To find the area of square $PQRS$ where points $T$ and $U$ lie on sides $PS$ and $QR$ respectively, and given that $QT = TU = SU = 40 \, \text{cm}$, we can proceed as follows:

1. Identify Variables:

   • Let the side length of the square $PQRS$ be $s$.
   • Since $QT = TU = SU = 40 \, \text{cm}$, $\triangle QTU$ is an isosceles right triangle (right angle at $T$ or $U$ due to symmetry).

2. Using the Properties of the Square and Triangle:

   • Since $QT = TU = SU$, the distances form an isosceles right triangle, suggesting that the segment $SU$ is the hypotenuse of $\triangle QTU$.
   • In a 45°-45°-90° triangle, the sides are in the ratio $1:1:\sqrt{2}$, meaning: $QT = TU = x \quad \text{and} \quad SU = x\sqrt{2}.$

3. Solve for $x$:

   • Given $SU = 40 \, \text{cm}$, we can set up the equation: $x\sqrt{2} = 40.$
   • Solving for $x$, we get: $x = \frac{40}{\sqrt{2}} = 20\sqrt{2} \, \text{cm}.$

4. Relate $s$ to $x$:

   • Since $QT = x = 20\sqrt{2} \, \text{cm}$ and represents half the side length $s$ (as $T$ and $U$ lie symmetrically), $s = 2 \times 20\sqrt{2} = 40\sqrt{2} \, \text{cm}.$

5. Calculate the Area of $PQRS$:

   • The area $A$ of square $PQRS$ is: $A = s^2 = (40\sqrt{2})^2 = 1600 \times 2 = 3200 \, \text{cm}^2.$

Final Answer

The area of square $PQRS$ is $\mathbf{3200 \, \text{cm}^2}$.

Would you like further details or have any questions?

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Follow-up Questions:

1. What is the perimeter of square $PQRS$ based on this side length?
2. How would the area change if $QT = 50 \, \text{cm}$ instead of $40 \, \text{cm}$?
3. Can you find the coordinates of points $T$ and $U$ if $P$ is at $(0,0)$?
4. What is the length of diagonal $PR$ in terms of $s$?
5. How does this setup demonstrate properties of 45°-45°-90° triangles in geometry?

Tip:

For problems involving isosceles right triangles, remember the $1:1:\sqrt{2}$ ratio for quick calculations of side lengths based on one known side.