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`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. AB + ACB. 0C. 2 (AB + BC)D. AC + BD |
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Answer» Correct Answer - B 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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