10,11 |
| :---: | :---: | :---: |
|  | [ Design helps solve the task] |  |
|  | $\left[\begin{array}{l}\text { Theorem of three perpendiculars } \\ {[\text { Pythagorean Theorem (direct and inverse).] }}\end{array}\right]$ |  |

Given a cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge 1. On its edges $A B, B C, C^{\prime} D^{\prime}$, and $D^{\prime} A^{\prime}$, points $K, L, M$, and $N$ are marked respectively such that $K L M N$ is a square.

Find its area.
The projection of the square onto the plane of face $ABCD$ is a parallelogram, and by the theorem of three perpendiculars, it is even a rectangle $KLM'N'$ (see the left figure).
![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-51.jpg?height=696&width=1708&top_left_y=0&top_left_x=173)

Since the opposite vertices of the rectangle lie on opposite sides of the square, the center of the rectangle lies at the intersection of the midlines of the square, which coincides with the center $O$ of the square $ABCD$.

A circle with center $O$ and diameter equal to the diagonal of the rectangle intersects each side of the square in no more than two points. Clearly, there are four inscribed rectangles in the square with these vertices: in two of them, the diagonals of the rectangle are parallel to the diagonals of the square, and in the other two, the rectangle is a square (see the right figure). In our case, the latter is impossible, since $KN' < KN = KL$.

Let $KL = KN = a$, $AK = AN' = x$, then $BK = 1 - x$ (see the left figure). From the right triangle $KNN'$, we have $a^2 = 2x^2 + 1$, and from the right triangle $KBL$, we have $a^2 = 2(1 - x)^2 = 2 - 4x + 2x^2$. Therefore, $2 - 4x = 1$, so $x = 1/4$. Thus, $S_{KLMN} = a^2 = 9/8$.

## Answer

$1.125$.