Question

Determine whether the triangle having sides $(a-1)cm$, $2\sqrt{a}cm$ and $(a+1)cm$ is a right angled triangle or not.

Answer 1

Let, in $△ABC$,

$AB=(a-1)cm$

$BC=2\sqrt{a}cm$

and, $AC=(a+1)cm$

$A{B}^2=(a-1{)}^{2}c{m}^2$

$=a^2-2a+1c{m}^2$

$B{C}^2=(2\sqrt{a}{)}^{2}c{m}^2$

$=4ac{m}^2$

$A{C}^2=(a+1{)}^{2}c{m}^2$

$=a^2+2a+1c{m}^2$

Now,

$A{B}^2+B{C}^2=a^2-2a+1+4ac{m}^2$

$=a^2+2a+1c{m}^2$

$=A{C}^2$

So, $A{B}^2+B{C}^2=A{C}^2$

Hence, $ABC$ is a right angled triangle.