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Q: 4     In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
               (i)  \small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)

              [Hint : Through P, draw a line parallel to AB.]


Answers (1)


We have a ||gm ABCD and AB || CD, AD || BC.  Through P, draw a line parallel to AB
 Now, \DeltaAPB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.
Therefore, ar (\DeltaAPB) = 1/2 . ar(ABEF)...............(i)
Similarly, ar (\DeltaPCD ) = 1/2 . ar (EFDC) ..............(ii)
Now, by adding both equations, we get
\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)

Hence proved.


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