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Shri Mahavir Jain Aradhana Kendra
www.kobatirth.org
Acharya Shri Kailassagarsuri Gyanmandir
OHAPTER VIIMEASUREMENT OF AREAS.
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subtracted. Then, on carrying out the process of sankramana 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.
An example in illustration thereof. 1304. 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 :
1314--133. (The larger numbers in the given 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,)
1311 to 1.33. If a and y represent the two adjacent sides of the first rectangle, and a and t 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:--
(1) x+y=a+b:
y = 2ab. (2) 2(x + y) = a +b :
æy = ab. (3) 2 (æ+ y) = a +b:
y = 2ab. The solution given in the role seems to be correot only for the particular cases given in the problems in stanzas 134 to 136.
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