Geometry haberdasher problem - square to equilateral triangle variation

Question

Let me remind the haberdasher's problem, proposed in 1907 by the puzzle composer Henry Dudeney. Dissect an equilateral triangle to a square, with only three cuts.

I would like to propose the variation of haberdasher's problem. Imagine the square to be made of two colored paper. Like this - one side red and the other yellow:

Here is my variation of the riddle, refined after comments
Cut the square with the least number of cuts to form equilateral triangle, provided that:

1. At least one element is flipped to the other side.
2. At least one element remains on its original side.
3. The flipped element is asymmetric.

(I would most welcome propositions to reword the riddle in less words.)

So say we have square painted with red on one side and yellow on the other. Starting with completely red square we build red-yellow triangle. Third condition prohibits flipping isosceles triangles or squares. In other words, the flipped element may not remain not flipped.

I suspect that there exists a solution:

1. for four elements with 1 element flipped
2. for four elements with 2 elements flipped

These solutions I would like to find. I would like to exclude all trivial solutions based on asymmetric mirror shaped figures like the letters db.

Update. Please be observant. The shapes in original Henry Dudeney solution are not symmetric. Exact measures are presented here:

Image grabbed from here

Answer 1

Here's one 5 piece solution:

5 pieces. The blue pieces are the construction: In this case, the yellow shows a rectangle that is $\sqrt{3}\times 1$, and the red shows a square that is $\sqrt{\sqrt{3}}\times \sqrt{\sqrt{3}}$.

To make this into a square:

Simply slide a and e down to the right and move b into the remaining space.

To make it into an equilateral triangle:

Flip a, b, and c vertically as a unit and move them to the right.

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Here's a different 5 piece solution:

Here's an answer with

5 pieces.

You will see that

Both b and c need to be flipped (obviously would be better if only one needed to be flipped). The construction is to take the original green piece and flip it upside down (to make b). Then c is equal to the piece that's missing out of the original yellow piece (but also flipped - unless that's isosceles, which I doubt))

However, which is still a problem, it's possible to just solve it the original way without any flips. To solve this, you can simply take a symmetrical sliver out of the blue piece and have it so that this must be filled by a matching extra bit on piece b. This then enforces the need to flip the piece::