imc

Area of the circle inscribed inside the semicircle $=\pi r^2\Rightarrow \pi (2^2)=4\pi .$ Area of the larger circle (semicircle's area x 2)$=\pi r^2\Rightarrow \pi (4^2)=16\pi$ (4, or the diameter of the inscribed circle is the same thing as the radius of the semicircle). Thus, the area of the semicircle is $\frac{1}{2}(16\pi )\Rightarrow 8\pi .$ Part of the semicircle that is unshaded is $\frac{4\pi }{8\pi }=\frac{1}{2}$ Therefore, the shaded part is $1-\frac{1}{2}=\frac{1}{2}$ Thus the answer is $\frac{1}{2}\Rightarrow \boxed{\frac{1}{2}}$