Question

pls ans above question ​

Answer 1

Answer:

I hope it you got your answer

Answer 2

$\large\underline{\sf{Solution-}}$

In order to show that, 3 sides formed the sides of right angle triangle, its enough to show that square of longest side is equals to sum of the squares of remaining two sides.

So, we have given three sides as

$\red{\rm :\longmapsto\: {a}^{2} - {b}^{2}, \: 2ab, \: {a}^{2} + {b}^{2} }$

Let assume that

$\red{\rm :\longmapsto\:x = {a}^{2} - {b}^{2}}$

$\red{\rm :\longmapsto\:y = 2ab}$

$\red{\rm :\longmapsto\:z = {a}^{2} + {b}^{2}}$

Now, Consider

$\red{\rm :\longmapsto\: {x}^{2} + {y}^{2}}$

$\rm = \: {\bigg[ {a}^{2} - {b}^{2} \bigg]}^{2} + {(2ab)}^{2}$

$\rm = \: {( {a}^{2} )}^{2} + {( {b}^{2} )}^{2} - 2 {a}^{2} {b}^{2} + 4 {a}^{2} {b}^{2}$

$\rm = \: {( {a}^{2} )}^{2} + {( {b}^{2} )}^{2} + 2 {a}^{2} {b}^{2}$

$\rm = \: {\bigg[ {a}^{2} + {b}^{2} \bigg]}^{2}$

$\rm = \: {z}^{2}$

$\rm \implies\:\boxed{ \tt{ \: {x}^{2} + {y}^{2} = {z}^{2} \: }}$

Hence, Three sides of a triangle

$\red{\rm :\longmapsto\: {a}^{2} - {b}^{2}, \: 2ab, \: {a}^{2} + {b}^{2} \: are \: measure \: of \: right \: angle \: \triangle }$

Hence, Verified