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Re: DS : TRIANGLE (m09q07) [#permalink]
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?????? _________________

Re: DS : TRIANGLE (m09q07) [#permalink]
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 ?? _________________

Re: DS : TRIANGLE (m09q07) [#permalink]
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

Re: DS : TRIANGLE (m09q07) [#permalink]
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.

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.

Re: DS : TRIANGLE (m09q07) [#permalink]
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.

Re: DS : TRIANGLE (m09q07) [#permalink]
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.

gmatclubot

Re: DS : TRIANGLE (m09q07)
[#permalink]
23 Feb 2010, 06:40