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286
GANITASI KASANGRAHA.
An example in illustration thereof. 1694. In the case of a certain triangular figure with unequa sides, it has been pointed out that 2 constitutes the accurate measure of its area and 3 is the optionally choson quantity What is the value of the base as well as of the sides of that triangle)?
Again, another rule for arriving, after knowing the exact numerical measure of a (given) area, at a triangular figure with unequal sides having that same (acourately measured) area (as ite own) :
1601-1615. The square root of the measure of the given area as multiplied by eight and as increased by the square of an option. ally chosen number is obtained. This and the optionally chosen number are divided by each other. The larger (of these quotients) is diminished by half of the smaller (quotient). The remainder (thus obtained) and (this) half of the smaller (quotient) are respeotively multiplied by the above-noted square root and the optionally chosen number. On carrying out, in relation to the products (thus obtained), the process of sankramana, the values of the base and of one of the sides are arrived at. Half of the optionally chosen number happens to be the measure of the other side in a triangular figure with unequal sides.
An example in illustration thereof. 1621. In the case of a triangle with unequal sides, the aoourate measure of the area is 2, and the optionally chosen quantity is 3. O friend who know the secret of caloulation, give out the measure of the base as well as of the sides.
The rule for arriving, after knowing the accurate measure of a (given) area, at a regularly circular figure having that acourately measured area (as its own) :
1634. The accurate measure of the area is multiplied by four and is divided by the square root of ten. On getting at the square
10.
1684. The rule in this stans is derived from the formule, aros = Where d in the diameter of the oirole.
