## Task B-3.6.

Inside the square $A B C D$ there is a point $M$ such that $|M A|=7 \mathrm{~cm}, |M B|=13 \mathrm{~cm}$ and $|M C|=17 \mathrm{~cm}$. Calculate the area of the square $A B C D$.
## Solution.

Let $a$ be the length of the side of the square. Set up the coordinate system as shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_30_c3f5968a3447b3c0b4b2g-13.jpg?height=564&width=745&top_left_y=580&top_left_x=663)

For the accurate image with all points labeled with coordinates:

(2 points)

If $(x, y)$ are the coordinates of point $M$, then:

$$
\begin{aligned}
& x^{2}+y^{2}=49 \\
& (x-a)^{2}+y^{2}=169 \\
& (x-a)^{2}+(y-a)^{2}=289
\end{aligned}
$$

After simplification, we get:

$$
\begin{aligned}
& a^{2}-2 a y=120 \\
& a^{2}-2 a x=120
\end{aligned}
$$

Thus, $x=y$. Therefore, point $M$ lies on the diagonal of the square.

The length of the diagonal of the square is $7+17=24$.

The area of the square is $P=\frac{1}{2} d^{2}=\frac{1}{2} 24^{2}=288$.

## Note:

The problem can also be solved without a coordinate system. Using the Pythagorean theorem, a similar system of equations can be derived. Then, grading is done similarly: 5 points for setting up the system and 5 points for solving it, including calculating the area.