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DS : TRIANGLE (m09q07)

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DS : TRIANGLE (m09q07) [#permalink] New post 28 Oct 2008, 23:53
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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}

[Reveal] Spoiler: OA
A

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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 00:54
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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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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 01:12
scthakur wrote:
A for me.

Simplifying stmt2 will give a > 0...insufficient.


how do we simplify stmt 2 ??

Thanks :)
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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 02:24
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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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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 05:32
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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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 07:59
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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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 08:10
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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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 15:58
Nice work..i almost fell for C..

I agree A is the ans..
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Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 21:36
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] New post 05 Oct 2009, 17:08
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] New post 06 Oct 2009, 21:54
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Re: DS : TRIANGLE (m09q07) [#permalink] New post 22 Feb 2010, 05:24
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] New post 22 Feb 2010, 06:45
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] New post 22 Feb 2010, 07:58
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] New post 22 Feb 2010, 10:22
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] New post 22 Feb 2010, 10:23
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] New post 22 Feb 2010, 11:46
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] New post 22 Feb 2010, 11:50
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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}

[Reveal] Spoiler: OA
A

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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, 00:02, edited 2 times in total.
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Re: DS : TRIANGLE (m09q07) [#permalink] New post 23 Feb 2010, 02:56
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] New post 23 Feb 2010, 06:40
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.
Re: DS : TRIANGLE (m09q07)   [#permalink] 23 Feb 2010, 06:40
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