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i)ar(PQRS) = ar(ABRS)

ii) ar(AXS) = \(\frac{1}{2} \) ar(PQRS)

Answer :

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.

- In Figure, ABCD is a parallelogram, \(AE\perp{DC}\) and \(CF\perp{AD}\). If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
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