Does rectangle A have a greater perimeter than rectangle B?

Does rectangle A have a greater perimeter than rectangle B?

Let the sides of rectangle A be \(s\) and \(t\) and the side of rectangle B \(m\) and \(n\).

Question: is \(2(s+t)>2(m+n)\)? --> or is \(s+t>m+n\)?


(1) The length of a side of rectangle A is twice the length of a side of rectangle B --> \(s=2m\), clearly insufficient as no info about the other side of rectangles.


(2) The area of rectangle A is twice the area of rectangle B --> \(st=2mn\), also insufficient as if \(s=t=2\), \(m=1\) and \(n=2\) then the answer would be YES, but if \(s=t=2\), \(m=\frac{1}{2}\) and \(n=4\) then the answer would be NO.


(1)+(2) \(s=2m\) and \(st=2mn\) --> substitute \(s\): \(2m*t=2mn\), so \(t=n\). Thus as \(s=2m\) and \(t=n\): \(s+t=2m+n\) which is obviously more than \(m+n\). Sufficient.


Answer: C.