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.

Answer 1

Construction : 

Draw DN⊥ AB and PM⊥ AB. 

Proof : 

Area of parallelogram ABCD = AB x DN 

Area (Δ APB) = \(\frac{1}{2}\)(AB x PM) 

= AB x PM < AB x DN 

= \(\frac{1}{2}\)(AB x PM) < (AB x DN) 

= Area (Δ APB) < Area of parallelogram ABCD