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

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


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

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


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

Regards,
Fabio.
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