In a square PQRS, point T lie on side PS and point U lies on side QR, so that QT = TU = SU = 40 cm. Find the area of square PQRS.

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.

In a square PQRS, point T lies on side PS and point U lies on side QR, so that QT = TU = SU = 40 cm. Find the area of square PQRS.

Learn how to find the area of square PQRS where points T and U divide sides PS and QR, respectively. Using the properties of isosceles right triangles, this solution involves key geometry concepts and the 45°-45°-90° triangle theorem. The area is calculated as 3200 cm², suitable for students in Grades 8-10.

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