In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR.

Question. In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) $\operatorname{ar}(\mathrm{PQRS})=\operatorname{ar}(\mathrm{ABRS})$

(ii) $\operatorname{ar}(A X S)=\frac{1}{2} \operatorname{ar}(P Q R S)$



Solution:

(i) It can be observed that parallelogram PQRS and ABRS lie on the same base SR

and also, these lie in between the same parallel lines SR and PB.

$\therefore$ Area $(P Q R S)=$ Area (ABRS) ... (1)

(ii) Consider $\triangle A X S$ and parallelogram $A B R S$.

As these lie on the same base and are between the same parallel lines $A S$ and $B R$,

$\therefore$ Area $(\triangle A X S)=\frac{1}{2}$ Area (ABRS) ... (2)

From equations (1) and (2), we obtain

Area $(\triangle A X S)=\frac{1}{2}$ Area $(P Q R S)$

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