16. As shown in the figure, the square $\mathrm{ABCD}$ with side length 1 is rotated counterclockwise by $30^{\circ}$ around point $\mathrm{A}$ to the position of square $\mathrm{AB}^{\prime} \mathrm{C}^{\prime} \mathrm{D}^{\prime}$, then the area of the shaded part in the figure is ( )
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{3}}{3}$
(C) $1-\frac{\sqrt{3}}{3}$
(D) $1-\frac{\sqrt{3}}{4}$
Answer: $(\mathrm{C})$
Solution: As shown in the figure, let the intersection of $\mathrm{B}^{\prime} \mathrm{C}^{\prime}$ and $\mathrm{CD}$ be $\mathrm{E}$, and connect $\mathrm{AE}$,
In Rt $\triangle \mathrm{AB}^{\prime} \mathrm{E}$ and Rt $\triangle \mathrm{ADE}$,
$$
\left\{\begin{array}{l}
\mathrm{AE}=\mathrm{AE} \\
\mathrm{AB}^{\prime}=\mathrm{AD}
\end{array},\right.
$$

Therefore,
Rt $\triangle \mathrm{AB}^{\prime} \mathrm{E} \cong \mathrm{Rt} \triangle \mathrm{ADE}$,

We can get
$$
\angle \mathrm{DAE}=\angle \mathrm{B}^{\prime} \mathrm{AE},
$$

Since the rotation angle is $30^{\circ}$,
Therefore,
$$
\begin{array}{c}
\angle \mathrm{DAB}^{\prime}=60^{\circ}, \\
\angle \mathrm{DAE}=\frac{1}{2} \times 60^{\circ}=30^{\circ}, \\
\mathrm{DE}=1 \times \frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{3},
\end{array}
$$

Thus,
The area of the shaded part $=1 \times 1-2 \times\left(\frac{1}{2} \times 1 \times \frac{\sqrt{3}}{3}\right)=1-\frac{\sqrt{3}}{3}$. Therefore, the answer is (C).