1.

If ABCD is a rombus whose diagonals cut at the origin O, then OA + OB + OC + OD equals to

Answer»

AB + AC
0
2 (AB + BC)
AC + BD

Solution : since, the diagonals of a rhombus bisect each other OA = - OC and OB =- OD OA + OB +OC + OD = - OC - OD + OC + OD = 0


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