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 $\displaystyle a$. Code:

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

Using the Distance Formula, we have:

. . $\displaystyle \begin{array}{ccccc}PA = 3: & x^2 + (y-a)^2 & = & 9 & {\bf (1)} \\

PB = 4: & (x-a)^2 + \,(y-a)^2 & = & 16 & {\bf(2)} \\

PC = 5: & x^2 + \,y^2 & = & 25 & {\bf(3)}\end{array}$

Subtract **(1)** from **(3)**: .$\displaystyle y^2 - (y-a)^2 \:=\:16$

. . which simplifies to: .$\displaystyle y \:=\:\frac{a^2+16}{2a}$ . **(4)**

Subtract **(1)** from **(2)**: .$\displaystyle (x-a)^2-x^2 \:=\:7$

. . which simplifies to: .$\displaystyle x \:=\:\frac{a^2-7}{2a}$ .**(5)**

Substitute **(4)** and **(5)** into **(3)**: .$\displaystyle \left(\frac{a^2-7}{2a}\right)^2 + \left(\frac{a^2+16}{2a}\right)^2\:=\:25$

. . which simplifies to: .$\displaystyle 2a^4 - 82a^2 + 305 \:=\:0$ . . . a quadratic in $\displaystyle a^2$.

Quadratic Formula: .$\displaystyle a^2 \;=\;\frac{-(\text{-}82) \pm\sqrt{(\text{-}82)^2 - 4(2)(305)}}{2(2)} \;=\;\frac{82 \pm \sqrt{4284}}{4}\;=\;\frac{41 \pm\sqrt{1071}}{2} $

Then: .$\displaystyle a \;=\;\sqrt{\frac{41 \pm\sqrt{1071}}{2}}\;=\;\begin{Bmatrix}6.07149637 \\ 2.033944893 \end{Bmatrix}$

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

Therefore, the side of the square is about $\displaystyle 6.07$ units.

Edit: corrected errors in last four lines.

.