Is the length of the diagonal of the rectangle greater than 6^(1/2)

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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Good question where you simply need to avoid super calculations and be simple as the GMAT demands.

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Official Solution:


Is the length of the diagonal of the rectangle greater than \(\sqrt{6}\) centimeters ?

(1) The shorter side of the rectangle measures 2 centimeters.

Since the length of the shorter side of the rectangle is 2 centimeters, the length of the diagonal must be greater than \(\sqrt{2^2 + 2^2} = \sqrt{8}\) centimeters. Therefore, the length of the diagonal of the rectangle is indeed greater than \(\sqrt{6}\) centimeters. Sufficient.

(2) The longer side of the rectangle measures 3 centimeters.

With the longer side of the rectangle measuring 3 centimeters, the length of the diagonal must be greater than \(\sqrt{0^2 + 3^2} = \sqrt{9}\) centimeters. Therefore, the length of the diagonal of the rectangle is also greater than \(\sqrt{6}\) centimeters. Sufficient.


Answer: D
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