पृष्ठम्:Ganita Sara Sangraha - Sanskrit.djvu/४२५

CHAPTER VII-MEASUREMENT OF AREAS. 227 137. The squares (of the ratio-values) of the perimeters (of the required isosceles triangles) are multiplied by (the ratio-values of) the areas (of those triangles) in alternation. (Of the two products so obtained), (the larger one is) divided by the smaller; and (the resulting quotient) is multiplied by six and (is also separately multiplied) by two. The smaller (of the two products so obtained) is diminished by one. The larger product and the diminished smaller product constitute the two bijus (in relation to the longish quardrilateral figure) from which one (of the re- quired triangles) is to be obtained. The difference between these (two bijas above noted) and twice the smaller one (of those bijas) constitute the tijas (in relation to the longish quadrilateral figure) from which the other (required triangle) is to be obtained. (From the two longish quadrilateral figures formed with the aid of their respective bijas), the sides and the other things (relating to the required triangles) are to be arrived at as (explained) before. 137. When a b is the ratio of the perimeters of the two isosceles triangles, and d the ratio of their areas, then, according to the rule, bec 6 2b² c a² d C 1 ad and - 2 are the two sets of bigas, with the help of which +1 and 46² c a d the values of the various requned elements of the two isosceles tiangles may be arrived at. The measures of the sides and the altitudes, calculated from these bijas according to stanza 108 in this chapter, when multiplied respectively by a and b, (the quantities occurring in the ratio of the perimeters), give the required measures of the sides and the altitudes of the two isosceles triangles They are as follow- I and 482 c a² d Equal side a x II Base a x 2 x 2 x Altitude ax Equal sideb x 6² 2 c { (*№* *^)² + (²*³** −−¹ )*} a² d ad Altitudeb x 600×(122-1) X (2 6² c a d 16 b Baseb x 2 x 2 x { a d a d 4 b² c a d 4 6² c a d +1 + 4 c ¹ ) ² + ( ² ² 2 ² - ² ) ² } d -2 ¹) X (4 b² c a d +1 + ¹)² - ( ²4 ³² 2² - ² ) ² } d Now it may be easily proved from these values that the ratio of the perime- ters is a b, and that of the areas is c. d, as taken for granted at the beginning.