In Delta XYZ,m angle Y = 90^{circ} and XY : X | vedclass

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(D) Given that $\Delta XYZ$ is a right-angled triangle with $m \angle Y = 90^{\circ}$.According to the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2$.Let $X

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$20$

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(D) Given that $\Delta XYZ$ is a right-angled triangle with $m \angle Y = 90^{\circ}$.According to the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2$.Let $XY = 15k$ and $XZ = 17k$ for some constant $k > 0$.Substituting the values into the theorem: $(15k)^2 + 4^2 = (17k)^2$.$225k^2 + 16 = 289k^2$.$16 = 289k^2 - 225k^2$.$16 = 64k^2$.$k^2 = \frac{16}{64} = \frac{1}{4}$.Therefore,$k = \frac{1}{2} = 0.5$.Now,calculate the side lengths:$XY = 15 	imes 0.5 = 7.5$.$XZ = 17 	imes 0.5 = 8.5$.$YZ = 4$.The perimeter of $\Delta XYZ = XY + YZ + XZ = 7.5 + 4 + 8.5 = 20$.

In $\Delta XYZ$,$m \angle Y = 90^{\circ}$ and $XY : XZ = 15 : 17$. If $YZ = 4$,find the perimeter of $\Delta XYZ$.

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