The correct answer is (A):
We can see that area of parallelogram ABCD = 2×Area of triangle ACD
48 = 2×Area of triangle ACD
Area of triangle ACD = 24
(1/ 2) × CD × DA × sinADC = 24
AD × sinADC = 6
We know that sinθ≤1, Hence, we can say that AD≥6
Let ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and $ (\vec{AB}-\vec{BC})+(\vec{AD}-\vec{DC})=k \vec{FE} $, then k is equal to