Let OA= a, OB=12 a+4 b, OC= b, where O is the origin.

Question

Let $\vec{OA}=\vec a$, $\vec{OB}=12\vec a+4\vec b$, $\vec{OC}=\vec b$, where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $OA$ and $OC$, then $\dfrac{	ext{area of quadrilateral }OABC}{	ext{area of }S}$ is equal to

Accepted Answer

Using the shoelace formula with position vectors $O=0,\ A=\vec a,\ B=12\vec a+4\vec b,\ C=\vec b$:
$$\text{Area }OABC = \frac12\big|\,\vec{OA}\times\vec{OB}+\vec{OB}\times\vec{OC}+\vec{OC}\times\vec{OO}+\vec{OO}\times\vec{OA}\,\big|$$
$$\vec a\times(12\vec a+4\vec b) = 4(\vec a\times\vec b),\qquad (12\vec a+4\vec b)\times\vec b = 12(\vec a\times\vec b)$$
$$\text{Area }OABC = \frac12|16(\vec a\times\vec b)| = 8|\vec a\times\vec b|$$

Area of $S = |\vec a\times\vec b|$. So the ratio is $\dfrac{8|\vec a\times\vec b|}{|\vec a\times\vec b|} = 8$.

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