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16 Sep 2014, 01:26
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D
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Difficulty:
95%
(hard)
Question Stats:
35% (02:07) correct
65%
(02:00)
wrong
based on 150
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Is the perimeter of a triangle with sides \(a\), \(b\) and \(c\) greater than 30?
(1) \(a-b=15\).
(2) The area of the triangle is 50.
(1) \(a-b=15\).
(2) The area of the triangle is 50.
16 Sep 2014, 01:26
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Official Solution:
This is a 700+ question.
(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.
(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.
Answer: D
This is a 700+ question.
(1) \(a-b=15\). Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.
(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.
Answer: D
General Discussion
01 Aug 2016, 05:19
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I think this is a high-quality question and I agree with explanation.
