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amitdgr · **Posted:** Wed Oct 29, 2008 12:53 am

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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Bunuel · **Posted:** Mon Feb 22, 2010 12:50 pm

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

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.

Bunuel · **Posted:** Thu Dec 02, 2010 12:56 pm

MathMind wrote:

Bunuel,

Great explanation - I am not sure about your conclusion "Hence a>1 is not true as when we decrease angle ABC, a will decrease as well and will become less than 1"

I think this may lead to "As the angle ABC decreases, since ht=a^2 when base = 2a (ht/base ratio = a/2), the base will become broader => a will increase

=> This triangle will have area = 1 for ABC = 90*

=> This triangle will have area > 1 for ABC < 90*

=> This triangle will have area < 1 for ABC > 90* "

Thx, JS

First of all base=2/a (and not 2a) and height/base=a^3/2.

Though I did have a typo there. It should be: 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.

scthakur · **Posted:** Wed Oct 29, 2008 1:54 am

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.

amitdgr · **Posted:** Wed Oct 29, 2008 2:12 am

scthakur wrote:

A for me.

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

how do we simplify stmt 2 ??

Thanks

scthakur · **Posted:** Wed Oct 29, 2008 3:24 am

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.

LiveStronger · **Posted:** Wed Oct 29, 2008 6:32 am

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 ?

GMAT TIGER · **Posted:** Wed Oct 29, 2008 8:59 am

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.

GMAT TIGER · **Posted:** Wed Oct 29, 2008 9:10 am

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.

FN · **Posted:** Wed Oct 29, 2008 4:58 pm

Nice work..i almost fell for C..

I agree A is the ans..

scthakur · **Posted:** Wed Oct 29, 2008 10:36 pm

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?

tejal777 · **Posted:** Mon Oct 05, 2009 6:08 pm

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??????

tejal777 · **Posted:** Tue Oct 06, 2009 10:54 pm

flyingbunny · **Posted:** Mon Feb 22, 2010 6:24 am

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 ??

jusjmkol740 · **Posted:** Mon Feb 22, 2010 7:45 am

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?

deepakdewani · **Posted:** Mon Feb 22, 2010 8:58 am

Please explain:

How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?

SudiptoGmat · **Posted:** Mon Feb 22, 2010 11:22 am

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

SudiptoGmat · **Posted:** Mon Feb 22, 2010 11:23 am

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.

nvgroshar · **Posted:** Mon Feb 22, 2010 12:46 pm

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.

kaptain · **Posted:** Tue Feb 23, 2010 3:56 am

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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