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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??????
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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