In Delta XYZ,mangle Y = 90^{circ}. If XY = a^ | vedclass

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(A) In a right-angled triangle $\Delta XYZ$,where $m\angle Y = 90^{\circ}$,the side $XZ$ is the hypotenuse.According to the Pythagorean theorem,$(XZ)^

Vedclass · Accepted Answer

$a^{2} + b^{2}$

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(A) In a right-angled triangle $\Delta XYZ$,where $m\angle Y = 90^{\circ}$,the side $XZ$ is the hypotenuse.According to the Pythagorean theorem,$(XZ)^{2} = (XY)^{2} + (YZ)^{2}$.Substituting the given values: $(XZ)^{2} = (a^{2} - b^{2})^{2} + (2ab)^{2}$.Expanding the terms: $(XZ)^{2} = (a^{4} - 2a^{2}b^{2} + b^{4}) + 4a^{2}b^{2}$.Simplifying the expression: $(XZ)^{2} = a^{4} + 2a^{2}b^{2} + b^{4}$.This is a perfect square: $(XZ)^{2} = (a^{2} + b^{2})^{2}$.Taking the square root of both sides,we get $XZ = a^{2} + b^{2}$.

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

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