Let OA → = a → , OB → = 12 a → + 4 b → and OC → = b → , where O is the origin. If S is the parallelogram…

Area of parallelogram, $\mathrm{S}=|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|$

Area of quadrilateral = Area $\left(∆\mathrm{OAB}\right)+$Area $\left(∆\mathrm{OBC}\right)$

$=\frac{1}{2}\left\{\right|\overrightarrow{\mathrm{a}}\times (12\overrightarrow{\mathrm{a}}+4\overrightarrow{\mathrm{b}})|+|\overrightarrow{\mathrm{b}}\times (12\overrightarrow{\mathrm{a}}+4\overrightarrow{\mathrm{b}})\left|\right\}$

$=\frac{1}{2}\left\{\right|4\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|+|12\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}\left|\right\}$

$=\frac{1}{2}\left\{4\right|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|+12|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\left|\right\}$

$=\frac{16}{2}|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|$

$=8\left|\right(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\left)\right|$

Ratio $=\frac{8\left|\right(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\left)\right|}{\left|\right(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\left)\right|}=8$