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Re M1704 [#permalink]
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16 Sep 2014, 00:00



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Re: M1704 [#permalink]
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02 Oct 2014, 02:53
Bunuel wrote: Official Solution:
(1) \(2AB = 3BC = 4AC\). We have the ratio of the sides: \(AB:BC:AC=6:4:3\). Now, ALL triangles with this ratio are similar and have the same fixed angles. No matter what these angles actually are, the main point is that we can get them and thus answer the question. Sufficient. (2) \(AC^2 + AB^2 \gt BC^2\). This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at \(B\) or \(C\) (so in this case one angle will be 90 degrees, for example consider a right triangle: \(AC=5\), \(AB=4\) and \(BC=3\)). Not sufficient.
Answer: A Is it right to say that the triangle is acute (and not right triangle) if \(BC^2 < AC^2 + AB^2\) and BC is the longest side of the triangle ? Also, I did not understand Statement (1). How can we get the angles from sides? Is there a formula?



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Re: M1704 [#permalink]
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02 Oct 2014, 02:58
b2bt wrote: Bunuel wrote: Official Solution:
(1) \(2AB = 3BC = 4AC\). We have the ratio of the sides: \(AB:BC:AC=6:4:3\). Now, ALL triangles with this ratio are similar and have the same fixed angles. No matter what these angles actually are, the main point is that we can get them and thus answer the question. Sufficient. (2) \(AC^2 + AB^2 \gt BC^2\). This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at \(B\) or \(C\) (so in this case one angle will be 90 degrees, for example consider a right triangle: \(AC=5\), \(AB=4\) and \(BC=3\)). Not sufficient.
Answer: A Is it right to say that the triangle is acute (and not right triangle) if \(BC^2 < AC^2 + AB^2\) and BC is the longest side of the triangle ? Also, I did not understand Statement (1). How can we get the angles from sides? Is there a formula? If the lengths of the sides of a triangle are a, b, and c, where the largest side is c, then: For a right triangle: \(a^2 +b^2= c^2\). For an acute (a triangle that has all angles less than 90°) triangle: \(a^2 +b^2>c^2\). For an obtuse (a triangle that has an angle greater than 90°) triangle: \(a^2 +b^2<c^2\). As for your other question: you don't need to know any formula. It's enough to know that you CAN find the angles because the triangle will be fixed.
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Re: M1704 [#permalink]
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02 Oct 2014, 12:47
Hi Bunuel, I did not understand how this condition is true for a right triangle. For right angle triangle it should be AC^2+AB^2=BC^2. Am i missing anything here? (2) AC^2 + AB^2 \gt BC^2. This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at B or C (so in this case one angle will be 90 degrees, for example consider a right triangle: AC=5, AB=4 and BC=3). Not sufficient. Thank you Bunuel wrote: Official Solution:
(1) \(2AB = 3BC = 4AC\). We have the ratio of the sides: \(AB:BC:AC=6:4:3\). Now, ALL triangles with this ratio are similar and have the same fixed angles. No matter what these angles actually are, the main point is that we can get them and thus answer the question. Sufficient. (2) \(AC^2 + AB^2 \gt BC^2\). This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at \(B\) or \(C\) (so in this case one angle will be 90 degrees, for example consider a right triangle: \(AC=5\), \(AB=4\) and \(BC=3\)). Not sufficient.
Answer: A
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Re: M1704 [#permalink]
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02 Oct 2014, 14:29
sunita123 wrote: Hi Bunuel, I did not understand how this condition is true for a right triangle. For right angle triangle it should be AC^2+AB^2=BC^2. Am i missing anything here? (2) AC^2 + AB^2 \gt BC^2. This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at B or C (so in this case one angle will be 90 degrees, for example consider a right triangle: AC=5, AB=4 and BC=3). Not sufficient. Thank you Bunuel wrote: Official Solution:
(1) \(2AB = 3BC = 4AC\). We have the ratio of the sides: \(AB:BC:AC=6:4:3\). Now, ALL triangles with this ratio are similar and have the same fixed angles. No matter what these angles actually are, the main point is that we can get them and thus answer the question. Sufficient. (2) \(AC^2 + AB^2 \gt BC^2\). This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at \(B\) or \(C\) (so in this case one angle will be 90 degrees, for example consider a right triangle: \(AC=5\), \(AB=4\) and \(BC=3\)). Not sufficient.
Answer: A AC^2+AB^2=BC^2 is true for a right triangle where BC is hypotenuse. Think what would it be if for example, AC is hypotenuse.
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Re: M1704 [#permalink]
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17 Nov 2014, 13:06
Hi Sunita, If you still do not understand, you can refer to another explanation here: Statement 1. Consider 2AB=3BC=4AC=12K WHERE K IS CONSTANT NOW 2AB=12K;AB=6K SIMILARLY BC=4K AND AC=3K SO AB:BC:AC=6:4:3 THIS IMPLIES ANGLE ACB IS MORE THAN 90 DEG(THIS ANGLE WOULD HAVE BEEN 90DEG IF THE RATIO WAS 5:4:3 AND LESS THAN 90DEG IF RATIO FIRST FIGURE WOULD HAVE BEEN LESS THAN 5) hence statement 1 is sufficient Statement 2. AC^2 + AB^2 > BC^2 consider AC=AB=BC, given condition satisfied.All angle less than 90deg now consider AC=7,AB=3,BC=5 given condition satisfied ,Angle B will be more than 90Deg So Insufficient Hope it helps! Source: http://www.beatthegmat.com/gmatclubm17dst50625.htmlsunita123 wrote: Hi Bunuel, I did not understand how this condition is true for a right triangle. For right angle triangle it should be AC^2+AB^2=BC^2. Am i missing anything here? (2) AC^2 + AB^2 \gt BC^2. This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at B or C (so in this case one angle will be 90 degrees, for example consider a right triangle: AC=5, AB=4 and BC=3). Not sufficient. Thank you Bunuel wrote: Official Solution:
(1) \(2AB = 3BC = 4AC\). We have the ratio of the sides: \(AB:BC:AC=6:4:3\). Now, ALL triangles with this ratio are similar and have the same fixed angles. No matter what these angles actually are, the main point is that we can get them and thus answer the question. Sufficient. (2) \(AC^2 + AB^2 \gt BC^2\). This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at \(B\) or \(C\) (so in this case one angle will be 90 degrees, for example consider a right triangle: \(AC=5\), \(AB=4\) and \(BC=3\)). Not sufficient.
Answer: A



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Re: M1704 [#permalink]
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12 Jan 2017, 02:21
We can solve this in another way
1. Lets write all the sides with respect to one side 2AB=3BC=4AC AB=3/2BC AC=3/4BC Now use AB^2 + AC^2 = BC^2 and substitute in this u will get LHS is greater than BC^2 So it is obviously obtuse triangle.
2. AC^2+AB^2>BC^2 As per this statement we have no information about which side is hypotenuse. so This is clearly insufficient.



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Re: M1704 [#permalink]
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29 Jun 2017, 09:45
The explanation says that we can find the angles no matter what they are. I get that  when given the ratios of three sides all the triangles with that ratio are similar. What I don't understand is why we don't need to know the angles. The question asks weather all than angles are less than 90 degrees. Here's an example:
Ratio of three sides 1:1:1
All angles are 60 degrees ANSWER YES
Ratio of three sides 3:4:5
One angle is 90 degrees ANSWER NO
With trigonometry we could find find the actual angles, but without it we couldn't tell decisively if one of the angles is equal to or greater than 90 degrees. I'm pretty sure that I'm missing an identity somewhere, but don't know what it is!



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Re: M1704 [#permalink]
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29 Jun 2017, 10:46
spence11 wrote: The explanation says that we can find the angles no matter what they are. I get that  when given the ratios of three sides all the triangles with that ratio are similar. What I don't understand is why we don't need to know the angles. The question asks weather all than angles are less than 90 degrees. Here's an example:
Ratio of three sides 1:1:1
All angles are 60 degrees ANSWER YES
Ratio of three sides 3:4:5
One angle is 90 degrees ANSWER NO
With trigonometry we could find find the actual angles, but without it we couldn't tell decisively if one of the angles is equal to or greater than 90 degrees. I'm pretty sure that I'm missing an identity somewhere, but don't know what it is! The ratio of the sides can only be \(AB:BC:AC=6:4:3\). It can not be 1:1:1 or 3:4:5. Does those ratio satisfy \(2AB = 3BC = 4AC\)? No. ALL triangles with this ratio (\(AB:BC:AC=6:4:3\)) are similar and have the same fixed angles. No matter whether they are 6, 4, and 3 or 12, 8, and 6, etc They all will have the same angles.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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