Equations And Inequalities Question 5
Question: The area and perimeter of a rectangle are A and P respectively. Then P and A satisfy the inequality.
Options:
A) $ P+A>PA $
B) $ P^{2}\le A $
C) $ A-P<2 $
D) $ P^{2}\ge 16A $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let x be the length and y be the breadth of a rectangle. $ A=xy,,P=2(x+y) $ Since $ {{(x+y)}^{2}}\ge 4xy $
$ \Rightarrow {{( \frac{P}{2} )}^{2}}\ge 4A\Rightarrow P^{2}\ge 16A. $
BETA
