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OA, OB, OC given as combinations of a and b: area of quadrilateral OABC compared with the parallelogram on OA and OC

The same question, asked 2 times: 29 Jan 2024*, 9 Apr 2024. Every wording below, oldest first.

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  1. Question 1

    29 Jan 2024

    158
    Let OA→=a→,OB→=12a→+4b→\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}= 12\overrightarrow{a}+ 4\overrightarrow{b} and OC→=b→\overrightarrow{OC}=\overrightarrow{b}, where OO is the origin. If SS is the parallelogram with adjacent sides OAOA and OCOC, then  area of the quadrilateral OABC area of S\frac{\text{~area of the quadrilateral~}OABC}{\text{~area of~}S} is equal to

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  2. Question 2

    9 Apr 2024

    70
    Let OA→=2a→,OB→=6a→+5b→\overrightarrow{OA}= 2\overrightarrow{a},\overrightarrow{OB}= 6\overrightarrow{a}+ 5\overrightarrow{b} and OC→=3b→\overrightarrow{OC}= 3\overrightarrow{b}, where OO is the origin. If the area of the parallelogram with adjacent sides OA→\overrightarrow{OA} and OC→\overrightarrow{OC} is 15 sq. units, then the area (in sq. units) of the quadrilateral OABCOABC is equal to:

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