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since parallelogram pqrs and abrs are on the same base rs and b/w the same parall sr and pb so ar(abrs)=ar(pqrs )
I don't know 2nd part sorry
I don't know 2nd part sorry
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in first case ,we have to prove
ar (PQRS) = ar (ABRS)
Now, it is given that PQRS and ABRS are parallelogram so, opposite sides are parallel
and, the parallelogram PQRS and ABRS are on the same base SR and between same parallel lines PQ//SR//AQ
SO, it is proved that ar (PQRS) = ar ( ABRS)
IN SECOND CASE
we can see that ∆ AXS and ABRS on the same base AS and between same parallel lines AS//BR
SO, WE KNOW THAT IF A TRIANGLE AND A QUADRILATERAL ARE ON THE SAME BASE BETWEEN SAME PARALLEL LINES THEN THE AREA OF TRIANGLE IS EQUAL TO HALF THE AREA OF QUADRILATERAL
Therefore, ar (∆AXS) = 1/2 ar ( ABRS )
but we proved above that ar ( PQRS) = ar ( ABRS)
SO, BY THIS WE CONCLUDE THAT.
ar (∆AXS) = 1/2 ar ( PQRS).
HENCE PROVED
HOPE IT WILL HOPE UU
ar (PQRS) = ar (ABRS)
Now, it is given that PQRS and ABRS are parallelogram so, opposite sides are parallel
and, the parallelogram PQRS and ABRS are on the same base SR and between same parallel lines PQ//SR//AQ
SO, it is proved that ar (PQRS) = ar ( ABRS)
IN SECOND CASE
we can see that ∆ AXS and ABRS on the same base AS and between same parallel lines AS//BR
SO, WE KNOW THAT IF A TRIANGLE AND A QUADRILATERAL ARE ON THE SAME BASE BETWEEN SAME PARALLEL LINES THEN THE AREA OF TRIANGLE IS EQUAL TO HALF THE AREA OF QUADRILATERAL
Therefore, ar (∆AXS) = 1/2 ar ( ABRS )
but we proved above that ar ( PQRS) = ar ( ABRS)
SO, BY THIS WE CONCLUDE THAT.
ar (∆AXS) = 1/2 ar ( PQRS).
HENCE PROVED
HOPE IT WILL HOPE UU
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