11) Given a square $A B C D$, we connect the midpoints of the sides having a common vertex, forming a new square $E F G H$. We repeat the same operation for $E F G H$ and obtain a new square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. What is the ratio between the area of $A B C D$ and the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$?
(A) 2,
(B) $2 \sqrt{2}$, (C) 4,
(D) $4 \sqrt{2}$,
(E) 8.
11. The answer is $(\mathbf{C})$.

The triangles $A H E, E B F, F C G$ and $G D H$ have the same area, therefore,

$$
\begin{aligned}
\operatorname{area}(E F G H) & =\operatorname{area}(A B C D)-4 \text { area }(A E H)= \\
& =\overline{A B}^{2}-4 \frac{\overline{A E} \times \overline{A H}}{2}=\overline{A B}^{2}-\frac{1}{2} \overline{A B}^{2}= \\
& =\frac{1}{2} \operatorname{area}(A B C D) .
\end{aligned}
$$

At each step of the construction, the value of the area of the square is halved, i.e., area $\left(A^{\prime} B^{\prime} C^{\prime} D^{\prime}\right)=\frac{1}{2} \operatorname{area}(E F G H)=$ $\frac{1}{4}$ area $(A B C D)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_3d973c0a184ce9fc7311g-3.jpg?height=414&width=402&top_left_y=407&top_left_x=1501)