The horizontal and vertical distances between adjacent points equal 1 unit.  What is the area of triangle $ABC$?

$\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/2 \qquad \text{(C)}\ 3/4 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 5/4$
$\triangle ADC$ takes up half of the 4x3 grid, so it has area of $6$.
$\triangle ABD$ has height of $BF = 2$ and a base of $AD = 4$, for an area of $\frac{1}{2}\cdot 2 \cdot 4 = 4$.
$\triangle CBD$ 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 $\triangle ABC$ can be found by taking $\triangle ADC$, and subtracting off $\triangle ABD$ and $\triangle CBD$.
Thus, the area of $\triangle ABC = 6 - 4 - \frac{3}{2} = \frac{1}{2}$, and the answer is $\boxed{B}$.
There are other equivalent ways of dissecting the figure; right triangles $\triangle ABF, \triangle BCE$ and rectangle $\square BEDF$ can also be used. You can also use $\triangle{BEC}$ and trapezoid $ADBE$.