If Delta XYZ has XY = 7 , cm and YZ = 10 , cm | vedclass

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(B) In any triangle,the sum of the lengths of any two sides must be greater than the length of the third side (Triangle Inequality Theorem).Let the si

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$20$

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(B) In any triangle,the sum of the lengths of any two sides must be greater than the length of the third side (Triangle Inequality Theorem).Let the sides of $\Delta XYZ$ be $XY = 7 \, cm$,$YZ = 10 \, cm$,and $XZ = z \, cm$.According to the triangle inequality theorem:$XY + YZ > XZ \implies 7 + 10 > z \implies z $XY + XZ > YZ \implies 7 + z > 10 \implies z > 3$$YZ + XZ > XY \implies 10 + z > 7 \implies z > -3$ (which is always true for a length).Thus,the third side $z$ must satisfy $3 The perimeter $P$ of the triangle is given by $P = XY + YZ + XZ = 7 + 10 + z = 17 + z$.Since $z > 3$,we have $P = 17 + z > 17 + 3 = 20$.Therefore,the perimeter of $\Delta XYZ$ is greater than $20 \, cm$.

If $\Delta XYZ$ has $XY = 7 \, cm$ and $YZ = 10 \, cm$,the perimeter of $\Delta XYZ$ is greater than how many centimeters?

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