Q11 of 25 Page 9

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.

Construction: Draw DN⊥ AB and PM⊥ AB.



Proof: Area of parallelogram ABCD = AB × DN


Area (Δ APB) = (AB × PM)


From figure we can see PM<DN.


= AB × PM < AB × DN


= (AB ×PM) < (AB ×DN)


= Area (Δ APB) < Area of parallelogram ABCD


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