Given a triangle $\Delta$ of unit area, which proportion of how much area can always at least be covered by $k$ isosceles triangles contained in $\Delta$ and intersecting at most at their boundaries?

The answer is easy for $k=1$. Without error of my part, the worst case is given by a triangle with sides proportional to $1+\epsilon,2+\epsilon,3+\epsilon$ where one can cover roughly $2/3$ of the total area.

Is there an elegant way to compute the answer for $k=2,3,\dots$?

Given a triangle $\Delta$ of unit area, which proportion of area can always at least be covered by $k$ isosceles triangles contained in $\Delta$ and intersecting at most at their boundaries?

The answer is easy for $k=1$. Without error of my part, the worst case is given by a triangle with sides proportional to $1+\epsilon,2+\epsilon,3+\epsilon$ where one can cover roughly $2/3$ of the total area.

Is there an elegant way to compute the answer for $k=2,3,\dots$?