# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४२०

एतत् पृष्ठम् परिष्कृतम् अस्ति
224
GAṆITASĀRASAṄGRAHA.

operation of saṅkramaṇa along with half the perimeter, the (required) base and also the perpendicular-side are arrived at.

An example in illustration thereof.

126${\tfrac {1}{2}}$ . The perimeter in this case is 34; and the diagonal is seen to be 13. Give out, after calculating, the measures of the perpendicular-side and the base in relation to this derived figure.

The rule for arriving at the numerical values of the base and the perpendicular-side when the area of the figure and the value of the diagonal are known :-

127${\tfrac {1}{2}}$ . Twice the measure of the area is subtracted from the square of the diagonal. It is also added to the square of the diagonal. The square roots (of the difference and of the sum so obtained) give rise to the measures of the (required) perpendicular-side and the base, if the larger (of the square roots) is made to undergo the process of saṅkramaṇa in relation to the smaller (square root).

An example in illustration thereof.

128${\tfrac {1}{2}}$ In the case of a longish quadrilateral figure, the measure of the area is 60, and the measure of its diagonal is 13. I wish to hear (from you) the measures of the perpendicular-side and the base.

The rule for arriving at the numerical values of the base and the perpendicular-side in relation to a longish quadrilateral figure, when the numerical value of the area of the figure and the numerical value of the perimeter (thereof) are known:-

129${\tfrac {1}{2}}$ . From the quantity representing the square of half the perimeter, the measure of the area as multiplied by four is to be

127${\tfrac {1}{2}}$ . Adopting the same symbols as in the note to stanza 126${\tfrac {1}{2}}$ , we have the following formula to represent the rule here given :-

${\Bigg \{}{\sqrt {\left(a^{2}+b^{2}\right)^{2}+2ab}}\pm {\sqrt {\left(a^{2}+b^{2}\right)^{2}-2ab}}{\Bigg \}}+2=a{\text{ or }}b$ , as the case may be.

129${\tfrac {1}{2}}$ . Here we have ${\Bigg \{}{\tfrac {2a+2b}{2}}\pm {\sqrt {{\tfrac {2a+2b}{2}}-2ab}}{\Bigg \}}+2=a{\text{ or }}b$ , as the case may be. 