Question

2) 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

Proved Using Pythagoras Theorem

Explanation:

• Let ABC be any Right Triangle such that AB is perpendicular, $AB = m^2 - n^2$

• BC is Base of Triangle = 2 mn

• Let CA be Hypotenuse = $m^2 + n^2$

If ABC is Right triangle, then as per Pythagoras Theorem,

$Hypotenuse^2 = Perpendicular^2 + Base^2$

Let's see if $AB^2 + BC^2$ equals $CA^2$

So, $AB^2 + BC^2$

= $(m^2 - n^2)^2 + (2mn)^2$

= $m^4 -2m^2n^2 + n^4 + 4m^2n^2$

= $m^4 + n^4 - 2m^2n^2 + 4m^2n^2$

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

Answer 2

use Pythagoras theorem considering the largest value as hypotenuse (can also check by hit-and-trial)

Explanation:

Given:

Three sides of a triangle,

$m^2-n^2$

$2mn$

$m^2+n^2$

Now, using Pythagoras theorem:

$\rm perpendicular^2+base^2=hypotenuse^2$

• As we know that the hypotenuse is the largest side of a right angled triangle.

$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$

$m^4+n^4-2m^2n^2+4m^2n^2=m^4+n^4+2m^2n^2$

$m^4+n^4+2m^2n^2=m^4+n^4+2m^2n^2$    which is true.

Hence above values are the sides of right angled triangle with $(m^2+n^2)$ as the hypotenuse.

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TOPIC: Pythagoras theorem

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