jmc

$△ADC$ takes up half of the 4x3 grid, so it has area of $6$. $△ABD$ has height of $BF=2$ and a base of $AD=4$, for an area of $\frac{1}{2}\cdot 2\cdot 4=4$. $△CBD$ has height of $BE=1$ and a base of $CD=3$, for an area of $\frac{1}{2}\cdot 1\cdot 3=\frac{3}{2}$ Note that $△ABC$ can be found by taking $△ADC$, and subtracting off $△ABD$ and $△CBD$. Thus, the area of $△ABC=6-4-\frac{3}{2}=\frac{1}{2}$, and the answer is $\boxed{1/2}$. There are other equivalent ways of dissecting the figure; right triangles $△ABF,△BCE$ and rectangle $\square BEDF$ can also be used. You can also use $△BEC$ and trapezoid $ADBE$.