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 O is the origin. If $S$ is the parallelogram with adjacent sides OA and OC , then $\frac{\text { area of the quadrilateral } \mathrm{OABC}}{\text { area of } \mathrm{S}}$ is equal to
Select the correct option:
A
6
B
10
C
7
D
8
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Area of parallelogram, $S=|\vec{a} \times \vec{b}|$
Area of quadrilateral $=\operatorname{Area}(\triangle \mathrm{OAB})+\operatorname{Area}(\triangle \mathrm{OBC})$
$$
\begin{aligned}
& =\frac{1}{2}\{|\overrightarrow{\mathrm{a}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{~b}})|+|\overrightarrow{\mathrm{b}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{~b}})|\} \\
& =8|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|
\end{aligned}
$$
$$
\text { Ratio }=\frac{8|(\vec{a} \times \bar{b})|}{|(\vec{a} \times \bar{b})|}=8
$$
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