Calculating area of a parallelogram using length of two diagonals Is there a formula for calculating area of a parallelogram using only the length of diagonals? if so what is it?                                                                                                                                
 A: No.  Imagine two fixed diagonals that are attached and can rotate in the center.  As they pivot, different parallelograms are induced; if they are almost parallel the area will be close to zero.
A: Imagine the two diagonals are sticks of wood, with a nail sort of holding them together around the middle, except that you can rotate one of the diagonals around the nail.  By making the angle between the diagonals small, you can make the area as small as you wish. By "opening up" the angle so that it is $90^\circ$, you can maximize the area.  
Remark: The area of the parallelogram is $\frac{1}{2}pq\sin\theta$, where $p$ and $q$ are the lengths of the diagonals, and $\theta$ is the angle between the diagonals. But we do not have to know that to see that the area is not determined by the lengths. 
A: If the diagonals are given in vector form, then as @RecklessReckoner said, one diagonal is vector sum $a+b$ and the other is vector sum $a-b$. So you have two equations and two variables, you can find out vectors $a$ and $b$, then simply find the magnitude of their cross product.
If the diagonals are vectors $p$ and $q$, the area can be represented as the magnitude of the cross product: $\frac 14 (p+q) \times (p-q)$.
