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In the Given Figure Mn|| Bc and Am: Mb= 1: 2 - Mathematics

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Question

In the given figure MN|| BC and AM: MB= 1: 2 

 

find    ` (area(ΔAMN))/(area(ΔABC))` 

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Solution

We have
AM : MB = 1 : 2 

⇒ AM:MB=1:2 

⇒ `(MB)/(AM)=2/1` 

Adding 1 to both sides, we get 

⇒`( MB)/(AM)+1=2/1+1` 

⇒`(MB+AM)/(AM)=(2+1)/1` 

⇒ `(AB)/(AM)=3/1` 

Now, In ΔAMN and ΔABC
∠๐ด๐‘€๐‘ = ∠๐ด๐ต๐ถ (๐ถ๐‘œ๐‘Ÿ๐‘Ÿ๐‘’๐‘ ๐‘๐‘œ๐‘›๐‘‘๐‘–๐‘›๐‘” ๐‘Ž๐‘›๐‘”๐‘™๐‘’๐‘  ๐‘–๐‘› ๐‘€๐‘ โˆฅ ๐ต๐ถ)
∠๐ด๐‘๐‘€ = ∠๐ด๐ถ๐ต (๐ถ๐‘œ๐‘Ÿ๐‘Ÿ๐‘’๐‘ ๐‘๐‘œ๐‘›๐‘‘๐‘–๐‘›๐‘” ๐‘Ž๐‘›๐‘”๐‘™๐‘’๐‘  ๐‘–๐‘› ๐‘€๐‘ โˆฅ ๐ต๐ถ)
By AA similarity criterion, ΔAMN ~ Δ ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.  

∴`( area (Δ AMN))/(area(ΔABC))=((AM)/(AB))^2=(1/3)^2=1/9` 

 

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Chapter 4: Triangles - Exercises 5

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 4 Triangles
Exercises 5 | Q 23
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