Rational Triangles: Difference between revisions
Line 28: | Line 28: | ||
<math>{\displaystyle =a^2(\sqrt{n})^2+a^2\left (\frac{n-1}{2} \right )^2= a^2\left ( \frac{n+1}{2} \right )^2}</math> | <math>{\displaystyle =a^2(\sqrt{n})^2+a^2\left (\frac{n-1}{2} \right )^2= a^2\left ( \frac{n+1}{2} \right )^2}</math> | ||
put m<sup>2</sup> for n in order to make the sides of the right angled triangle without the radical, we have | put ''m<sup>2</sup>'' for ''n'' in order to make the sides of the right angled triangle without the radical, we have | ||
<math>{\displaystyle =m^2a^2+a^2\left (\frac{m^2-1}{2} \right )^2= a^2\left ( \frac{m^2+1}{2} \right )^2}</math> | <math>{\displaystyle =m^2a^2+a^2\left (\frac{m^2-1}{2} \right )^2= a^2\left ( \frac{m^2+1}{2} \right )^2}</math> which gives the rational solution of (1). |
Revision as of 22:16, 29 March 2022
A rational triangle can be defined as one having all sides with rational length.
Rational Right Triangles - Early Solutions
In Śulba solution for the equation -------(1) is available. Baudhāyana (c. 800 B.C.), Āpastamba and Kātyāyana (c. 500 B.C.) gave a method for the transformation of a rectangle into a square, which is the equivalent of the algebraical identity.
where m, n are any two arbitrary numbers. Thus we get
substituting p2,q2 for m, n respectively in order to eliminate the irrational quantities, we get
which gives the rational solution of (1).
Kātyāyana gives a very simple method for finding a square equal to the sum of a number of other squares of the same size which gives us with another solution of the rational right triangle.
Kātyāyana says: "As many squares (of equal size) as you wish to combine into one, the transverse line will be (equal to) one less than that; twice a side will be (equal to) one more than that; (thus) form (an isosceles) triangle. Its arrow (i.e., altitude) will do that."
<Triangle picture to be added here>
For combining n squares of sides a each we form the isosceles triangle ABC such that and
Then which gives the formula
put m2 for n in order to make the sides of the right angled triangle without the radical, we have
which gives the rational solution of (1).