Three congruent circles with centers $P$, $Q$, and $R$ are tangent to the sides of rectangle $ABCD$ as shown.  The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R$.  The area of the rectangle is

$\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$
If circle $Q$ has diameter $4$, then so do congruent circles $P$ and $R$.  Draw a diameter through $P$ parallel to $AD$.  The diameter will be congruent to $AD$, and thus $AD = 4$, which is the height of the rectangle.
Draw a horizontal line $PQR$ that extends to the sides of the rectangle.  This line is $2$ diameters long, so it has length $4\cdot 2 = 8$.  It is parallel and congruent to $AB$, so the width of the rectangle is $8$.
Thus, the area of the rectangle is $lw = 4\cdot 8 = 32$, and the answer is $\boxed{C}$.