Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a [quadrilateral](https://artofproblemsolving.com/wiki/index.php/Quadrilateral). Its [area](https://artofproblemsolving.com/wiki/index.php/Area) in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \  } 5 \qquad \mathrm{(D) \  } 5.5 \qquad \mathrm{(E) \  }6$
Solution 1
We see that the number of interior points is $5$ and the number of boundary points is $4$. Therefore, by [Pick's Theorem](https://artofproblemsolving.com/wiki/index.php/Pick%27s_Theorem), the area is $5+\frac{4}{2}-1=6, \boxed{\text{E}}$.

Solution 2

Draw in the perimeter of the rectangle and label the points as shown. We have $[ABCD]=(3)(4)=12$, $[AEH]=\frac{1}{2}(1)(2)=1$, $[EBF]=\frac{1}{2}(3)(2)=3$, $[FCG]=\frac{1}{2}(1)(1)=\frac{1}{2}$, and $[GDH]=\frac{1}{2}(3)(1)=\frac{3}{2}$. Therefore, $[EFGH]=[ABCD]-([AEH]+[EBF]+[FCG]+[GDH])$ $=12-(1+3+\frac{1}{2}+\frac{3}{2})=6, \boxed{\text{E}}$.