Question

`ABCD` is a parallelogram whose diagonals meet at P. If O is a fixed point, then `bar(OA)+bar(OB)+bar(OC)+bar(OD)` equals :
A. OP
B. 2OP
C. 3OP
D. 4OP

Answer 1

Correct Answer - D
We know that, P will be the mid-point of AC and BD.

`thereforeOA+OC=2OP` . . . (i)
and `OB+OD=2OP`
On adding Eqs. (i) and (ii), we get
`OA+OB+OC+OD=4OP`