The area of a parallelogram whose diagonals are the vectors $2 \bar{a}-\bar{b}$ and $4 \bar{a}-5 \bar{b}$, where $\bar{a}$ and $\overline{\mathrm{b}}$ are unit vectors forming an angle of $45^{\circ}$ is

The diagonals of the parallelogram are given by $\bar{d}_1 = 2\bar{a} - \bar{b}$ and $\bar{d}_2 = 4\bar{a} - 5\bar{b}$, where $\bar{a}$ and $\bar{b}$ are unit vectors separated by $45^\circ$.

The area of a parallelogram with diagonals $\bar{d}_1$ and $\bar{d}_2$ is $\frac{1}{2}|\bar{d}_1 \times \bar{d}_2|$.

The cross product evaluates to:

$\bar{d}_1 \times \bar{d}_2 = (2\bar{a} - \bar{b}) \times (4\bar{a} - 5\bar{b}) = -10(\bar{a} \times \bar{b}) + 4(\bar{a} \times \bar{b}) = -6(\bar{a} \times \bar{b})$

Taking magnitude gives $|\bar{d}_1 \times \bar{d}_2| = 6|\bar{a} \times \bar{b}|$.

Since $|\bar{a} \times \bar{b}| = |\bar{a}||\bar{b}|\sin(45^\circ) = 1 \times 1 \times \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$, it follows that $|\bar{d}_1 \times \bar{d}_2| = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.

The area is therefore $\frac{1}{2} \times 3\sqrt{2} = \frac{3\sqrt{2}}{2}$.

This matches option B.