Perimeter comparison for equal areas:\nA square and a (non-square) rectangle have equal areas, with perimeters p and q respectively. Determine the correct relation between p and q.

Introduction / Context:
(Recovery–First applied) The square minimizes perimeter for a given area among all rectangles. For equal areas, a non-square rectangle must have strictly larger perimeter than the square. We repair options to reflect the correct inequality, choosing the strict form when the rectangle is not a square.

Given Data / Assumptions:

• Square area = Rectangle area.
• Rectangle is not a square (otherwise perimeters would be equal).

Concept / Approach:
For fixed area A, by AM–GM (or calculus), the rectangle with minimum perimeter is the square. Thus p (square) is minimal, and any non-square rectangle has q > p.

Step-by-Step Solution (idea):

Verification / Alternative check:
Pick A = 100. Square 10×10 has p = 40. Rectangle 5×20 has q = 50 > 40.

Why Other Options Are Wrong:
p > q contradicts the extremal property; p = q holds only if the rectangle is also a square; “none” is unnecessary since the strict inequality is known.

Common Pitfalls:
Assuming equal areas imply equal perimeters; perimeter depends on side distribution, not area alone.

Final Answer:
p < q