smc

Drop the altitude $AD$ from the vertex $A$ to base $BC$, and note that it intersects the centroid $M$. Then $BCM$ is an isosceles right triangle, so $MB=1$, so $MD=\frac{\sqrt{2}}{2}$. By centroid properties, $AD=3MD=\frac{3\sqrt{2}}{2}$. Then the area of $ABC$ is $\frac{1}{2}\sqrt{2}\frac{3\sqrt{2}}{2}=1.5\to \boxed{1.5}$.