# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४२१

एतत् पृष्ठम् परिष्कृतम् अस्ति
225
CHAPTER VII--MEASUREMENT OF AREAS.

subtracted. Then, on carrying out the process of saṅkramaṇa with the square root (of this resulting difference) in relation to half the measure of the perimeter, the values of the (required) base and the perpendicular-side are indeed obtained.

130${\displaystyle {\tfrac {1}{2}}}$. In a derived longish quadrilateral figure, the measure of the perimeter is 170; the measure of the given area is 1,500. Tell me the values of the perpendicular-side and the base (thereof).

The rule for arriving at the respective pairs of (required) longish quadrilateral figures, (1) when the numerical measures of the perimeter are equal, and the area of the first figure is double that of the second; or, (2) when the areas of both the figures are equal, and the numerical measure of the perimeter of the second figure is twice the numerical measure of that of the first figure; or, (3) (again) when, in relation to the two required figures, the numerical measure of the perimeter of the second figure is twice the numerical measure of the perimeter of the first figure, and the area of the first figure is twice the area of the second figure:

131${\displaystyle {\tfrac {1}{2}}}$-133. (The larger numbers in the give ratios of) the perimeters as also (of) the areas (relating to the two required longish quadrilateral figures,) are divided by the smaller (numbers) corresponding to them. (The resulting quotients) are multiplied (between themselves) and (then) squared. (This same quantity,)

131${\displaystyle {\tfrac {1}{2}}}$ to 133. If x and y represent the two adjacent sides of the first rectangle, and a and b the two adjacent sides of the second rectangle, the conditions mentioned in the three kinds of problems proposed to be solved by this rule may be represented thus:-

{\displaystyle {\begin{aligned}&(1)\quad x+y=a+b:\\&\qquad \quad xy=2ab\\&(2)\quad 2(x+y)=a+b:\\&\qquad \quad xy=ab\\&(3)\quad 2(x+y)=a+b:\\&\qquad \quad xy=2ab\end{aligned}}}

The solution given in the rule seems to be correct only for the particular cases given in the problems in stanzas 134 to 136