We will show a variant of Pick's Theorem for equilateral triangular lattices. Henceforth, all work will be done in the context of an infinite equilateral triangular lattice where every individual equilateral triangle has area $1$ (scale it, WLOG). Indeed, we will show that:
Claim. For any lattice polygon, if $A, I, B$ denote its area, the number of lattice points in its strict interior, and the number of lattice points on its boundary, then we have that $A = 2I + B - 2.$
Proof. We will first prove it for triangles. We'll prove it for triangles by induction on $A$, its area. If the triangle has a point on its boundary which isn't one of the three vertices, then we can split it into two triangles and use induction on these two smaller triangles. If the triangle has a point in its interior, then we can split it into three smaller triangles and induct down. Otherwise, we only need to resolve the case where $I = 0$ and $B = 3$, we desire to show that $A = 1$ in this case. To do this, let our triangle be $P_1P_2P_3$. Then, let $P_2P_3$ be the smallest side, WLOG. If it has length $1$ then we are clearly done. Otherwise, let it have length $\ge \sqrt{3}.$ Let $v$ be the vector which goes from $P_2$ to $P_3$, and consider the points $P_1 + kv,$ for $k \in \mathbb{Z}.$ It's clear that there must exist one such point $P = P_1 + kv$ so that $\angle PP_2P_3$ and $\angle PP_3P_2$ are both at most $90$, and it's easy to verify that this triangle $PP_2P_3$ has the same area. Furthermore, this triangle has strictly less perimeter. Hence, we can continually apply this operation until eventually some side is of length $1$, and we'd be done then.
Now, we'll solve it for general polygons. In this case, just take a triangulation and use induction on the number of sides of the polygon, removing an "ear" for the inductive step.
$\blacksquare$
Just apply the above claim to the problem, we're done.
$\square$
