Expression of the sides of Right-angled Triangles, fyc. 69 



Let m, and n, be any rational numbers whatever ; then m 2 +n% 

 m* — n 2 , and 2mn y will be sides of a right-angled triangle. For, 



(m 2 +n' 2 ) 2 =(m 2 -n 2 ) 2 + (2mn) 2 . 



If m remains constant for any set of examples, and n be succes- 

 sively increased by unity; the hypothenuse will increase by the fol- 

 lowing series, 3, 5, 7, &c. the base will decrease by the same series, 

 and the perpendicular will increase constantly by 2m. 



In the following examples m remains constantly equal to 10, and n 

 increases by successive units. The required triangular sides may 

 be extended ad infinitum by the process above explained, without 

 changing the value of m. If m be changed, other series equally un- 

 limited will arise. 



m n hyp. base. perp. 



10 1 - 101 - 99 20 



2 - 104 96 40 • 



3 109 91 - 60 



4 116 - 84 80 



5 - 125 - 75 - • . . 100 



6 - 136 64 - 120 



7 149 - 51 - 140 



8 164 36 - 160 



9 181 19 - 180 



10 - 200 - - 200 



11 - 221 - 21 - 220 



12 - 244 - 44 - 240 



13 - 269 - 69 - 260 



14 - 296 96 - 280 





15 - 325 - 125 - 300 



16 - 356 - 156 - . 320 



17 - 389* - 189 340 



18 - 424 - 224 - 360 



19 - 461 - 261 - 380 



20 - 500 - 300 - 400 &c. 

 30 - 1000 - 800 600 



I 



40 - 1700' - 1500 - 800 



50 - 2600 - 2400 - 1000 &c. 

 From this set of examples, it will be seen, 



h That when- in =n, then the hypothenuse and perpendicular are 

 equal, and the base vanishes. 



2. That as n increases above m, the bases are negative, that is, 

 they lie on the contrary side of the perpendicular. 



