Question

In fig 2 MN || BC and AM:MB=1:2, then
ar(∆AMN)
________=___________.
ar(∆ABC)​

Answer 1

The required ratio is 1:9.

Step-by-step explanation:

Since we have given that

AM:MB = 1:2

So, AB = 1+2 = 3

Since MN || BC

So, Using "Area similarity theorem", we get that

$\dfrac{ar(\Delta AMN)}{ar(\Delta ABC)}=\dfrac{AM^2}{AB^2}=\dfrac{1}{3^2}=\dfrac{1}{9}$

Hence, the required ratio is 1:9.

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12. In fig. 2, MN || BC and AM: MB = 1:2, then

ar(A AMN)

ar(A ABC)

Fig.-2​

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Answer 2

Answer:

We have

AM : MB = 1 : 2

Adding 1 to both sides, we get

Now, In △AMN and △ABC

∠AMN = ∠ABC (Corresponding angles in MN∥BC)

∠ANM = ∠ACB (Corresponding angles in MN∥BC)

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.