Let $\overline{a}, \overline{b}$ be vectors $\left|\overline{a}\right|=\left|\overline{b}\right|=5$ and $\left(\overline{a}, \overline{b}\right)={45}^0$ Find area of triangle having $\overline{a}-2\overline{b}, 3\overline{a}+2\overline{b}$ as two of its sides.

Given $\left|\overline{a}\right|=\left|\overline{b}\right|=5$ and $(\overline{a},\overline{b})={45}^0$

area of the $\mathrm{\Delta }$ having  $\overline{a}-2\overline{b},3\overline{a}+2\overline{b}$ as two of its sides.$=\frac{1}{2}\left|\right(\overline{a}-2\overline{b})\times (3\overline{a}+2\overline{b}\left)\right|$

$=\frac{1}{2}\left|3\right(\overline{a}\times \overline{a})+2(\overline{a}\times \overline{b})-6(\overline{b}\times \overline{a})-4(\overline{b}\times \overline{b}\left)\right|$

$=\frac{1}{2}\left|8\right(\overline{a}\times \overline{b}\left)\right|=\frac{8}{2}\left|\overline{a}\right|\left|\overline{b}\right|\sin (\overline{a},\overline{b})=4\left(5\right)\left(5\right)\sin {45}^{\circ }$

$=4\left(25\right)\frac{1}{\sqrt{2}}=25\left(2\sqrt{2}\right)=50\sqrt{2}\text{ sq units. }$