Pentagonal Puzzle with Pythagorus

Label the regions in the diagram as follows:

At the vertex of the $54°$ angle, the two regular polygon angles are $108°$ (from regions $G1$ and $G2$ ), which makes the missing angle on $O1$ equal to $360° - 54° - 2 \cdot 104° = 90°$ , so that $O1$ is a right triangle.

Each of the three different regular pentagons also share one side of right triangle $O1$ , and by the Pythagorean Theorem, these three sides have the relation $a^2 + b^2 = c^2$ , and since the ratio of the areas of similar polygons is equal to the ratio of the square of its sides, the area of the small pentagon plus the area of the medium pentagon is equal to the area of the big pentagon, or:

$(T1 + G1) + (G2 + T2) = (O1 + O2 + O3 + G1 + G2)$

which solves to:

$T1 + T2 = O1 + O2 + O3$

In other words, the sum of the areas of the teal regions is equal to the sum of the areas of the orange regions, which makes the ratio of their areas equal to $\boxed{1}$ .

Focus on the orange leftmost triangle. It is a right triangle, considering that the given angle of 54° combines with two pentagonal internal angles of 108° to make up a total of 270°. Then obviously the two legs of it are the sides of two pentagonal shapes containing the teal patches and its hypothenuse is the sides of regular pentagon with orange patches. The uncoloured (or grey?) spaces are actually the shared / overlapping area of the two kind of pentagons (teal legs vs orange hypothenuse), so clearly we can dismiss the commonality.

1) RT equation : a² + b² = c²
2) Area with 2D vs length / distance of 1D
3) all the shapes are regular pentagons, therefore using the exact same area formula with the uniform side lengths (all regulars) as the variable
4) area of largest pentagon : total area of the smaller two
= Orange area + Grey area : Teal area + Grey area

Orange area : Teal area
= c² – G : (a² + b²) – G
= c² – G : c² – G
= 1 : 1

Important note : G isn't the exact grey area