Consider the following statements: Every square is a rectangle. Eve

From statement 1 and 2, we can form a chain: Square -> Rectangle -> Parallelogram.
This implies that every square is a parallelogram.

Now let’s analyze the conclusions:
A) All parallelograms are either squares or rectangles. This is false. A rhombus is a parallelogram but is not necessarily a square or a rectangle.
B) A non-parallelogram figures cannot be either a square or a rectangle. This is true.
From Rectangle -> Parallelogram, the contrapositive is (Not Parallelogram) -> (Not Rectangle).
From Square -> Rectangle, the contrapositive is (Not Rectangle) -> (Not Square).
Combining these, if a figure is not a parallelogram, then it is not a rectangle. If it is not a rectangle, then it is not a square. Therefore, if a figure is a non-parallelogram, it cannot be a rectangle, and thus cannot be a square. This conclusion is logically derived from the first two statements.
C) All rectangles are either squares or parallelograms. This statement is technically true because all rectangles *are* parallelograms (Statement 2), and the set of squares is a subset of rectangles. So, a rectangle is always in the set of parallelograms, and sometimes in the set of squares. However, the wording “either…or” often implies mutual exclusion or at least that being a parallelogram is not always the case for a rectangle, which contradicts statement 2. Conclusion B is a more direct and stronger inference from the combined statements.
D) Squares and rectangles are the only parallelograms. This is false. Rhombuses and other non-rectangular parallelograms exist.

Conclusion B is the only valid and direct conclusion that can be drawn based on the transitive property implied by the statements.