Ganitasara Sangraha

SR No.: 020800
Original Sutra Author: N/A
Author: Mahaviracharya, M Rangacharya
Publisher: Government of Madras
Publication Year: 1912
Total Pages: 523
Language: Hindi, Sanskrit
Classification: Book_Devnagari
File Size: 22 MB

Page 424

CHAPTER VII-MEASUREMENT OF AREAS.
II
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 bijas (in relation to the longish quardrilateral figure) from which one (of the required triangles) is to be obtained. The difference between these (two bajas above noted) and twice the smaller one (of those biījas) 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.
+ 1 and
137. When a b is the ratio of the perimeters of the two isosceles triangles, and 662 c c: d the ratio of their areas, then, according to the rule, 26 c -1 and a2 d a2 d 462 c 2 are the two sets of bijas, with the help of which a2 d
and
462 c a2 d
the vales of the various required elements of the two isosceles triangles may be arrived at. The measures of the sides and the altitudes, calculated from these bījas 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
Equal side a x
Altitude α x
Base= ax 2 x 2 x
{(663)2 + (2,69 € 1)°}
c a d
-1
6b2 d 8 62 c a 2 d b2 c 2
{(25) -
d
Equal sideb x
Altitudeb x
Base b x 2 x 2 x
X
4 c a d
62
(2022 - 1)
d
2 4 b2 c
c a2 d
{ (12 + 1 ) + ( 122 )"}
-2
d
c
(2-1)}
a d
(1380 + 1) x (+382)
{(+)-(-)"}
+1
c a2 d
2
227
Now it may be easily proved from these values that the ratio of the perimeters is a b, and that of the areas is c: d, as taken for granted at the beginning.