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

(D) Given that in $\Delta ABC$,$\overline{MN} \parallel \overline{BC}$.According to the Basic Proportionality Theorem (Thales Theorem),if a line is dr

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Given that in $\Delta ABC$,$\overline{MN} \parallel \overline{BC}$.According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.Therefore,$\frac{AM}{AB} = \frac{AN}{AC}$.Given $\frac{AM}{AB} = \frac{2}{3}$ and $AC = 15$.Substituting the values,we get $\frac{2}{3} = \frac{AN}{15}$.$AN = \frac{2 	imes 15}{3} = 10$.Since $A-N-C$,we have $AC = AN + NC$.$15 = 10 + NC$.$NC = 15 - 10 = 5$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM : AB = 2 : 3$ and $AC = 15$,then $NC = \ldots$

Similar Questions

In a square,the length of a diagonal is $8$. Find the length of the sides of the square. (in $\sqrt{2}$)

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YP}$ is a median. If $YP = 6$,then $XZ = \ldots$

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 6$,$MB = 12$ and $AN = 8$,find $AC$.

Using Theorem $6.1,$ prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

$BL$ and $CM$ are medians of a triangle $ABC$ right-angled at $A$. Prove that $4(BL^2 + CM^2) = 5BC^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module