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 required operations step by step.

### Step 1: Define the Sets
- Let \( S \) be the set of all squares.
- Let \( R \) be the set of all rectangles.

### Step 2: Understand the Union of Sets
- The union of two sets \( A \) and \( B \), denoted \( A \cup B \), is the set containing all elements from both sets without duplication.
- Since all squares are rectangles, we can say that \( S \subseteq R \). Therefore, the union of \( S \) and \( R \) can be expressed as:
\[ S \cup R = R \]
because adding squares to rectangles does not introduce any new elements outside of rectangles.

### Step 3: Calculate \( S \cup S \)
- The union of a set with itself is simply the set itself. Thus:
\[ S \cup S = S \]

### Step 4: Perform the Set Difference
- We need to find the difference between the union of \( S \) and \( R \) and the union of \( S \) with itself:
\[ (S \cup R) - (S \cup S) = R - S \]
- This means we are looking for all elements that are in \( R \) (rectangles) but not in \( S \) (squares).

### Step 5: Interpret the Result
- The result \( R - S \) represents the set of all rectangles that are not squares.

### Final Answer
Thus, the final answer is:
\[ R - S \text{ (the set of all rectangles that are not squares)} \]