imc

We see that side $BE$, which we know is $1$, is also the shorter leg of one of the four right triangles (which are congruent, I won't prove this). So, $AH=1$. Then $HB=HE+BE=HE+1$, and $HE$ is one of the sides of the square whose area we want to find. So: $$1^2+(HE+1{)}^2={\sqrt{50}}^2$$ $$1+(HE+1{)}^2=50$$ $$(HE+1{)}^2=49$$ $$HE+1=7$$ $$HE=6$$ So, the area of the square is $6^2=\boxed{36}$