In Delta ABC,A-M-B,A-N-C and overline{MN} par | vedclass

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(B) Given that $\overline{MN} \parallel \overline{BC}$ in $\Delta ABC$.By the Basic Proportionality Theorem or properties of similar triangles,$\Delta

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) Given that $\overline{MN} \parallel \overline{BC}$ in $\Delta ABC$.By the Basic Proportionality Theorem or properties of similar triangles,$\Delta AMN \sim \Delta ABC$.Therefore,the ratio of corresponding sides is equal:$\frac{AM}{AB} = \frac{MN}{BC}$.We are given $AM = 2$ and $MB = 5$.Thus,$AB = AM + MB = 2 + 5 = 7$.Substituting the values into the ratio:$\frac{2}{7} = \frac{4}{BC}$.Solving for $BC$:$2 	imes BC = 4 	imes 7$$2 	imes BC = 28$$BC = \frac{28}{2} = 14$.Thus,the length of $BC$ is $14$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2$,$MB = 5$ and $MN = 4$,find $BC$.

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