In a right angled triangle, the area of the square on hypotenuse is equal to the sum of the areas of squares on the remaining sides.

[color=#ff0000][b][size=85][size=150]If a, b, c are natural numbers and a>b, then [(a[sup]2[/sup]+b[sup]2[/sup]), (a[sup]2[/sup]-b[sup]2[/sup]), (2ab)] is Pythagorean triplet.[/size][/size][/b] [br][/color](a[sup]2[/sup]+b[sup]2[/sup])[sup]2 [/sup]= (a[sup]2[/sup])[sup]2[/sup] + 2a[sup]2[/sup]b[sup]2[/sup] + (b[sup]2[/sup])[sup]2[/sup] [br](a[sup]2[/sup]+b[sup]2[/sup])[sup]2 [/sup]= a[sup]4[/sup]+ 2a[sup]2[/sup]b[sup]2[/sup] + b[sup]4[/sup]   ............................(I)[br](a[sup]2 [/sup]- b[sup]2[/sup])[sup]2[/sup]= (a[sup]2[/sup])[sup]2 [/sup] - 2a[sup]2[/sup]b[sup]2[/sup] + (b[sup]2[/sup])[sup]2[/sup]  [br](a[sup]2[/sup]- b[sup]2[/sup])[sup]2 [/sup]= a[sup]4 [/sup]- 2a[sup]2[/sup]b[sup]2[/sup] + b[sup]4[/sup]  ............................(II)[br]Subtracting (II) from (I), we get[br](a[sup]2[/sup]+b[sup]2[/sup])[sup]2[/sup] - (a[sup]2[/sup]- b[sup]2[/sup])[sup]2[/sup] = 4a[sup]2[/sup]b[sup]2[br][/sup](a[sup]2[/sup]+b[sup]2[/sup])[sup]2[/sup] - (a[sup]2[/sup]- b[sup]2[/sup])[sup]2[/sup] = (2ab)[sup]2[br][/sup][size=100][size=150](a[sup]2[/sup]+b[sup]2[/sup])[sup]2[/sup] = (a[sup]2[/sup]- b[sup]2[/sup])[sup]2[/sup]+ (2ab)[sup]2[br][/sup]Therefore [(a[sup]2[/sup]+b[sup]2[/sup]) [sup][/sup], (a[sup]2[/sup]- b[sup]2[/sup])[sup] [/sup]and (2ab)] is Pythagorean Triplet. [br]Taking different values of a & b we can find more and more triplets.[sup][/sup][/size][/size][br][sup][/sup]

[b]Property of 30[sup]0[/sup]-60[sup]0[/sup]-90[sup]0 [/sup]Triangle[br][/b]If acute angles of a right angled triangle are 30[sup]0[/sup],60[sup]0[/sup] and 90[sup]0[/sup], then the side opposite 30[sup]0 [/sup]angle is half of the hypotenuse and the side opposite to 60[sup]0[/sup] angle is[math]\frac{\sqrt{3}}{2}[/math] times the hypotenuse.

[b]Property of 45[sup]0[/sup]-45[sup]0[/sup]-9[sup]0[/sup]triangle[br][/b]If the acute angles of right angled triangles are 45[sup]0[/sup] each, then each of the perpendicular side is[math]\frac{1}{\sqrt{2}}[/math] times the hypotenuse.