Question

The area and perimeter of a rectangle are \(A\) and \(P\) respectively. Then \(P\) and \(A\) satisfy the inequality

• A. \( P + A > PA \)
• B. \( P^2 \le A \)
• C. \( A - P < 2 \)
• D. \( P^2 \le 4A \)
• E. \( P^2 \ge 16A \)

Hint:

For rectangles, minimum perimeter for fixed area occurs when it becomes a square.

Answer 1

Answer: (E)

Concept: For a rectangle with length \(l\) and breadth \(b\): \[ A = lb, \quad P = 2(l+b) \] We use inequality: \[ (l+b)^2 \ge 4lb \] which comes from AM-GM inequality.

Step 1: Express perimeter in terms of sides
\[ P = 2(l+b) \Rightarrow P^2 = 4(l+b)^2 \]

Step 2: Apply AM-GM inequality
\[ l+b \ge 2\sqrt{lb} \] Squaring both sides: \[ (l+b)^2 \ge 4lb \]

Step 3: Substitute area
\[ (l+b)^2 \ge 4A \] \[ 4(l+b)^2 \ge 16A \]

Step 4: Replace with \(P\)
\[ P^2 \ge 16A \]

Step 5: Interpretation
Equality holds when: \[ l = b \] i.e., rectangle becomes a square.

Step 6: Final Answer
\[ \boxed{P^2 \ge 16A} \]