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Let's analyze and solve the problem.

We are given:

- The perimeter of square \( S \) is 40.
- Square \( T \) is inscribed in square \( S \), meaning the vertices of square \( T \) touch the midpoints of the sides of square \( S \).
- We need to find the least possible area of square \( T \).

### Step 1: Find the side length of square \( S \)
The perimeter of square \( S \) is 40. Since the perimeter \( P \) of a square is given by:

\[
P = 4 \times \text{side length of square S}
\]

Let the side length of square \( S \) be \( s_S \). So, we have:

\[
4 \times s_S = 40
\]

Solving for \( s_S \):

\[
s_S = \frac{40}{4} = 10
\]

So, the side length of square \( S \) is 10 units.

### Step 2: Find the relationship between square \( T \) and square \( S \)
Since square \( T \) is inscribed in square \( S \), the diagonal of square \( T \) is equal to the side length of square \( S \). The diagonal \( d_T \) of a square is related to its side length \( s_T \) by the formula:

\[
d_T = s_T \sqrt{2}
\]

Since the diagonal of square \( T \) equals the side length of square \( S \) (which is 10 units), we have:

\[
s_T \sqrt{2} = 10
\]

Solving for \( s_T \), the side length of square \( T \):

\[
s_T = \frac{10}{\sqrt{2}} = \frac{10 \times \sqrt{2}}{2} = 5\sqrt{2}
\]

### Step 3: Find the area of square \( T \)
The area \( A_T \) of a square is given by:

\[
A_T = s_T^2
\]

Substituting \( s_T = 5\sqrt{2} \):

\[
A_T = (5\sqrt{2})^2 = 25 \times 2 = 50
\]

### Conclusion:
The least possible area of square \( T \) is \( \boxed{50} \).

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. What is the diagonal of square \( S \)?
2. If the side length of square \( S \) was 20, what would be the area of the inscribed square?
3. How does the area of square \( T \) change if the perimeter of square \( S \) is doubled?
4. What is the ratio of the area of square \( T \) to the area of square \( S \)?
5. How would the area of square \( T \) change if the diagonal of square \( S \) was 14?

**Tip:** Whenever dealing with inscribed shapes, the relationship between their diagonals and side lengths is often key to solving area-related problems!

The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible area of square T?

Learn how to find the least possible area of a square inscribed in another square with a perimeter of 40. The solution covers key geometric concepts such as the properties of squares, calculating diagonals, and the use of the Pythagorean Theorem. Suitable for students in Grades 8-10.

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