# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४३२

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236
GAṆITASĀRASAṄGRAHA

An example in illustration thereof

159${\tfrac {1}{2}}$ . In the case of a certain triangular figure with unequal sides, it has been pointed out that 2 constitutes the accurate measure of its area and 8 is the optionally chosen quantity. What is the value of the base as well as of the sides (of that triangle)?

Again, another rule for arriving, after knowing the exact numerical measure of a (given) area, at a triangular figure with unequal sides having that same (accurately measured) area (as its own):-

160${\tfrac {1}{2}}$ -161${\tfrac {1}{2}}$ . The square root of the measure of the given area as multiplied by eight and as increased by the square of an optionally chosen number is obtained. This and the optionally chosen number are divided by each other. The larger (of these quotients) is diminished by half of the smaller (quotient). The remainder (thus obtained) and (this) half of the smaller (quotient) are respectively multiplied by the above-noted square root and the optionally chosen number. On carrying out, in relation to the products (thus obtained), the process of saṅkramaṇa, the values of the base and of one of the sides are arrived at. Half of the optionally chosen number happens to be the measure of the other side in a triangular figure with unequal sides.

An example in illustration thereof.

162${\tfrac {1}{2}}$ . In the case of a triangle with unequal sides, the accurate measure of the area is 2, and the optionally chosen quantity is 3. O friend who know the secret of calculation, give out the measure of the base as well as of the sides.

The rule for arriving, after knowing the accurate measure of a (given) area, at a regularly circular figure having that accurately measured area (as its own):-

168${\tfrac {1}{2}}$ . The accurate measure of the area is multiplied by four and is divided by the square root of ten. On getting at the square

163${\tfrac {1}{2}}$ . The role in this stanza is derived from the formula, area $={\tfrac {d^{2}}{4}}\times {\sqrt {10}}$ , where d is diameter of the circle. 