पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४३६

240

GAṆITASĀRASAṄGRAHA.

of area equal to the accurate measure of the area of the given quadrilateral figure with two equal sides:-

173${\tfrac {1}{2}}$. If the square of the value of the perpendicular (in the given quadrilateral figure with two equal sides) is used along with the given optional number in carrying out the process of viṣamasaṅkramaṇa the larger (of the two results obtained) becomes the measure of either of the equal sides (in the required quadrilateral figure with two equal sides). Half of the sum of the values of the top-side and the base (in the given quadrilateral figure with two equal sides), on being respectively increased and decreased by the smaller (of the two results in the viṣamasaṅkramaṇa process above-mentioned), give rise to the values of the base and the top-side in the (required) quadrilateral figure with two equal sides

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173${\tfrac {1}{2}}$. The problem contemplated in this rule is to construct a quadrilateral with two equal sides that shall be equal in area to a given quadrilateral with two equal sides and shall also have the same perpendicular distance from the top-side to the base. Let a and c be the equal sides of the given quadrilateral, and a and d be the top side and the base thereof respectively; and let the value of the perpendicular distance be p. If $a_{1},b_{1},c_{1},d_{1}$ be taken to be the corresponding sides of the required quadrilateral, then, since the area and the perpendicular are the same in the case of both the quadrilaterals, we have,

$d_{1}+b_{1}=d+b(1)$

and $a_{1}^{2}-\left({\tfrac {d_{1}-b_{1}}{2}}\right)^{2}=p^{2}$(II):

that is, $\left(a_{1}+{\tfrac {d_{1}-b_{1}}{2}}\right)\left(a_{1}-{\tfrac {d_{1}-b_{1}}{2}}\right)=p^{2}$

Let $a_{1}-{\tfrac {d_{1}-b_{1}}{2}}=N;{\text{ then }}a_{1}+{\tfrac {d_{1}-b_{1}}{2}}={\tfrac {p^{2}}{N}}$

$\left(a_{1}+{\tfrac {d_{1}-b_{1}}{2}}\right)+\left(a_{1}-{\tfrac {d_{1}-b_{1}}{2}}\right)={\tfrac {p^{2}}{N}}+N$.

$\therefore {\tfrac {{\tfrac {p^{2}}{N}}+N}{2}}=a_{1}$(III);

and ${\tfrac {d+b}{2}}\pm {\tfrac {{\tfrac {p^{2}}{N}}-N}{2}}={\tfrac {d_{1}+b_{1}}{2}}\pm {\Bigg \{}{\tfrac {\left(a_{1}+{\tfrac {d_{1}-b_{1}}{2}}\right)-\left(a_{1}-{\tfrac {d_{1}-b_{1}}{2}}\right)}{2}}{\Bigg \}}=d_{1}{\text{ or }}b_{1};$(IV)

Here N is what is called इष्ट or the optionally given number in the rule; and formulas III and IV are those that are given in the rule for the solution of the problem.