Starting from the right angle, the vertices are labelled in CC order as $ABCDE$
Draw segments $BE$ and $BD$
$BD=\sqrt{3}$ by LOC, so $\angle DBC=30$
This means $\angle EBD=75$
So $[ABCDE]=[ABE]+[BCD]+[EBD]=\frac{1}{2}+\frac{\sqrt{3}}{2}+\frac{\sqrt{6}}{2}\sin 75$
$=\frac{5+3\sqrt{3}}{2}$