$A, B, C, D$ are mid points of sides of parallelogram $P Q R S$ . If $a r (P Q R S) = 36 c m^{2}$ then $a r (A B C D)$

24 c m^{2}

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18 c m^{2}

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30 c m^{2}

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36 c m^{2}

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Solution

The correct option is B $18 c m^{2}$
$A, B, C, D$ are point of sides of parallelogram $P Q R S$ area $(P Q R S) = 36 c m^{2} a r (A B C D) = ?$
$A P = \frac{1}{2} S P, B P = \frac{1}{} Q R$
$P S = Q R, \frac{1}{2} P S = \frac{1}{2} Q R$
area $(△ A B C) = \frac{1}{2} (△ P Q C Q)$
area $(△ A D C) = \frac{1}{2} a r (A C R S)$
area $△ A B C + a r e a (△ A D C) = \frac{1}{2} a r e a (A P C Q) = \frac{1}{2} a r e a (A C R S)$
$a r e a (A D C D) = \frac{1}{2} a r e a (A P C Q) + (A C R S) a r e a$
$a r e a (A B C D) = \frac{1}{2} a r e a (P Q R S)$
$a r e a (A B C D) = \frac{1}{2} (36)$
$= 18 c m^{2}$

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