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DS : TRIANGLE (m09q07) [#permalink]
 29 Oct 2008, 00:53
On the picture below, is the area of the triangle ABC greater than 1? 1. \angle ABC < 90^\circ 2. Perimeter of the triangle ABC is greater than \frac{4}{a}Source: GMAT Club Tests - hardest GMAT questions
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A for me.
Area of triangle = 1/2 * 2/a * a^2 = a. Hence, we need to find out whether a > 1/
From stmt1: AB^2 + BC^2 > AC^2 or 2*AB^2 > AC^2
or 2*(a^4 + 1/a^2) > 4/a^2
or, (a^6 + 1) > 2 or, a^6 > 1 that means a > 1. Hence, sufficient.
Simplifying stmt2 will give a > 0...insufficient.
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scthakur wrote: A for me.
Simplifying stmt2 will give a > 0...insufficient. how do we simplify stmt 2 ?? Thanks 
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amitdgr wrote: scthakur wrote: A for me.
Simplifying stmt2 will give a > 0...insufficient. how do we simplify stmt 2 ?? Thanks Perimeter = AB + BC + AC = 2AB + AC = 2*sqrt((1/a^2)+ a^4) + 2/a = 2*(sqrt(1+a^6))/a + 2/a > 4/a or, sqrt(1+a^6) + 1 > 2 or, sqrt(1+a^6) > 1 or a > 0.
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scthakur wrote: A for me.
Area of triangle = 1/2 * 2/a * a^2 = a. Hence, we need to find out whether a > 1/
From stmt1: AB^2 + BC^2 > AC^2 or 2*AB^2 > AC^2
or 2*(a^4 + 1/a^2) > 4/a^2
or, (a^6 + 1) > 2 or, a^6 > 1 that means a > 1. Hence, sufficient.
Simplifying stmt2 will give a > 0...insufficient. scthakur : Can you explain how you simplified stmt1 And from a^6 > 1, how did you come up with a > 1, a could be -2 also right ?, or is it because 1/a is a point on Positive X-axis ?
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LiveStronger wrote: scthakur wrote: A for me.
Area of triangle = 1/2 * 2/a * a^2 = a. Hence, we need to find out whether a > 1/
From stmt1: AB^2 + BC^2 > AC^2 or 2*AB^2 > AC^2
or 2*(a^4 + 1/a^2) > 4/a^2
or, (a^6 + 1) > 2 or, a^6 > 1 that means a > 1. Hence, sufficient.
Simplifying stmt2 will give a > 0...insufficient. scthakur : Can you explain how you simplified stmt1 And from a^6 > 1, how did you come up with a > 1, a could be -2 also right ?, or is it because 1/a is a point on Positive X-axis ? nice work by scthakur. since a is a measurement of length of sides, a cannot be -ve. therefore a has some +ve value.
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scthakur wrote: amitdgr wrote: scthakur wrote: A for me.
Simplifying stmt2 will give a > 0...insufficient. how do we simplify stmt 2 ?? Thanks Perimeter = AB + BC + AC = 2AB + AC = 2*sqrt((1/a^2)+ a^4) + 2/a = 2*(sqrt(1+a^6))/ a + 2/a > 4/a or, sqrt(1+a^6) + 1 > 2 or, sqrt(1+a^6) > 1 or a > 0. that should be a^2. if so: 2*(sqrt(1+a^6))/ a^2 + 2/a > 4/a or, sqrt(1+a^6)/a + 1 > 2 or, sqrt(1+a^6) > a or, (1+a^6) > a^2 or, a^6 - a^2 > -1 or, a^2 (a^4 - 1) > -1 1: a^2 > -1 a > -1 2: a^4 - 1 > -1 a^4 > 0 . in either case a is only +ve but not sure whether it is >1. so nsf. A is correct.
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Nice work..i almost fell for C..
I agree A is the ans..
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GMAT TIGER wrote: scthakur wrote: Perimeter = AB + BC + AC = 2AB + AC = 2*sqrt((1/a^2)+ a^4) + 2/a
= 2*(sqrt(1+a^6))/a + 2/a > 4/a
or, sqrt(1+a^6) + 1 > 2
or, sqrt(1+a^6) > 1 or a > 0. that should be a^2. if so: 2*(sqrt(1+a^6))/ a^2 + 2/a > 4/a or, sqrt(1+a^6)/a + 1 > 2 or, sqrt(1+a^6) > a or, (1+a^6) > a^2 or, a^6 - a^2 > -1 or, a^2 (a^4 - 1) > -1 1: a^2 > -1 a > -1 2: a^4 - 1 > -1 a^4 > 0 . in either case a is only +ve but not sure whether it is >1. so nsf. A is correct. GT, I think my expression is correct as a is out of sqrt. Within sqrt, it will be a^2. Am I missing something?
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Re: DS : TRIANGLE (m09q07) [#permalink]
05 Oct 2009, 18:08
Hey guys..not clear..could somebody explain it in more detail? I understood that area of the triangle=a so we have to see whther a>1.. Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ?? and then 2*(a^4 + 1/a^2) > 4/a^2??????
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Re: DS : TRIANGLE (m09q07) [#permalink]
06 Oct 2009, 22:54
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Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 06:24
tejal777 wrote: Hey guys..not clear..could somebody explain it in more detail?
I understood that area of the triangle=a
so we have to see whther a>1..
Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ??
and then 2*(a^4 + 1/a^2) > 4/a^2?????? got the same question: How does it imply AB^2 + BC^2 > AC^2 ??
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Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 07:45
flyingbunny wrote: tejal777 wrote: Hey guys..not clear..could somebody explain it in more detail?
I understood that area of the triangle=a
so we have to see whther a>1..
Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ??
and then 2*(a^4 + 1/a^2) > 4/a^2?????? got the same question: How does it imply AB^2 + BC^2 > AC^2 ?? Although I'm still confused with the problem (actually the significance of angleABC < 90-degree), still "trying to force" a logic for the above algebric inequality. Since the AD (y-axis) is perpendicular to BC (x-axis), AB^2 = AD^2 + BD^2. Thus, AB^2 = (a^6 + 1)/ a^2 = AC^2 ------- (i) Now for triangle ABC, length of 2-sides must be greater than that of the 3rd side. So, AB + AC > BC. => 2 AB > BC => sqrt (2(a^6 + 1)/ a^2) > 2/a ------- (ii) i.e. 2(a^6 + 1)/ a^2 > 4/a^2 => a^6 + 1 > 2 => a^6 > 1. Since "a" (measure of side of a triangle) cannot be negative, "a^6 > 1" implies a > 1 Can someone help with OE/ OA pl?
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Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 08:58
Please explain:
How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?
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Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 11:22
St 1: If Ang (ABC) <90 then Ang (ABO) < 45 So, Ang (BAO) > 45 as Ang (ABO) = 90 OB/OA > 1 as tan (BAO) > 1 a^2/(1/a) > 1 or a^3 > 1 and a>0 as it is a length so a>1 but st 2: a>0 as analysed by others. So answer is A
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Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 11:23
deepakdewani wrote: Please explain:
How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"? Its rule. Just memorise it.
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Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 12:46
1.The area S = a (proven in the above posts)..
tanA=tanC=BO/AO=a^2/(1/a)=a^3.
If angle ABC<90 then A =(180-ABC)/2 > 45, -> tanA>1 -> a^3>1 -> Area S=a >1.
2. Perimeter P = (2/a)(1 + sqrt(a^6+1) > 4/a.
1 + sqrt(a^6+1) > 2
sqrt(a^6+1) > 1 for any a not equal to 0.
Rhe Stmnt 2 alone is not suff.
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Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 12:50
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amitdgr wrote: On the picture below, is the area of the triangle ABC greater than 1? 1. \angle ABC < 90^\circ 2. Perimeter of the triangle ABC is greater than \frac{4}{a}Source: GMAT Club Tests - hardest GMAT questions Given isosceles triangle ABC with base=AC=\frac{2}{a} and height=a^2. Question: is area=\frac{1}{2}*base*height=\frac{1}{2}*\frac{2}{a}*a^2=a>1. So we see that basically the question aska: is a>1 true? (1) \angle ABC < 90^\circ --> assume \angle ABC=90^\circ then hypotenuse is AC=\frac{2}{a}, as ABC becomes isosceles right triangle ( 45-45-90=1-1-\sqrt{2}) then the leg=BC=AB=\frac{2}{a\sqrt{2}}=\frac{\sqrt{2}}{a}. But BC=\frac{\sqrt{2}}{a} also equals to \sqrt{(\frac{1}{a})^2+(a^2)^2}, so we have \frac{\sqrt{2}}{a}=\sqrt{(\frac{1}{a})^2+(a^2)^2} --> 2=1+a^6 --> a^6=1 --> a=1 (as a per diagram is positive). Now, if we increase a then the base \frac{2}{a} will decrease and the height a^2 will increase thus making angle ABC smaller than 90 and if we decrease a then the base \frac{2}{a} will increase and the height a^2 will decrease thus making the angle ABC greater than 90. So, as angle ABC is less than 90, a must be more than 1. Sufficient. (2) Perimeter of the triangle ABC is greater than \frac{4}{a} --> P=BC+AB+AC=2BC+AC=2\sqrt{(\frac{1}{a})^2+(a^2)^2}+\frac{2}{a}>\frac{4}{a} --> \sqrt{\frac{1+a^6}{a^2}}>\frac{1}{a} --> a^6>0 --> a>0. Hence we don't know whether a>1 is true. Not sufficient. Answer: A.
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Last edited by Bunuel on 23 Apr 2012, 01:02, edited 2 times in total.
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Re: DS : TRIANGLE (m09q07) [#permalink]
23 Feb 2010, 03:56
Bunuel,
For (1), how did you deduce that ABC is an isosceles right triangle ? As angle ABC is less than 90 deg, it is only isosceles at the most. We need not have 45-45 each at the lower angles as well. It could be a 50-50-80 (adding to 180) triangle but never a right triangle. Or am I missing something ??
Also, what do you think of AB^2 + BC^2 > AC^2 ??? I think the rule is right but never quite heard of it before.
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Re: DS : TRIANGLE (m09q07) [#permalink]
23 Feb 2010, 07:40
SudiptoGmat wrote: deepakdewani wrote: Please explain: How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"? Quote: Its rule. Just memorise it. While I agree that eventually remebering this rule for GMAT will be the best bet, I am hoping that you can provide the underlying rationale/logic for this rule. Haven't quite come across this rule in the strategy guides...though i am sure this rule can be quite handy in geometry questions involving some variation of inequalities.
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Re: DS : TRIANGLE (m09q07)
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23 Feb 2010, 07:40
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