A square sheet was folded twice along its diagonals, resulting in a triangle. Then, a straight cut was made on the folded sheet, parallel to the larger side of this triangle, passing through the midpoints of the other sides, as shown in the illustration.
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-003.jpg?height=158&width=798&top_left_y=1998&top_left_x=710)

Unfolding the sheet, a square hole was obtained in the middle of the sheet. The area of the hole corresponds to which fraction of the area of the entire original square sheet?
(a) $\frac{1}{2}$
(b) $\frac{1}{6}$
(c) $\frac{3}{8}$
(d) $\frac{3}{4}$
(e) $\frac{1}{4}$
The correct option is (e).

Solution 1: Let $\triangle A B C$ be the triangle obtained after folding the original square along its two diagonals, and let $M N$ be the cut along the mid-base of this triangle, parallel to the side $B C$, which is one of the sides of the original square. The area of the original square is
$(B C)^{2}$. Unfolding the sheet, we see that the hole is a square with side $M N$ and, since $M N=\frac{1}{2} B C$, its area is

$$
(M N)^{2}=\left(\frac{1}{2} B C\right)^{2}=\frac{1}{4}(B C)^{2}
$$

Thus, the hole has one quarter of the area of the original square.
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-040.jpg?height=408&width=1264&top_left_y=617&top_left_x=434)

Solution 2: The cut is made along the mid-base of the triangle, removing a small triangle similar to the original, with a similarity ratio of $1 / 2$. Thus, the area of the removed triangle is one quarter of the area of the original triangle. Unfolding the sheet, we see this situation repeated four times, so the hole has one quarter of the area of the original square.