A triangular corner with side lengths $DB=EB=1$ is cut from equilateral triangle ABC of side length $3$. 
The perimeter of the remaining quadrilateral is

$\text{(A)} \ 6 \qquad  \text{(B)} \ 6\frac{1}{2} \qquad  \text{(C)} \ 7 \qquad  \text{(D)} \ 7\frac{1}{2} \qquad  \text{(E)} \ 8$
$\triangle DBE$ is similar to $\triangle ABC$ by AA, so $\overline{DE}$ = 1 by similarity, and $\overline{CE} = \overline{AD} = 2$, by subtraction. Thus the perimeter is $3+2+2+1 = 8$, or $\boxed{E}$. -slackroadia