In a parallelogram PQRS, L and M are the mid-point of QR and B RS respectively Prove that: $a r (Δ P L M) = \frac{3}{8} a r P Q R S)$

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Solution

$L M = \frac{1}{2} S Q$ [ $∵$ L and M are the mid-points]
Area of $Δ R L M = \frac{1}{4} (a r e a o f Δ S Q R)$
[ $∵$ The base LM is $\frac{1}{2} S Q$ and altitude of $Δ R L M$ is $\frac{1}{2}$ of altitude of $Δ S R Q$ ]
In $Δ P R Q$ , PL in the median
$\Rightarrow a r (Δ P R L) = a r (Δ P L Q) = \frac{1}{4} a r (P Q R S)$
In $Δ P S R$ , PM is the median
$\Rightarrow a r (Δ P S M) = a r (Δ P M R) = \frac{1}{4} a r (P Q R S)$
$A r (P M L) = a r (Δ P R L) + a r (Δ P M R) - a r (Δ M R L)$
$= \frac{1}{4} a r (P Q R S) + \frac{1}{4} a r (P Q R S) - \frac{1}{4} (a r Δ S Q R)$
$= \frac{1}{4} a r (P Q R S) + \frac{1}{4} a r (P Q R S) - \frac{1}{4} (a r Δ P Q R S)$
$= (\frac{1}{4} + \frac{1}{4} - \frac{1}{8}) a r (P Q R S)$
$= (\frac{4 - 1}{8}) a r (P Q R S)$
$A r (Δ P M L) = \frac{3}{8} a r (P Q R S)$

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