After our exchange, I have understood you mean the type of "rotation" given in the figure below. Caution : it isn't a rotation in the usual sense of the word because this transformation hasn't a center (which should be an invariant point) : see the second interpretation in the "Comment" below.

Fig. 1 : The sides of the moving rectangle "envelope" curves. The North-East curve generated by line segment $[B_tC_t]$ is a piece of astroid : see the classical "sliding ladder image" for example here. The North curve is divided into two parts : an arc of the black curve with equation given below in (1) and an arc of the red curve which is an ellipse with parametric equations given in (2).
Using the technique of envelopes it is possible to get the equation of the envelope of line $A_tB_t$ (black curve on fig. 1). Here is the way I have proceeded :
I start from the equation of line $A_tB_t$ which is :
$$x \cos t - y \sin t = -w (\sin t)^2 \tag{a}$$
I differentiate (1) wrt parameter $t$, considering variables $x$ and $y$ as constant, giving, after a sign change :
$$x \sin t + y \cos t = 2 w \sin t \cos t \tag{a'}$$
Solving system (a) + (a') gives the parametric equations of the envelope (black line) of moving line $A_tB_t$ :
$$\begin{cases}x&=&\tfrac12 w \sin(t)\sin(2t)\\y&=&\tfrac12 w \sin(t)(3+\cos(2t))\end{cases}\tag{1}$$
But in fact, we deal with a moving segment which is not the same as a moving line ; as a consequence, there is a kind of "competition" (I haven't found a better word) with the (small) elliptic arc issued from the locus of $B_t$, itself a quarter of an ellipse (red curve) with equation :
$$\begin{cases}x&=&h \sin(t)\\y&=& w \sin(t)+ h \cos(2t)\end{cases}\tag{2}$$
Please note that in (2), both parameters $w$ and $h$ are present whereas in (1), only $w$ is present.
In fact, we have to do the same kind of calculations for $B_tC_t$. Can you take it from here ?
Fig. 1 has been generated by the following Matlab program :
clear all;close all;hold on;
axis([0,2.4,0,2.4]);
axis equal tight
w=2;h=1; % rectangle's height and width
for t=0:pi/50:pi/2
b=w*cos(t);c=i*w*sin(t);d=h*i*exp(-i*t);
R=[b,b+d,c+d,c,b]; % rectangle : complex coord.
c='g';
if t==0;c='b';end;
plot(real(R),imag(R),'color','c','linewidth',2);
end
t=0:pi/50:pi/2;
plot(w*(sin(t).^2).*cos(t),w*sin(t).*(1+cos(t).^2),'k','linewidth',8);
plot(h*sin(t),w*sin(t)+h*cos(t),'r','linewidth',8)
text([0.,0.,w,w,h+0.05],[0.1,1.1,1.1,0.1,w],{'A_0','B_0','C_0','D_0','B_1'})
Comment : here is a different interpretation of your question with a true rotation around center $(w/2,w/2)$ :

Fig. 2 : Axes are in black. The center of rotation is materialized by a little circle.
The swept area can be computing rather easily because it can be decomposed into circular sectors.
(in fact, I don't think it is the kind of transformation you had in mind).
Matlab program having generated figure 2 :
clear all;close all;hold on;axis equal;
w=3;h=1; % length and width of rectangle
C=[w/2;w/2]; % center
plot(C(1),C(2),'ok')
R=[0,w,w,0,0
0,0,h,h,0]; % rectangle's vertices
for t=0:-pi/50:-pi/2
M=[cos(t),-sin(t);sin(t),cos(t)]; % rot. matrix
X=M*R+((eye(2)-M)*C)*ones(1,5); using transl. by C
plot(X(1,:),X(2,:),'linewidth',4);
end
plot([0,w],[0,0],'linewidth',8,'color','k');
plot([0,0],[0,w],'linewidth',8,'color','k'); % axes' plots