(This question is based on a question posed in a math riddle post on Reddit.)
Let $P$ be a convex polygon. Let the non-triangularity of $P$ be the minimum area of the symmetric difference (shown with infix operator $\oplus$) over all triangles and $P$, normalized to the volume of $P$: $$ \inf_{T \subset \mathbb{R}^2,\text{ a triangle}} \left\{ \frac{\operatorname{area}(P \oplus T)}{\operatorname{area}(P)} \right\} $$ In the picture below, the polygon $P$ is a rectangle, and we're trying to choose a triangle $T$ such that the area in orange is minimized.
Triangles have a non-triangularity of $0$, by "approximating" the triangle by itself. It's easy to construct an $n$-gon with very small non-triangularity: just make all but three of the sides arbitrarily small.
Also, all polygons $P$ have a triangularity less than $1$, just take a triangle $T$ in the interior, so that $\operatorname{area}(P \oplus T) = \operatorname{area}(P) - \operatorname{area}(T) < \operatorname{area}(P)$.
Question
What is the most non-triangular convex polygon? (Or, in the limit, the most non-triangular convex set?)
Or if this question is too hard, what are some non-trivial bounds on how non-triangular a convex polygon can be? (Perhaps constructive lower bounds would be useful to see, for example, computing the non-triangularity of the square or the circle.)
If this question is too easy, what is the most non-triangular $n$-gon for each $n$?