A square is inscribed in a circle.  A smaller square has one side coinciding with a side of the larger square and has two vertices on the circle, as shown.  What percent of the area of the larger square is the area of the smaller square?

[asy]
draw(Circle((0,0),1.4142));
draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle);
draw((0.2,1)--(0.2,1.4)--(-0.2,1.4)--(-0.2,1));
[/asy]
[asy]
draw(Circle((0,0),1.4142));
draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle);
draw((0.2,1)--(0.2,1.4)--(-0.2,1.4)--(-0.2,1));
label("$O$",(0,0),S);
label("$A$",(0,1.4),N);
label("$B$",(0.2,1.4),NE);
dot((0,0)); dot((0,1.4)); dot((0.2,1.4));
draw((0,0)--(0,1.4)--(0.2,1.4)--cycle,red);
[/asy]

We label the points as shown.  $A$ is the midpoint of the top side of the square, and $B$ is a vertex of the square.  We look at right triangle $\triangle OAB$.  We seek a ratio of areas, which remains constant no matter the side lengths, so for simplicity, we let the big square have side length $2$ and the small square have side length $2x$.  Then, $OA=1+2x$, $AB=x$, and $OB$ is a radius of the circle, which has length $\sqrt{2}$ by 45-45-90 triangles.  Then, the Pythagorean theorem states that $OA^2+AB^2=OB^2$, or \[(1+2x)^2 + x^2 = (\sqrt{2})^2.\]   Simplifying the equation yields \begin{align*}
& 1+4x+4x^2 + x^2 = 2 \\
\Longleftrightarrow\ & 5x^2 + 4x-1 =0 \\
\Longleftrightarrow\ & (5x-1)(x+1).
\end{align*} Thus, $x=-1$ or $x=1/5$.  Lengths are clearly positive, so the valid solution is $x=1/5$.  Then the small square has side length $2x=2/5$, and area $(2/5)^2 = 4/25$.  The large square has area $2^2=4$, so the small square has \[\frac{4/25}{4}=1/25=\boxed{4\%}\] the area of the large square.