1967 AMC 12 Problem 33

Subtract the two small semicircle areas from the large one

In right triangle $ADB,$ the altitude theorem gives $CD^2=AC\cdot CB$

Let $AC=u$ and $CB=v.$ The shaded area is the large semicircle minus the two smaller ones: $$\frac{\pi}{8}\bigl((u+v)^2-u^2-v^2\bigr) =\frac{\pi uv}{4}.$$ Since $D$ lies on the semicircle with diameter $AB,$ triangle $ADB$ is right, and its altitude satisfies $CD^2=uv.$ A circle of radius $CD$ therefore has area $\pi uv.$ The required ratio is $1:4.$

Therefore, the correct answer is D.