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# In Figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show thati)ar(PQRS) = ar(ABRS)ii) ar(AXS) = $$\frac{1}{2}$$ ar(PQRS)

i) It can be observed that paralleogram PQRS and ABRS lie on the same base SR and also, these lie in between the same parallel lines SR and PB.

$$\therefore$$ Area (PQRS) = Area (ABRS) ...(i)

ii) Consider $$\triangle{AXS}$$ and paralleogram ABRS.

As these lie on the same base and are between the same parallel lines AS and BR

$$\therefore$$ Area ($$\triangle{AXS}$$) = $$\frac{1}{2}$$ Area (ABRS) ...(ii)
From equations (i) and (ii), we get,
ar(AXS) = $$\frac{1}{2}$$ ar(PQRS)
Hence, proved.