If S is the set of squares and R is the set of rectangles, then `(SuuR)-(SuuS)` is

Options

A. S
B. R
C. set of squares but not rectangles
D. set of rectangles but not squares

Answer

To solve the problem, we need to analyze the sets involved and perform the operations as described in the question.

### Step-by-Step Solution:

1. **Identify the Sets**:
- Let \( S \) be the set of squares.
- Let \( R \) be the set of rectangles.

2. **Union of Sets**:
- The union of two sets \( A \) and \( B \), denoted as \( A \cup B \), is the set of elements that are in \( A \), in \( B \), or in both.
- Therefore, \( S \cup R \) is the set of all squares and rectangles.

3. **Union of Set with Itself**:
- The union of a set with itself, \( S \cup S \), is simply \( S \). This is because the union does not add any new elements; it just combines the elements of the same set.
- Thus, \( S \cup S = S \).

4. **Subtracting the Sets**:
- We need to compute \( (S \cup R) - (S \cup S) \).
- Since \( S \cup S = S \), we can rewrite this as \( (S \cup R) - S \).

5. **Understanding the Result**:
- The operation \( (S \cup R) - S \) means we take the set of all squares and rectangles and remove all squares from it.
- What remains is the set of rectangles that are not squares. This is simply the set of rectangles \( R \) minus any elements that are also in \( S \).

6. **Final Result**:
- Therefore, \( (S \cup R) - (S \cup S) = R - S \), which is the set of rectangles that are not squares.

### Conclusion:
The final result is the set of rectangles that are not squares, which can be denoted as \( R - S \).