The small horizontal rectangle has dimensions $1.20613\times1$. That first value is the square root of the largest root of $-1 - 5 x + x^2 + 2 x^3 = 0$. The discriminant of that polynomial is 1101. All the various roots used in the other rectangles can be equated to this discriminant either directly or through squaring the root. We will say that this rectangle has discriminant 1101.
A few rules that this rectangle follows:
1. At least one rectangle has an edge of length 1.
2. If two rectangles are congruent, they have different orientations.
3. The outer rectangle is similar to the component rectangles.
4. All component rectangles are similar.
5. The outer rectangle is divided into more than 1 component rectangles.
Here's a similar rectangle dissection with discriminant 5, corresponding to the golden ratio. If a discriminant cannot be reached with fewer rectangles, the dissection is minimal.
If Rule 3 is relaxed, a square can be divided into rectangles with discriminant -23, which indicates the plastic constant.
I'm particularly looking for solutions (without relaxation of rule3) giving the Pisot numbers, which have discriminants
-23 (plastic constant),
-31 (Narayana Cow constant),
-283 ($-1 - x^3 + x^4 = 0$),
1609 ($-1 + x^2 - x^3 - x^4 + x^5=0$),
3857 ($-1 - x - x^2 - x^3 + x^5=0$),
4477 ($-1 - x^2 - x^4 + x^5=0$),
29077 ($-1 + x - x^2 + x^4 - 2 x^5 + x^6=0$) and
37253 ($-1 + x^2 - x^4 - x^5 + x^6=0$)
Other solutions I'd like to see are those where all component rectangles appear twice in different orientations and those where the ratio of component sizes is a simple series, and all possible discriminants for 6 and fewer component rectangles.
Here's one where I did not use an edge of length 1 because I wanted to show rectangle areas. If the 40 was 36 instead, we would have squares of 1 to 7 as areas of similar rectangles. But not quite.
Here's an amazing find, now that I'm understanding these better. This has the same aspect ratio as A4 paper.
These are related to New Substitution Tilings Using 2, φ, ψ, χ, ρ.