Can a convex polygon with rational (integer) sides and rational area always be placed on a lattice?

Question

I am thinking on a problem whether a convex polygon with rational (integer) sides and rational area can be placed on a lattice, but could not find an answer or conterexample in a literature. Sure, for a (Heron) triangle it is true. How about other polygons? It likely seems it is true, however any reviews will be helpful. The first question was about all polygons, not exactly convex, therefore for that polygons a right conterexample was given in answer 1 below. By the "lattice" here we mean an existence of an (orthogonal) lattice in which the vertices of the polygon would have rational coordinates.

Answer 1

Ops, i think i got an answer. Consider an isosceles triangle with one side equal to $$AB=2*(\sqrt7+\sqrt2)/\sqrt2$$ and the height to side AB equal to $$h=(\sqrt7-\sqrt2)/\sqrt2$$ Then we have other sides of the triangle equal to $3$ and the area equal to $5/2$. Then we reflect the triangle and get the necessary conterexample quadrilateral (deltoid with integer sides and area). It is not than hard to prove that if the Area of a polygon divided by it's squared side (or diagonal) is irrational, than that polygon cannot be put on a lattice.

Answer 2

This was posted before the question asked for a convex polygon. Consider the figure below. $BCD$ and $GHI$ are equilateral triangles of side $2$. All sides have length $2$ and the area is $12$. We know an equilateral triangle cannot be put on a lattice.