पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४१८

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GAṆITASĀRASAṄGRAHA.

117${\tfrac {1}{2}}$. In the case of a longish quadrilateral figure, (the numerical measures of) twice the diagonal, three times the base and four times the perpendicular-side being taken, the measure of the perimeter is added to them. Twice (this sum) is the (numerical) measure of the area. (Find out the measure of the base.)

118${\tfrac {1}{2}}$. In the case of a longish quadrilateral figure, the (numerical) measure of the perimeter is 1. Tell me quickly, after calculating, what the measure of its perpendicular side is, and what that of the base.

119${\tfrac {1}{2}}$. In the case of a longish quadrilateral figure, the (numerical measures of twice the diagonal, three times the base, and fours times the perpendicular, on being added to the (numerical) measure of the perimeter, become equal to 1. (Find out the measure of the base.)

Another rule regarding the process of arriving at the number the bījas in relation to the derived longish quadrilateral figure :-

120${\tfrac {1}{2}}$.^[1] The operation to arrive at the generating (bījas) in relation to a longish quadrilateral figure consists in getting at the square roots of the two qnantities represented by (1) half of the diagonal as diminished by the perpendicular-side and (2) the difference between this quantity and the diagonal.

An example in illustration thereof.

121${\tfrac {1}{2}}$. In the case of a longish quadrilateral figure, the perpendicular-side is 55, the base is 48, and then the diagonal is 73, What are the bījas here ?

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120${\tfrac {1}{2}}$.^  The rule in stanza 95${\tfrac {1}{2}}$ of this chapter relates to the method of arriving at the bījas from the base or the perpendicular or the diagonal of a longish quadrilateral. But the role in this stanza gives a method for finding out the bījas from the perpendicular and the diagonal of a longish quadrilateral. The process described is based on the following identities:-

${\sqrt {\tfrac {a^{2}+b^{2}-(a^{2}-b^{2})}{2}}}=b{\text{; and }}{\sqrt {a^{2}+b^{2}{\tfrac {a^{2}+b^{2}-(a^{2}-b^{2})}{2}}}}=a$,

where $a^{2}+b^{2}$ is the measure of the diagonal, and $a^{2}-b^{2}$ is the measure of the perpendicular-side of a longish quadrilateral, a and b being the required bījas.