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(B) Given that $DE \parallel BC$ in $	riangle ABC$,by the Basic Proportionality Theorem (Thales Theorem),$	riangle ADE \sim 	riangle ABC$.Therefore

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$3$

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(B) Given that $DE \parallel BC$ in $	riangle ABC$,by the Basic Proportionality Theorem (Thales Theorem),$	riangle ADE \sim 	riangle ABC$.Therefore,the ratio of corresponding sides is equal: $\frac{AD}{AB} = \frac{DE}{BC}$.Given $AD = 2 \, cm$ and $BD = 3 \, cm$,we have $AB = AD + BD = 2 + 3 = 5 \, cm$.Substituting the values into the ratio: $\frac{2}{5} = \frac{DE}{7.5}$.Solving for $DE$: $DE = \frac{2}{5} 	imes 7.5$.$DE = 2 	imes 1.5 = 3 \, cm$.

$D$ and $E$ are points on the sides $AB$ and $AC$ of a triangle $ABC$ respectively,such that $AD = 2 \, cm$,$BD = 3 \, cm$,$BC = 7.5 \, cm$,and $DE \parallel BC$. Find the length of $DE$ (in $cm$).

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