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. $P^2\ge 16A$
• D. $P^2\le 4A$
• E. $A-P<2$

Answer 1

Answer: (C)

$P^2\ge 16A$

Let area and perimeter of a rectangle of sides $x$ and $y$ be $A$ and $P$.

$\therefore A=xy,P=2(x+y)$

Using $A.M.\ge G.M.$ inequality for $x$ & $y$

Since, ${(x+y)}^2\ge 4xy$

$\Rightarrow {\left(\frac{P}{2}\right)}^2\ge 4A$

$\Rightarrow P^2\ge 16A$