Eleven Square Theorem

If four squares are arranged as follows:

Then by the law of cosines on $\triangle ABC$ , $\cos \angle ACB = \cfrac{a^2 + b^2 - c^2}{2ab}$ .

And $\cos \angle A'CB' = \cos(180° - \angle ACB) = -\cos ACB$ .

And by the law of cosines on $\triangle A'B'C$ , $c'^2 = a^2 + b^2 - 2ab \cos A'CB' = a^2 + b^2 + 2ab \cos ACB = a^2 + b^2 + 2ab \cdot \cfrac{a^2 + b^2 - c^2}{2ab} = 2(a^2 + b^2) - c^2$ .

Using $c'^2 = 2(a^2 + b^2) - c^2$ on the red, blue, and yellow squares, the square to the immediate right has an area of $2(3 + 4) - 2 = 12$ , and the square to the immediate left has an area of $2(2 + 4) - 3 = 9$ . Then using it on the $2$ , $4$ , and $9$ squares, the other square on the left has an area of $2(4 + 9) - 2 = 24$ .

Now label some points as follows:

From the area of the known squares, $B'C = \sqrt{3}$ , $CA' = DA' = 2$ , $A'B' = A'F = 2\sqrt{3}$ , $ED = 3$ , and $EA' = 2\sqrt{6}$ .

By the law of cosines on $\triangle A'B'C$ , $\cos \angle CA'B' = \cfrac{A'C^2 + A'B'^2 - CB'^2}{2 \cdot A'C \cdot A'B'} = \cfrac{4 + 12 - 3}{2 \cdot 2 \cdot 2\sqrt{3}} = \cfrac{13\sqrt{3}}{24}$ .

By the law of cosines on $\triangle A'DE$ , $\cos \angle DA'E = \cfrac{A'D^2 + A'E^2 - DE^2}{2 \cdot A'D \cdot A'E} = \cfrac{4 + 24 - 9}{2 \cdot 2 \cdot 2\sqrt{6}} = \cfrac{19\sqrt{6}}{48}$ .

Using $\sin \theta = \sqrt{1 - \cos^2 \theta}$ , $\sin \angle CA'B' = \sqrt{1 - (\cfrac{13\sqrt{3}}{24})^2} = \cfrac{\sqrt{69}}{24}$ and $\sin \angle DA'E = \sqrt{1 - (\cfrac{19\sqrt{6}}{48})^2} = \cfrac{\sqrt{138}}{48}$ .

That means $\cos \angle EA'F = \cos(180° - (\angle CA'B' + \angle DA'E)) = -\cos(\angle CA'B' + \angle DA'E) = -\cos \angle CA'B' \cos \angle DA'E + \sin \angle CA'B' \sin \angle DA'E = -\cfrac{13\sqrt{3}}{24} \cdot \cfrac{19\sqrt{6}}{48} + \cfrac{\sqrt{69}}{24} \cdot \cfrac{\sqrt{138}}{48} = -\cfrac{7\sqrt{2}}{12}$ .

So by the law of cosines on $\triangle A'EF$ , $EF^2 = EA'^2 + A'F^2 - 2 \cdot EA' \cdot A'F \cdot \cos \angle EA'F = 24 + 12 + 2 \cdot 2 \sqrt{6} \cdot 2 \sqrt{3} \cdot \cfrac{7\sqrt{2}}{12} = 64$ , which is also the area of the purple square.

Using $c'^2 = 2(a^2 + b^2) - c^2$ on the $12$ , $24$ , and $64$ squares, the next square has an area of $2(12 + 24) - 64 = 8$ .

Using $c'^2 = 2(a^2 + b^2) - c^2$ on the $8$ , $12$ , and $24$ squares, the next square has an area of $2(8 + 12) - 24 = 16$ .

Using $c'^2 = 2(a^2 + b^2) - c^2$ on the $8$ , $12$ , and $16$ squares, the next square has an area of $2(12 + 16) - 8 = 48$ .

Using $c'^2 = 2(a^2 + b^2) - c^2$ on the $12$ , $16$ , and $48$ squares, the next square has an area of $2(16 + 48) - 12 = 116$ .

Therefore, the area sum of the purple and green square is $64 + 116 = \boxed{180}$ .