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

$$\vec{OA} = \vec{a}$$, $$\vec{OB} = 12\vec{a} + 4\vec{b}$$, $$\vec{OC} = \vec{b}$$, where $$O$$ is the origin.

$$S$$ has adjacent sides $$\vec{OA} = \vec{a}$$ and $$\vec{OC} = \vec{b}$$:

$$\text{Area of } S = |\vec{a} \times \vec{b}|$$

The area of quadrilateral $$OABC$$ with vertices in order $$O, A, B, C$$ can be computed by splitting into triangles $$OAB$$ and $$OBC$$:

$$\text{Area}(OABC) = \text{Area}(OAB) + \text{Area}(OBC)$$

$$\text{Area}(OAB) = \frac{1}{2}|\vec{OA} \times \vec{OB}| = \frac{1}{2}|\vec{a} \times (12\vec{a} + 4\vec{b})| = \frac{1}{2}|12(\vec{a} \times \vec{a}) + 4(\vec{a} \times \vec{b})| =  2|\vec{a} \times \vec{b}|$$

$$\text{Area}(OBC) = \frac{1}{2}|\vec{OB} \times \vec{OC}| = \frac{1}{2}|(12\vec{a} + 4\vec{b}) \times \vec{b}| = \frac{1}{2}|12(\vec{a} \times \vec{b}) + 4(\vec{b} \times \vec{b})| = 6|\vec{a} \times \vec{b}|$$

$$\text{Area}(OABC) = 2|\vec{a} \times \vec{b}| + 6|\vec{a} \times \vec{b}| = 8|\vec{a} \times \vec{b}|$$

$$\frac{\text{Area}(OABC)}{\text{Area}(S)} = \frac{8|\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|} = 8$$