A) \[P+A>PA\]
B) \[{{P}^{2}}\le A\]
C) \[A-P<2\]
D) \[{{P}^{2}}\ge 16A\]
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 \[{{(x+y)}^{2}}\ge 4xy\] \[\Rightarrow {{\left( \frac{P}{2} \right)}^{2}}\ge 4A\Rightarrow {{P}^{2}}\ge 16A.\]
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