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The figure shows parallelogram ABCD, and O is the intersection of two diagonals. OA = 4, OB

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The figure shows parallelogram ABCD, and O is the intersection of two diagonals. OA = 4, OB

by Max@Math Revolution » Wed Jun 10, 2020 2:34 am

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[GMAT math practice question]
6.10PS.png
The figure shows parallelogram ABCD, and O is the intersection of two diagonals. OA = 4, OB = 6, and P is a point on segment OC. What is the area of ABCD?

A. 45
B. 46
C. 47
D. 48
E. 49
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Re: The figure shows parallelogram ABCD, and O is the intersection of two diagonals. OA = 4, OB

by Max@Math Revolution » Fri Jun 12, 2020 12:43 am
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Since diagonals of a parallelogram bisect each other, we have OA = OC = 4 and OB = OD = 6.
Then triangles OBP and ODP are congruent.
Thus, we have ∠POB = ∠POD = 90°.
Since two diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.
Then the area of the rhombus is (1/2)·8·12 = 48.

Therefore, D is the answer.
Answer: D
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