Let $n,m$ be the lengths of the sides of the parallelogram. First we will count the unit triangles that are outside the parallelogram .The parallelogram will necessarily have 3 unit triangles on the 60 degrees corners and two on 120 degrees corner that have one of their vertices common with the vertice of the parallelogram . Now in the side of length $m$ we will have $2m-1$ unit triangles that have at least a common point with the Parallelogram (without counting the triangles that we counted on the vertices). Therefore we will have $3+3+2+2+2(2m-1) + 2(2m-1)=4(m+n) + 6 $ triangles that are outside the parallelogram . $(1)$ Now thinking similar as before we will count the triangles that are inside the parallelogram and have at least one common point with it . Counting clockwise the triangles in the interior we get that there are exactly $2(m-1)-1$ triangles on the side of size $m$ ( and $2n-1$ on the side of size n. To count this you can seperate the interior in trapezoids with their bases on the sides of the parallelogram) $(2)$ Adding $(1)$ and $(2)$ we get $4(m+n) +6 +4(m+n) - 8= 46 \leftrightarrow m+n=6$ $(3)$(this case works only when $m,n>1$) Now For $n=1$ we get $m=6$ and the area is 12 Using $(3)$ we conclude: For $n=2$ we get $m=4$ and the area is 16 For $n=3$ we get $m=3$ and the area is 18 By symmetry we don't have to check for $n>3$

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