Let $\gamma(t)=(x(t),y(t))$ be a positively-oriented simple closed curve in $\mathbb{R}^2$ with period $a$. The area of its interior is given by
$$A(\mathrm{int}(\gamma))=\int\int_{\mathrm{int}(\gamma)}dxdy=\frac{1}{2}\int_0^a(x\dot y-y\dot x)dt$$
The last line integral is obtained by applying Green’s Theorem to the second double integral. A rigid motion can be given by $M=T_b\circ R_{\theta}$ where $R_{\theta}=\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}$ is a rotation by an angle $\theta$ and $T_b$ is a translation by a constant vector $b=(b_1,b_2)$. So $\tilde\gamma=M(\gamma)$ is given by
$$\tilde\gamma=(\tilde x,\tilde y)=(x\cos\theta-y\sin\theta+b_1,x\sin\theta+y\cos\theta+b_2)$$
and
\begin{align*}\tilde x\dot{\tilde y}-\tilde y\dot{\tilde x}&=x\dot y-y\dot x+b_1(\dot x\sin\theta+\dot y\cos\theta)-b_2(\dot x\cos\theta-\dot y\sin\theta)\\&=x\dot y-y\dot x+b\cdot\begin{pmatrix}
\sin\theta & \cos\theta\\
-\cos\theta & \sin\theta
\end{pmatrix}\begin{pmatrix}
\dot x\\
\dot y
\end{pmatrix}\\&=x\dot y-y\dot x+b\cdot R_{\theta-\frac{\pi}{2}}(\dot\gamma)\end{align*}
\begin{align*}\int_0^a b\cdot R_{\theta-\frac{\pi}{2}}(\dot\gamma(t))dt&=b\cdot R_{\theta-\frac{\pi}{2}}\left(\int_0^a\dot\gamma(t)dt\right)\\
&=0\end{align*}
since $\int_0^a\dot\gamma(t)dt=\gamma(a)-\gamma(0)=0$. ($\gamma(t)$ is a simple closed curve with period $a$.) Therefore, $A(\mathrm{int}(\gamma))=A(\mathrm{int}(M(\gamma))$.