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# DS : TRIANGLE (m09q07)

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DS : TRIANGLE (m09q07) [#permalink]

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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 (m09q07) [#permalink]

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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

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, 00:02, edited 2 times in total.

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

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Area of the triangle : $$\frac{1}{2}*AC*OB$$
= $$\frac{1}{2}*\frac{2}{a}*a^2$$
= a

So we to find that a > 1 or not ??

Statement 1:

angle ABC < 90 => angle ABO < 45.
mean angle BAO is greater than angle ABO
that further implies that side OB > side OA
$$a^2 >\frac{1}{a}$$
=> $$a^3> 1$$ => $$a > 1$$ since length of side can not be negative

So A is sufficient

Statement 2:

Perimeter of the triangle is greater than 4/a

So fine perimeter, first we have to find AB

$$AB^2 = OB^2 + OA ^2$$

$$AB^2 = (a^2)^2 + \frac{1}{a^2}$$

$$AB^2 = \frac{(a^6 +1)}{a ^2}$$

$$AB =\sqrt{(a^6 +1)}/a$$

Perimeter of triangle = AB + BC +CA
since AB = BC,

Perimeter = $$2*AB + CA = 2 * \sqrt{(a^6 +1)} /a + \frac{2}{a}$$

$$2 * \sqrt{(a^6 +1)} /a + 2/a > 4/a$$

$$\sqrt{(a^6 +1)} + 1 > 2$$

$$\sqrt{(a^6 +1)} > 1$$

$$a^6 +1 > 1$$

$$a^6 > 0$$

$$a > 0$$

So, we can not say anything from statement 2 that a >1 or not.

Hence, correct answer is A.

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Re: DS : TRIANGLE [#permalink]

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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 (m09q07) [#permalink]

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02 Dec 2010, 11:56
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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.
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Re: DS : TRIANGLE [#permalink]

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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 (m09q07) [#permalink]

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18 Feb 2014, 00:49
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Bunuel's explanation is excellent, as usual.
Here's a different way that does not involve lots of algebra...

(1) Let's assume that a = 1, the simplest thing to try. Also, let's call the origin (0, 0) point D (got to name it so we can talk about it).

This gives us a triangle with height 1 (BD) and base 2 (AC), so the area is clearly 1.
It should be intuitively obvious, if you look at this triangle in your mind, that it's an isosceles right triangle (angle ABC = 90).
But if you're not sure, consider that if a = 1, BD = DC = 1, so triangle BDC is an isosceles right triangle.
This means that angle BCD is 45 degrees, and since angle BDC is 90 degrees, angle DBC is 45 degrees.
In the same way, we know that angle DBA is 45 degrees, and that therefore angle ABC is 90 degrees.

If we make a bigger, the triangle gets taller and narrower; try a = 2. Height 4, base 1, area 2.
Ah-hah! looks like as a gets bigger, area gets bigger.

One more quick check to validate: try $$a = \frac{1}{2}$$. Height $$\frac{1}{4}$$, base 4, area $$\frac{1}{2}$$.

So it's clear, without doing any algebra: as the triangle gets taller and narrower, its area increases.
As the triangle gets taller and narrower (a > 1), angle ABC gets smaller, i.e. less than 90 degrees. That's what Statement (1) says. Sufficient.

(2) Again, let's try and avoid algebra with some quick numbers. If a = 1, BD = DC = 1, triangle BDC is an isosceles right triangle.
It must have sides in the ratio $$1:1:\sqrt{2}$$; therefore BC = $$\sqrt{2}$$.
AB will also = $$\sqrt{2}$$, so the perimeter is $$2 + 2\sqrt{2}$$, which is clearly > 4.
(If a = 1, $$\frac{4}{a}=4$$).

So when a = 1, the perimeter is > $$\frac{4}{a}$$, but the area of triangle ABC is = 1 (see Statement (1) above) and therefore not > 1.
When a = 1, the answer to the overall question, whether the area > 1, is False.

Now if a becomes a larger number (a > 1), the perimeter will increase.

(Try a = 10. Now $$a^2$$ = BD = 100. Clearly BC must be still larger.
Therefore the perimeter is much, much larger than $$\frac{4}{a}$$, which is $$\frac{4}{10}$$, and much larger than 4 as well.)

We already know from Statement (1) above that when a > 1, the answer to the question is True (area ABC > 1).
Therefore, without doing any more work, both True and False answers are possible and Statement (2) is Insufficient.

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Re: DS : TRIANGLE [#permalink]

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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]

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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]

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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]

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29 Oct 2008, 08:10
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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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Re: DS : TRIANGLE [#permalink]

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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]

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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]

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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]

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06 Oct 2009, 21:54

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

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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]

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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]

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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]

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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]

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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] 22 Feb 2010, 10:23

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