Question:
The area of the semicircle in Figure A is half the area of the circle in Figure B. The area of a square inscribed in the semicircle, as shown, is what fraction of the area of a square inscribed in the circle? Express your answer as a common fraction.

[asy]
defaultpen(linewidth(0.8));

size(5cm,5cm);

draw((0,0)..(1,1)..(2,0)--(0,0));

draw((0.5,0)--(0.5,0.87)--(1.5,0.87)--(1.5,0));

draw(Circle((4,0),1));

pair A,B,C,D;

A=(3.3,0.7);
B=(3.3,-0.7);
D=(4.7,0.7);
C=(4.7,-0.7);

draw(A--B--C--D--A);

label("Figure A",(1,1.3));
label("Figure B",(4,1.3));

[/asy]

Answer:
Let $s$ be the side length of the square in Figure A.

Because the area of the semicircle in Figure A is half the area of the circle in Figure B, these two figures have the same radius, $r$. In figure A, if we draw a radius of the semicircle to a vertex of the inscribed square, we obtain a right triangle whose sides are $s/2$, $s$, and $r$. The Pythagorean Theorem tells us that $r^2 = s^2 + s^2/4$. After some manipulation, we see that $$s = \frac{2}{\sqrt{5}}r.$$ In Figure B, we see that the diameter of the circle makes up a diagonal of the square. Because the diagonal has length $2r$, it follows that the side length of the square is $2r/\sqrt{2} = r\sqrt{2}$.

To calculate the ratio of the areas, we square the ratio of the sides: $$\left(\frac{\frac{2r}{\sqrt{5}}}{r\sqrt{2}}\right)^2 = \left(\frac{2}{\sqrt{10}}\right)^2 = \frac{4}{10} = \boxed{\frac{2}{5}}.$$