# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४१४

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218
GAṆITASĀRASAṄGRAHA.

reference). The quotients (so obtained) are multiplied respectively by the perpendicular-side and the base of the smaller (oblong of reference). The (resulting) products give rise to the measures of the perpendiculars (in relation to the required quadrilateral). To these (two perpendiculars), the above values of the two sides (other than the base and the top-side) are (separately) added, (the larger side being added to the larger perpendicular and the smaller side to the smaller perpendicular). The differences between these perpendiculars and sides are also obtained (in the same order). The sums (above noted) are multiplied (respectively) by (these) differences. The square roots (of the products so obtained) give rise to the values of the segments (of the base in relation to the required quadrilateral). Half of the product of the diagonals (of the required quadrilateral) gives the value of (its) area.

The rule for arriving at an isosceles triangle with the aid of a single derived oblong (of reference).

108${\tfrac {1}{2}}$ . The two diagonals (of the oblong of reference constructed with the aid of the given bījas) become the two (equal) sides of the (required) isosceles triangle. The base (of the oblong of reference), multiplied by two, becomes the base (of the required triangle). The perpendicular-side (of the oblong of reference) is the perpendicular (of the required triangle from the apex to the base thereof). The area (of the required triangle) is the area (of the oblong of reference).

108${\tfrac {1}{2}}$ .^  The rationale of the rule may be made out thus:- Let ABCD be an oblong and let AD be produced to E so that AD=DE. Join EC. It will be seen that ACE is an isosceles triangle whose equal sides are equal to the diagonals of the oblong and whose area is equal to that of the oblong. 