We can easily prove that EFGH is a square

\(\Rightarrow EH=EF=\dfrac{HF}{\sqrt{2}}=\dfrac{AB}{\sqrt{2}}=\sqrt{2}\left(m\right)\)

The area of the sector EFH is : \(\dfrac{\left(\sqrt{2}\right)^2\pi}{4}=\dfrac{\pi}{2}\left(m^2\right)\)

\(S_{EFH}=\dfrac{\left(\sqrt{2}\right)^2}{2}=1\left(m^2\right)\)

The answer is : \(2\left(\dfrac{\pi}{2}-1\right)=\pi-2\approx1.14\)