Since $AB=BC=CD=1$ $\implies$ $AE=6$$\hspace{0.2cm}$ and $\hspace{0.2cm}$$DE=3$$\hspace{0.2cm}$ because$\hspace{0.2cm}$$ D$ $\hspace{0.2cm}$is the center of largest circle.
Area of circle with diameter $CE=4$$\hspace{0.2cm}$ is $\hspace{0.2cm}$$\pi r^2=\pi 2^2=4\pi$
Area of circle with diameter $BE=5$$\hspace{0.2cm}$ is $\hspace{0.2cm}$$\pi r^2=\pi (\frac{5}{2})^2=\frac{25}{4}\pi$
Hence the area of shaded region$=\frac{25}{4}\pi-4\pi=\frac{9\pi}{4}$
Since $AB=BC=CD=1$ $\implies$ $AE=6$$\hspace{0.2cm}$ and $\hspace{0.2cm}$$DE=3$$\hspace{0.2cm}$ because$\hspace{0.2cm}$$ D$ $\hspace{0.2cm}$is the center of largest circle.
Area of circle with diameter $CE=4$$\hspace{0.2cm}$ is $\hspace{0.2cm}$$\pi r^2=\pi 2^2=4\pi$
Area of circle with diameter $BE=5$$\hspace{0.2cm}$ is $\hspace{0.2cm}$$\pi r^2=\pi (\frac{5}{2})^2=\frac{25}{4}\pi$
Hence the area of shaded region$=\frac{25}{4}\pi-4\pi=\frac{9\pi}{4}$