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

• A

  $6$
• B

  $10$
• C

  $7$
• Correct answer
  D

  $8$

Hence, the correct answer is option: 'D'.