I already did some progress and labeled some similar segments.
The red triangle has area $1$ but it has base and height $x$ so the area will be $x^2/2=1$ so $x^2=2$ and finally $x=\sqrt 2$. By the Pythagorean theorem we also have that $y^2=x^2+4^2=4^2+\sqrt 2^2=18$. But $y^2$ is just the area of the larger square since it's the side lenght squared. So the area will be $18$