Question

Suppose m and n are any two numbers. If m2-n2,2 mn and m² + n° are the
three sides of a triangle, then show that it is a right angled triangle.​

Answer 1

Hence, it is proved. The three sides of  ($m^{2}$ - $n^{2}$), 2mn and ($m^{2}$ + $n^{2}$) "a right angle trianle".

Step-by-step explanation:

Given,

The three sides of tringle are ($m^{2}$ - $n^{2}$), 2mn and ($m^{2}$ + $n^{2}$).

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

$(2mn^{2} )^{2}$

In right angle triange,

Pythagotas Theorem,

∴ $h^{2}$ = $p^{2}$ + $b^{2}$

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

$(2mn^{2} )^{2}$, it is proved.

Hence, it is proved. The three sides of  ($m^{2}$ - $n^{2}$), 2mn and ($m^{2}$ + $n^{2}$) a right angle trianle.