If $\left|\overline{\mathrm{a}}\right|=\left|\overline{\mathrm{b}}\right|=5$ and angle between $\overline{\mathrm{a}},\overline{\mathrm{b}}$ is ${45}^0$ then area of parellelogram formed by $\overline{\mathrm{a}}-2\overline{\mathrm{b}}$ and $3\overline{\mathrm{a}}-2\overline{\mathrm{b}}$ as  adjcent sides is
 

$$\begin{gathered} |\overline{\mathrm{a}}\times \overline{\mathrm{b}}|=5\times 5\times \sin {45}^0=\frac{25}{\sqrt{2}} \\ \overline{\mathrm{OA}}=\overline{\mathrm{a}}-2\overline{\mathrm{b}} \overline{\mathrm{OB}}=3\overline{\mathrm{a}}-2\overline{\mathrm{b}} \\ \mathrm{area}=|\overline{\mathrm{OA}}\times \overline{\mathrm{OB}}|=\left|\right(\overline{\mathrm{a}}-2\overline{\mathrm{b}}\left)\right|\times (3\overline{\mathrm{a}}-2\overline{\mathrm{b}}) \\ =|-2 \overline{\mathrm{a}}\times \overline{\mathrm{b}}-6 \overline{\mathrm{b}}\times \overline{\mathrm{a}}|=4|\overline{\mathrm{a}}\times \overline{\mathrm{b}}| \\ =4\times \frac{25}{\sqrt{2}}=50\sqrt{2} \end{gathered}$$