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Question

Let OA=a,OB=10a+2band OC=b where O is origin and A,C are non-collinear points. Let p denotes the area of the quadrilateral OABC and q denotes the area of the parallelogram with OA and OC as adjacent sides. Then pq is equal to

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Solution

p is area of quadrilateral OABC=12|OB×AC|
p=12|(10a+2b)×(OCOA)|p=12|(10a+2b)×(ba)|p=12|10(a×b)+0+02(b×a)|p=12|10(a×b)+2(a×b)|p=6|(a×b)|
q= area of parallelogram with OA,OC as adjacent sides.
q=|OA×OC|q=|a×b|
Therefore pq=6|a×b||a×b|=6

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