In $ \Delta \mathrm{XYZ}, \quad \angle \mathrm{Y}=90^{\circ} $. If $ \mathrm{XY}=a^{2}-\mathrm{b}^{2} $ and $ \mathrm{YZ}=2 a b $, find $ \mathrm{XZ} .(a>b>0) $.

Given:

In \( \Delta \mathrm{XYZ}, \quad \angle \mathrm{Y}=90^{\circ} \).

\( \mathrm{XY}=a^{2}-\mathrm{b}^{2} \) and \( \mathrm{YZ}=2 a b \).

To do:

We have to find XZ.

Solution:  

In triangle XYZ, by Pythagoras theorem,

$XZ^2=XY^2+YZ^2$

$=(a^2-b^2)^2+(2ab)^2$

$=a^4+b^4-2a^2b^2+4a^2b^2$

$=a^4+b^4+2a^2b^2$

$=(a^2+b^2)^2$

$XZ=a^2+b^2$

Therefore, $XZ=a^2+b^2$.