Let PQRS be square piece of paper. P is folded onto R and then Q is folded onto S. The area of the resulting of the square piece of paper is 9 square inches. Find the perimeter of square PQRS
Let's label the points after folding as follows:
[asy]
unitsize(0.5 inch);
pair P, Q, R, S;
P = (4,4);
Q = (0,0);
R = (0,8);
S = (8,8);
draw(P--Q--R--S--cycle);
label("$P$", P, NE);
label("$Q$", Q, SW);
label("$R$", R, NW);
label("$S$", S, SE);
draw(P--R,dashed);
draw(Q--S,dashed);
[/asy]
Since $P$ is folded onto $R$, we know that $RP$ is a line of symmetry of square $PQRS$. Therefore, $RP$ is a diagonal of $PQRS$.
The same reasoning applies to $Q$ being folded onto $S$, so $QS$ is also a diagonal of $PQRS$.
Therefore, the area of the folded square is equal to half the area of $PQRS$. So we have
$\frac{1}{2} \cdot PQRS = 9$
$PQRS = 18$
We know that $PQRS$ is a square, so all four sides are equal in length.
Let's say each side has length $x$. Therefore, the area of $PQRS$ can also be expressed as $x^2$.
Setting $x^2 = 18$, we can solve for $x$:
$x^2 = 18$
$x = \sqrt{18}$
Therefore, each side of $PQRS$ has length $\sqrt{18}$.
The perimeter of $PQRS$ is then $4$ times that length:
$4 \cdot \sqrt{18} = \boxed{12 \sqrt{2}}$ inches.