A square and a rectangle have the same perimeter. Let a be the area of the square and b be the area of the rectangle. Which of the following relations between a and b is true?

Introduction:
This is a conceptual question involving optimization of area for a given perimeter. Among all rectangles with a fixed perimeter, the one with maximum area is a square. Therefore, when a square and a rectangle share the same perimeter, the area of the square is greater than or equal to the area of the rectangle, and it is strictly greater unless the rectangle is also a square. This problem tests whether students understand this classical result and can apply it symbolically using the variables a and b for areas.

Given Data / Assumptions:

• There is a square with area a. • There is a rectangle with area b. • The perimeter of the square equals the perimeter of the rectangle. • The shapes are general; the rectangle need not be a square. • We must determine the correct relation between a and b.

Concept / Approach:
Let the side of the square be s, so its perimeter is 4s. If the rectangle has sides L and W, its perimeter is 2(L + W). Given that the perimeters are equal, we have 4s = 2(L + W), or L + W = 2s. The area of the square is s^2, whereas the area of the rectangle is L * W. For a given sum L + W, the product L * W is maximized when L = W, that is, when the rectangle becomes a square. Thus, s^2 is at least as large as L * W, and strictly larger for any non-square rectangle. So a > b in general.

Step-by-Step Solution:
Step 1: Let the square have side length s. Then its perimeter = 4s and its area a = s^2. Step 2: Let the rectangle have length L and breadth W. Its perimeter = 2(L + W). Step 3: Use the condition of equal perimeters. 4s = 2(L + W) ⇒ L + W = 2s. Step 4: Express the rectangle’s area in terms of L and W. Rectangle area b = L * W. Step 5: Use the inequality for a fixed sum. For a fixed sum S = L + W, the product L * W is maximized when L = W = S / 2. Here S = 2s, so maximum L * W occurs when L = W = s, giving b_max = s^2. Step 6: Conclude relation between a and b. The maximum possible area for the rectangle is s^2, which equals the square’s area, and any other rectangle with unequal sides has a smaller area than s^2. Therefore, in general, a ≥ b, and for a non-square rectangle, a > b. Since typical exam questions consider a general rectangle, we select a > b.

Verification / Alternative check:
Consider a concrete example. Let the common perimeter be 40 units. Then square side s = 40 / 4 = 10, so square area a = 10^2 = 100. For a rectangle with the same perimeter 40, choose L = 12 and W = 8. Then L + W = 20, perimeter = 40, and area b = 12 * 8 = 96. Here a = 100 and b = 96, so a > b. Trying other combinations like 11 and 9 gives area 99, still less than 100. Only when L = W = 10 does the rectangle become a square and match the area of the square, not exceed it.

Why Other Options Are Wrong:
Option a < b: This would mean that a rectangle with given perimeter could always exceed the square’s area, which contradicts the maximum area property of the square. Option a = b: This holds only when the rectangle itself is a square. The question, however, refers to a general rectangle. Option none of these: This is incorrect because we have a definite relationship a > b under the usual interpretation. Option a + b is constant: This has no basis in the problem statement and mixes two unrelated areas.

Common Pitfalls:
Students may incorrectly assume that if two shapes have the same perimeter, their areas must also be equal. Others may know the result but misinterpret which shape has the larger area. It is important to remember the key fact: for a given perimeter, the square has the maximum area among all rectangles. Careful use of this principle avoids confusion.

Final Answer:
The correct relation is a > b, that is, the square’s area is greater than the rectangle’s area for the same perimeter.