Hello, hossein!

I have an approach . . . but is it acceptable?

Firstly the shape is a sqaure.

Lines have been drawn to 3 corners of the square from a point inside the square.

The smallest line is 3 units, the mid one is 4 units, and the largest is 5 units.

Place the square on a coordinate system.

Let the side of the square be . Code:

(0,a) (a,a)
A * - - - - - - - * B
| |
| * P |
| (x,y) |
| |
| |
| |
- C * - - - - - - - * D -
(0,0) (a,0)

Using the Distance Formula, we have:

. .

Subtract **(1)** from **(3)**: .

. . which simplifies to: . . **(4)**

Subtract **(1)** from **(2)**: .

. . which simplifies to: . .**(5)**

Substitute **(4)** and **(5)** into **(3)**: .

. . which simplifies to: . . . . a quadratic in .

Quadratic Formula: .

Then: .

. . The second root is too small be the side of the square.

Therefore, the side of the square is about units.

Edit: corrected errors in last four lines.

.