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Question
If P is any point in the interior of a parallelogram ABCD, then prove that area of the
triangle APB is less than half the area of parallelogram.
Solution
Draw DN ⊥ AB and PM ⊥ AB.
Now,
`Area (ΙΙ^(gm) ABCD) = AB xx DN , ar (ΔAPB ) = 1/2 (AB xx PM)`
Now , PM < DN
⇒ `AB xx PM < AB xx DN`
⇒ ` 1/2 (AB xx PM) < 1/2 (AB xx DN)`
⇒ `Area ( ΔAPB ) <1/2 ar ( Parragram ABCD)`
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