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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $BC$.

Similar Questions

In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $AP = 8x-7$,$PB = 5x-3$,$AQ = 4x-3$ and $QC = 3x-1$,find the value of $x$.

In $\Delta ABC$,$AB = 17$,$BC = 15$,and $AC = 8$. Then,the length of the median on the longest side of the triangle is $\ldots$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QS}$ is an altitude. If $PS - SR = 10$ and $PQ^2 - QR^2 = 260$,then $PR = \dots$

$\square XYZW$ is a rectangle. If $XY + YZ = 17$ and $XZ + YW = 26$,find $XY$ and $YZ$ (given $XY > YZ$).

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $\frac{AB}{5} = \frac{AC}{3}$ and $BD = 4.5$,find $BC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module