पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४२३

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CHAPTER VII--MEASUREMENT OF AREAS.

187. The squares (of the ratio-values) of the perimeters (of he required isosceles triangles) are multiplied by (the ratio-values of) the areas (of those triangles) in alternation. (Of the two products so obtained), (the larger one is) divided by the smaller; and (the resulting quotient) is multiplied by six and (is also separately multiplied) by two. The smaller (of the two products so obtained) is diminished by one. The larger product and the diminished smaller product constitute the two bījas (in relation to the longish quardrilateral figure) from which one (of the required triangles) is to be obtained. The difference between these (two bījas above, noted) and twice the smaller one (of those bījas constitute the ins (in relation to the longish quadrilateral figure) from which the other (required triangle) is to be obtained. (From the two longish quadrilateral figures formed with the aid of their respective bījas)the sides and the other things (relating to the required triangles) are to be arrived at as (explained) before.

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187. When a : b is the ratio of the perimeters of the two isosceles triangles, and c : d the ratio of their areas, then, according to the rule, ${\tfrac {6b^{2}c}{a^{2}d}}$ and ${\tfrac {2b^{2}c}{a^{2}d}}-1$ and ${\tfrac {4b^{2}c}{a^{2}d}}+1$ and ${\tfrac {4b^{2}c}{a^{2}d}}+2$ are the two sets of bījas, with the help of which the values of the various required elements of the two isosceles triangles may be arrived at. The measures of the sides and the altitudes, calculated from these bījas according to stanza 108${\tfrac {1}{2}}$ in this chapter, when multiplied respectively by a and b, (the quantities occurring in the ratio of the perimeters), give the required measures of the sides and the altitudes of the two isosceles triangles. They are as follow:-

I

Equal side $=a\times {\Big \{}\left({\tfrac {6b^{2}c}{a^{2}d}}\right)^{2}+\left({\tfrac {2b^{2}c}{a^{2}d}}-1\right)^{2}{\Big \}}$

Base $=a\times 2\times 2\times {\tfrac {6b^{2}c}{a^{2}d}}\times \left({\tfrac {2b^{2}c}{a^{2}d}}-1\right)$

Altitude $=a\times {\Big \{}\left({\tfrac {6b^{2}c}{a^{2}d}}\right)^{2}+\left({\tfrac {2b^{2}c}{a^{2}d}}-1\right){\Big \}}$

II

Equal side $=b\times {\Big \{}\left({\tfrac {4b^{2}c}{a^{2}d}}+1\right)^{2}+\left({\tfrac {4b^{2}c}{a^{2}d}}-2\right)^{2}{\Big \}}$

Base $=b\times 2\times 2\times \left({\tfrac {4b^{2}c}{a^{2}d}}+1\right)\times \left({\tfrac {4b^{2}c}{a^{2}d}}-2\right)$

Altitude $=b\times {\Big \{}\left({\tfrac {4b^{2}c}{a^{2}d}}+1\right)^{2}-\left({\tfrac {4b^{2}c}{a^{2}d}}-2\right)^{2}{\Big \}}$

Now it may be easily proved from these values that the ratio of the perimeters is a:b, and that of the areas is c:d, as taken for granted at the beginning.