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In the given figure MN|| BC and AM: MB= 1: 2
find ` (area(ΔAMN))/(area(ΔABC))`
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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
∠ЁЭР┤ЁЭСАЁЭСБ = ∠ЁЭР┤ЁЭР╡ЁЭР╢ (ЁЭР╢ЁЭСЬЁЭСЯЁЭСЯЁЭСТЁЭСаЁЭСЭЁЭСЬЁЭСЫЁЭССЁЭСЦЁЭСЫЁЭСФ ЁЭСОЁЭСЫЁЭСФЁЭСЩЁЭСТЁЭСа ЁЭСЦЁЭСЫ ЁЭСАЁЭСБ тИе ЁЭР╡ЁЭР╢)
∠ЁЭР┤ЁЭСБЁЭСА = ∠ЁЭР┤ЁЭР╢ЁЭР╡ (ЁЭР╢ЁЭСЬЁЭСЯЁЭСЯЁЭСТЁЭСаЁЭСЭЁЭСЬЁЭСЫЁЭССЁЭСЦЁЭСЫЁЭСФ ЁЭСОЁЭСЫЁЭСФЁЭСЩЁЭСТЁЭСа ЁЭСЦЁЭСЫ ЁЭСАЁЭСБ тИе ЁЭР╡ЁЭР╢)
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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