Question

Suppose m and n are any two numbers. If $m^2-n^2,2mn$ and $m^2+n^2$ are the three sides of a triangle, then show that it is a right angled triangle and hence write any two pairs of Pythagorean triplet

Answer 1

1.$3^2+4^2=5^2$

2.$5^2+{12}^2={13}^2$

Now to show,

$m^2-n^2,2mn$ and $m^2+n^2$ are the sidesof right angled triangle.

Sides of right angled triangle form pythagorean triplet.

$(m^2-n^2{)}^2+(2mn{)}^2=m^4+n^4-2m^2{n}^2+4m^2{n}^2$

$=m^4+n^4+2m^2{n}^2$

$=(m^2+n^2{)}^2$

$\therefore$ It form right angles triangles,

Examples of triplet:

1.$3^2+4^2=5^2$

2.$5^2+{12}^2={13}^2$