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

(C) In $\Delta AMN$ and $\Delta ABC$,since $\overline{MN} \parallel \overline{BC}$,the triangles are similar by the $AA$ similarity criterion.Therefor

Vedclass · Accepted Answer

$3:4$

Vedclass · Answer

(C) In $\Delta AMN$ and $\Delta ABC$,since $\overline{MN} \parallel \overline{BC}$,the triangles are similar by the $AA$ similarity criterion.Therefore,$\frac{AM}{AB} = \frac{MN}{BC}$.Given $MN = 3$ and $BC = 7$,we have $\frac{AM}{AB} = \frac{3}{7}$.Since $AB = AM + MB$,we can write $\frac{AM}{AM + MB} = \frac{3}{7}$.By cross-multiplication,$7 AM = 3 AM + 3 MB$.$4 AM = 3 MB$.Therefore,$\frac{AM}{MB} = \frac{3}{4}$.Thus,$M$ divides $\overline{AB}$ from $A$ in the ratio $3:4$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $MN = 3$ and $BC = 7$,then $M$ divides $\overline{AB}$ from $A$ in the ratio.......

Similar Questions

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2.5$,$MB = 3$ and $AN = 3.75$,find $AC$.

In Figure $(i)$ and $(ii),$ $DE || BC.$ Find $EC$ in $(i)$ and $AD$ in $(ii).$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$BM$ is the geometric mean of..............

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = x$,$MB = x-2$,$AN = x+2$ and $NC = x-1$,find the value of $x$.

In $\Delta ABC$,$m \angle A = 90^\circ$. If $b = 12$ and $c = 35$,then $a = \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module