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01 Nov 2010, 05:41
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Question Stats:
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43%
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Is the length of the diagonal of the rectangle greater than \(\sqrt{6}\) centimeters ?
(1) The shorter side of the rectangle measures 2 centimeters.
(2) The longer side of the rectangle measures 3 centimeters.
(1) The shorter side of the rectangle measures 2 centimeters.
(2) The longer side of the rectangle measures 3 centimeters.
Hi,
In Geomtry I, question 3 the DS question is:
Question: The OA assumes sides as integers, otherwise it should be another answer. Why is assuming integers and not decimals?
Thank you,
In Geomtry I, question 3 the DS question is:
Question: The OA assumes sides as integers, otherwise it should be another answer. Why is assuming integers and not decimals?
Thank you,
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01 Nov 2010, 06:02
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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).
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).
02 Nov 2018, 18:03
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vivaslluis
\(a \ge b > 0\,\,\,\,\,\left[ {{\rm{rectangle}}\,\,{\rm{dimensions}}} \right]\)
\({a^2} + {b^2}\,\,\mathop > \limits^? \,\,6\)
\(\left( 1 \right)\,\,a > b = 2\,\,\,\, \Rightarrow \,\,\,{a^2} + {b^2} > {2^2} + {2^2} = 8\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
\(\left( 2 \right)\,\,3 = a > b > 0\,\,\,\, \Rightarrow \,\,\,{a^2} + {b^2} > {3^2} = 9\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
