A square and a rectangle have equal areas. If their perimeters are p and q respectively, what is the correct relationship between p and q?

Introduction / Context:
This question compares a square and a rectangle that share the same area but may have different shapes. The focus is on understanding how perimeter behaves when the area is fixed. In geometry and optimization, a well known fact is that among all rectangles of a given area, the square has the smallest perimeter. This property is the key to answering the question about the relationship between p and q.

Given Data / Assumptions:

• A square has perimeter p and area A.
• A rectangle has perimeter q and the same area A as the square.
• The rectangle is not stated to be a square, so in general its sides are unequal.
• All figures are standard plane shapes with right angles.
• We assume positive side lengths and non degenerate rectangles and squares.

Concept / Approach:
For a fixed area, the square is the most compact rectangle, meaning that it has the minimum possible perimeter. Any other rectangle with the same area but unequal sides will require a longer total boundary to enclose the same area. Therefore, when the areas are equal, the perimeter of the square must be less than or at most equal to that of the rectangle. Since the rectangle is not forced to be a square, the strict inequality p < q is the correct general statement.

Step-by-Step Solution:
Let the side of the square be s. Then its area is A = s^2 and perimeter p = 4s.For the rectangle, let its sides be a and b. Then area A = a * b and perimeter q = 2(a + b).Given that the square and rectangle have equal areas, we have s^2 = a * b.For a rectangle with fixed product a * b = s^2, it is known that a = b = s gives the minimum value of a + b.Thus, any rectangle with unequal sides has a + b greater than 2s, which implies q = 2(a + b) > 4s = p.

Verification / Alternative check:
Take a numerical example. Suppose the square has side s = 4 units. Then area = 16 and perimeter p = 16 units. Consider a rectangle with area 16 but sides 2 and 8. Its perimeter is q = 2(2 + 8) = 20 units, which is greater than 16. Other combinations like 1 and 16 give an even larger perimeter of 34 units. In each case, as long as the rectangle is not a square, its perimeter is larger than that of the square with the same area, confirming p < q.

Why Other Options Are Wrong:
Option p > q contradicts the minimum perimeter property of the square. Option p = q would hold only if the rectangle also happens to be a square, which is not guaranteed by the problem statement. Option p ≥ q is too weak, because it allows p = q as the only possibility, and does not reflect the strict inequality for non square rectangles. Option p ≤ q is incomplete and ambiguous because equality only occurs in the special case when both shapes are squares, which is not the general scenario described.

Common Pitfalls:
Some learners confuse area minimization with perimeter minimization and mistakenly think that for equal area, all rectangles have the same perimeter. Others assume that because the rectangle can be very elongated, it might somehow reduce perimeter, which is the opposite of what actually happens. Forgetting the inequality property of arithmetic and geometric means also leads to confusion about which shape is more compact.

Final Answer:
The correct relationship is p < q, meaning that the square has a smaller perimeter than the rectangle when both have equal area.