$M$ is the mid-point of $B C$ of $△ A B C$ point $L$ division $A B$ is the ratio $2 : 1$ . That is $A L : L B = 2 : 1$ then area $(△ A M L) = \frac{2}{3}$ area of $(△ A M B)$

True

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False

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Solution

The correct option is A True
As M is the midpoint of AC
$∴ a r (Δ A M B) = \frac{1}{2} a r (Δ A B C) - (1)$

Also
$a r (Δ A M L) = \frac{1}{2} a r (Δ A L C) ‘ ‘ ‘ - (2)$

Since point l divides the line AB in ratio 2:1

$⟹ \frac{a r (Δ A L C)}{a r (Δ A B C)} = \frac{2}{3}$

eqs. (2) $⟹ a r (Δ A M L) = \frac{1}{3} a r (Δ A B C) - (3)$
From eqs.(1) and (3) ,
we get

$a r (Δ A M L) = \frac{2}{3} a r (Δ A M B)$

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M is the mid-point of BC of △ABC point L division AB is the ratio 2:1. That is AL:LB=2:1 then area (△AML)=23 area of (△AMB)

The correct option is A TrueAs M is the midpoint of AC∴ar(ΔAMB)=12ar(ΔABC) −(1)Also ar(ΔAML)=12ar(ΔALC) ‘‘‘−(2)Since point l divides the line AB in ratio 2:1⟹ar(ΔALC)ar(ΔABC)=23eqs. (2) ⟹ar(ΔAML)=13ar(ΔABC) −(3)From eqs.(1) and (3) ,we get ar(ΔAML)=23ar(ΔAMB)

$M$ is the mid-point of $B C$ of $△ A B C$ point $L$ division $A B$ is the ratio $2 : 1$ . That is $A L : L B = 2 : 1$ then area $(△ A M L) = \frac{2}{3}$ area of $(△ A M B)$