You don't need to assume sides as integers.
Statement (1) The shorter side of the rectangle is 2 which means the greater side will be more than 2. Let us say it is just a little more than 2, say 2.01.
The diagonal will be \(\sqrt{{2^2 + (2.01)^2}} = \sqrt{8.04}.\) This is definitely greater than \(\sqrt{6}\). So it doesn't matter what the greater side is, the diagonal will be greater than root 6. Statement (1) is sufficient.
Statement (2) The greater side of the rectangle is 3. The smaller side can be as small as possible, let's say a little more than 0. Still, the diagonal will be \(\sqrt{{3^2 + (0.01)^2}} = \sqrt{9.0001}.\) This is definitely greater than \(\sqrt{6}\). Therefore, statement (2) is sufficient.
Answer (D).
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Hi Karishma,

Once posted, I realized it was talking about the diagonal, while I was thinking in the area.
Apologize for my mistake.

Thanks!

Is this really 700+ question?
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maybe during exam condition it can be.

The difficulty level is calculated automatically based on the timer stats from the users which attempted the question.
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i'm sure it's not a 700 level question. I solved it just by applying logic...

1. if short one is 2, then we can minimize the longer one to be 2 for example too... a^2 + b^2 = c^2. so we have a=2, b=2, and c^2 = 8. c is sqrt(8). we can give only 1 answer so sufficient.
2. if the bigger one is 3, the other one doesn't matter, as 3^2 = 9. and 9+ smth squared will always be bigger than sqrt(6)... sufficient.

D is the answer.

\(a \ge b > 0\,\,\,\,\,\left[ {{\rm{rectangle}}\,\,{\rm{dimensions}}} \right]\)

\({a^2} + {b^2}\,\,\mathop > \limits^? \,\,6\)


\(\left( 1 \right)\,\,a > b = 2\,\,\,\, \Rightarrow \,\,\,{a^2} + {b^2} > {2^2} + {2^2} = 8\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\)

\(\left( 2 \right)\,\,3 = a > b > 0\,\,\,\, \Rightarrow \,\,\,{a^2} + {b^2} > {3^2} = 9\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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