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Let OA =a, OB...

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Let $- - \to O A = \to a, - - \to O B = 10 \to a + 2 \to b$ and $- - \to O C = \to b$ where $O$ is origin and $A, C$ are non-collinear points. Let $p$ denotes the area of the quadrilateral $O A B C$ and $q$ denotes the area of the parallelogram with $- - \to O A$ and $- - \to O C$ as adjacent sides. Then $\frac{p}{q}$ is equal to

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Solution

$p$ is area of quadrilateral $O A B C = \frac{1}{2} | - - \to O B \times - - \to A C |$
$p = \frac{1}{2} | (10 \to a + 2 \to b) \times (- - \to O C - - - \to O A) | p = \frac{1}{2} | (10 \to a + 2 \to b) \times (\to b - \to a) | p = \frac{1}{2} | 10 (\to a \times \to b) + 0 + 0 - 2 (\to b \times \to a) | p = \frac{1}{2} | 10 (\to a \times \to b) + 2 (\to a \times \to b) | p = 6 | (\to a \times \to b) |$
$q =$ area of parallelogram with $O A, O C$ as adjacent sides.
$q = | - - \to O A \times - - \to O C | q = | \to a \times \to b |$
Therefore $\frac{p}{q} = \frac{6 | \to a \times \to b |}{| \to a \times \to b |} = 6$

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