ or }}2ab{\text{ as the case may be }}.\\II.&\left\{{\tfrac {(2ab)^{2}}{2b^{2}}}\pm 2b^{2}\right\}\div 2=a^{2}+b^{2}{\text{ or }}a^{2}-b^{2}.\\III.&{\sqrt {(a^{2}+b^{2})^{2}-(2ab)^{2}}}=a^{2}-b^{2}.\end{aligned}}$

99${\tfrac {1}{2}}$.^  The problem solved in the role stated in this stanza is to construct with the aid of two given bījas a quadrilateral having two equal sides. The lengths of the sides, of the diagonals, of the perpendicular from the end-points of the top-side to the base, and of the segments thereof caused by the perpendicular are all derived from two rectangles constructed with the aid of the given bījas. The first of these rectangles is formed according to the rule given in stanza 90${\tfrac {1}{2}}$ above. The second rectangle is formed according to the same rule from two optionally chosen factors of half the length of the base of the first rectangle,