In Delta XYZ,mangle X = 90^{circ},XY = 8 and | vedclass

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(D) Given that $\Delta XYZ$ is a right-angled triangle with $m\angle X = 90^{\circ}$.Here,$YZ$ is the hypotenuse and $XY$ is one of the legs.Using the

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$60$

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(D) Given that $\Delta XYZ$ is a right-angled triangle with $m\angle X = 90^{\circ}$.Here,$YZ$ is the hypotenuse and $XY$ is one of the legs.Using the Pythagorean theorem: $XY^2 + XZ^2 = YZ^2$.Substituting the given values: $8^2 + XZ^2 = 17^2$.$64 + XZ^2 = 289$.$XZ^2 = 289 - 64 = 225$.$XZ = \sqrt{225} = 15$.The area of a right-angled triangle is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.$	ext{Area} = \frac{1}{2} 	imes XY 	imes XZ$.$	ext{Area} = \frac{1}{2} 	imes 8 	imes 15 = 4 	imes 15 = 60$.

In $\Delta XYZ$,$m\angle X = 90^{\circ}$,$XY = 8$ and $YZ = 17$. Then,the area of $\Delta XYZ$ is.............

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