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In the given figure MN|| BC and AM: MB= 1: 2. Find \frac{\text { Area }(\triangle A M N)}{\text { Area }(\triangle A B C)}
CBSE | Class 10 | Exercise 4E | Maths | RS Aggarwal | Triangles
In the given figure MN|| BC and AM: MB= 1: 2. Find \frac{\text { Area }(\triangle A M N)}{\text { Area }(\triangle A B C)}

 

 

 

Answer:

Given,

AM : MB = 1 : 2

\frac{M B}{A M}=\frac{2}{1}

Adding 1 to both sides,

\frac{M B}{A M}+1=\frac{2}{1}+1

\frac{M_{B}+A M}{A M}=\frac{2+1}{1}

\frac{A B}{A M}=\frac{3}{1}

In ∆AMN and ∆ABC

∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ???????????????????????? ???????? ???????? ∥ ????????)

∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ???????????????????????? ???????? ???????? ∥ ????????)

By AA similarity criterion,

∆AMN ~ ∆ ABC

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.

\begin{aligned} \frac{\text { area }(\triangle A M N)}{\text { area }(\triangle A B C)} &=\left(\frac{A M}{A B}\right)^{2} \\ & \Rightarrow\left(\frac{1}{3}\right)^{2} \\ & \Rightarrow \frac{1}{9} \end{aligned}

 

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