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# m07 #22

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m07 #22 [#permalink]  03 Sep 2008, 18:41
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Is area of triangle $$ABC$$ greater than area of triangle $$DEF$$ ?

1. The value of area of $$ABC$$ is less than that of perimeter of $$DEF$$.
2. Angles of $$ABC$$ = Angles of $$DEF$$.

[Reveal] Spoiler: OA
E

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Re: m07 #22 [#permalink]  03 Sep 2008, 20:23

Have you considered that both statements will give you the answer 'No'?

I tried using two 3,4,5 triangles: Area=6, and P=12 so satisfies constraint in Stmt 1

Stmt 2: tells us that triangles are similar.

Together, the answer is NO -as in- Area of traingle ABC is not Greater then EFG.

Alternatively, i tried using two 6,8,10 triangles, which gave me Area=24 and P=24 so didnt satisfy stmt 1.

Also, the triangle in your method should've been "1, 1, root2" or "10, 10, 10root2" (the isoscelese right triangle) Think of a square cut in half diagonally. If a square has side 10, then 10x10=100 -> in half is 50.

Hope this helps and i hope i'm right!
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Re: m07 #22 [#permalink]  04 Sep 2008, 03:38
Ok, so the 10-10-20 thing was a typo. I got confused with that...

If you know both the triangles are similar, the area could be less than or greater than the perimeter. I see my mistake, thanks ryan.
But this only gives you info on the Area of ABC relative to the perimeter of ABC...statement (1) relates area ABC to perimeter DEF which we know nothing about.

3-4-5 P=12 Area=6
6-8-10 P=24 Area=24
9-12-15 P=36 Area=54

I think it is E after all.
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Re: m07 #22 [#permalink]  04 Sep 2008, 05:37
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Let me clarify....

I considered the case that ABC and DEF are BOTH 3-4-5 triangles.

Look at that list you made a little more closely.

3-4-5 P=12 Area=6 ............area is smaller than P
6-8-10 P=24 Area=24 .......... area and P are equal
9-12-15 P=36 Area=54 ..........area is greater than P

Using two 3-4-5 triangles satisfies the constraint in Stmt1, and then when stmt 2 confirms that the triangles are similar, stmt 1&2 gives us the definitive answer 'NO' to the question if Area of ABC is greater then EFG - because the area is equal.

Another consideration.... let's take a 6-8-10 as ABC and 9-12-15 as DEF:
Stmt 1 is satisfied - Area ABC is less then P of DEF
Stmt 2 is satisfied - They are similar triangles
1&2 - together: Is area of ABC > DEF? .... NO it is not.

Make sense?
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Re: m07 #22 [#permalink]  23 May 2009, 19:59
I'm having some trouble understanding this question.
I see why it doesn't work by picking numbers, but I always have a hard time doing that on the test. I feel like I might miss one set of numbers which is why I try to solve algebraically.

So can someone conceptually explain this?

If the two triangles are similar, then their areas would be proportionate too. (right?)
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Re: m07 #22 [#permalink]  18 Nov 2010, 10:40
Would someone please take a moment and explain to me why this is E and not C? Once you know that the angles are the same AND that the perimeter is bigger don't you have enough information to say that ABC is in fact not bigger?
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Re: m07 #22 [#permalink]  18 Nov 2010, 12:23
mrcrescentfresh wrote:
Would someone please take a moment and explain to me why this is E and not C? Once you know that the angles are the same AND that the perimeter is bigger don't you have enough information to say that ABC is in fact not bigger?

The fact is that your reasoning works just for some types of triangles, while for others it doesn't. So, it must be E since both of them are too general.
When I see question like these, it's a good idea not to waste too much time in experiments...the best idea is choosing E.
Paolo
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Re: m07 #22 [#permalink]  18 Nov 2010, 13:49
1
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michael127 wrote:
Is area of triangle $$ABC$$ greater than area of triangle $$DEF$$ ?

1. The value of area of $$ABC$$ is less than that of perimeter of $$DEF$$.
2. Angles of $$ABC$$ = Angles of $$DEF$$.

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I narrowed it down to C & E and chose the wrong answer. After looking at this problem for some time (waaaay more than 2 min I have an explanation)

1) Let's first try a 3:4:5 triangle. The area of a 3:4:5 triangle is 6 and the perimeter is 12. Now we go up to the question stem and we get NO.

Now that we have proven ABC is smaller than DEF we have to look for an instance where ABC is larger than DEF that would make statement 1) insufficient. If we make ABC a 2:2:2\sqrt{2} isosceles right triangle and DEF a 1:1:\sqrt{2} the area of ABC is 2 and the perimeter of DEF is 2 + \sqrt{2} (this makes the statement true).

We need to go to the question stem and plug this information in. The answer will be YES making statement 1 insufficient.

2) Statement 2 doesn't give us any information we can use.

Now the selections are C & E. Together the statements are not sufficient. So the answer is E.
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Re: m07 #22 [#permalink]  18 Nov 2010, 18:52
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S1: Area of ABC = 0.5 (b) (h)
Perimeter of DEF = s1+s2+s3

0.5bh< s1+s2+s3
S2: not sufficient

Combining also there is no solution, so answer is E
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Re: m07 #22 [#permalink]  19 Nov 2010, 08:46
can someone give an example of where "The value of area of ABC is less than that of perimeter of DEF" and vice versa? Stmt1 is throwing me off!
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Re: m07 #22 [#permalink]  22 Nov 2011, 05:57
Assume: AB = BC = 2, Angle ABC = 90; DE = EF = 1, Angle DEF = ABC = 90.
Area ABC = .5 * 2 * 2 = 2
Perimeter EDF = 1 + 1 + sqrt(2) > 2, but obviously Area ABD > Area DEF = 0.5.

The area of any triangle is the largest when it is a right triangle with both legs equal!
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Re: m07 #22 [#permalink]  22 Nov 2011, 23:46
simple geometry...the trick is in st 1: area of ABC is compared to perimeter of DEF..which does not lead us anywhere
st2 itself does not help

go for E
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Re: m07 #22 [#permalink]  22 Nov 2012, 05:25
michael127 wrote:
Is area of triangle $$ABC$$ greater than area of triangle $$DEF$$ ?

1. The value of area of $$ABC$$ is less than that of perimeter of $$DEF$$.
2. Angles of $$ABC$$ = Angles of $$DEF$$.

[Reveal] Spoiler: OA
E

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B. can NEVER answer the question, because other two angles can vary and so do the sides, so -incorrect

A. In given situation when the question is not limited to any specific angle consider the middle angle to be a right one, now even if it's a right angle
1/2 BC X AB = DE+EF+DF
makes no sense ; incorrect

E wins
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Re: m07 #22 [#permalink]  22 Nov 2012, 18:55
Hi Bunuel,
How do you eliminate A and C in this question?
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Re: m07 #22 [#permalink]  22 Nov 2012, 20:15
Hi All,

If the triangles are similar, can we say...
The ratio of their area is (side of one triangle)^2 / (side of other triangle)^2

Regards,
Pritish
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Re: m07 #22 [#permalink]  23 Nov 2012, 02:21
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Expert's post
pritish2301 wrote:
Hi All,

If the triangles are similar, can we say...
The ratio of their area is (side of one triangle)^2 / (side of other triangle)^2

Regards,
Pritish

Yes, in two similar triangles, the ratio of their areas is the square of the ratio of their sides.
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Re: m07 #22 [#permalink]  23 Nov 2012, 06:24
I am such an idiot.
I kept thinking same angles;same area...of course we can have different areas with the same angles.This is what a job does to your brains.#rusted
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Re: m07 #22 [#permalink]  17 Sep 2013, 13:01
I don't understand why Statement 1 is Insufficient even after I read the posts from above and the official explanation.

3-4-5 P=12 Area=6 ............area is smaller than P
6-8-10 P=24 Area=24 .......... area and P are equal
9-12-15 P=36 Area=54 ..........area is greater than P

The question asks: "Is area of triangle $$ABC$$ greater than area of triangle $$DEF$$ ?"

Statement 1's condition must be satisfy first before you can apply to the question. The perimeter must be GREATER than the area. After you satisfy St1 condition, then you can evaluate whether the area of triangle ABC is greater then area of triangle DEF. In ALL the cases in which the area is less than the perimeter, the area is also less than or equal to area of the triangle. There are no cases in which the area is less than the perimeter but have greater area.

St1: The value of area of $$ABC$$ is less than that of perimeter of $$DEF$$

Here is the official explanation:

Statement (1) by itself is insufficient. Let's pick numbers: if the sides of $$ABC$$ are 1, 1 and $$\sqrt{2}$$ (a half of a square with sides equal to 1), the area equals $$0.5$$ and t perimeter is $$2+\sqrt{2}$$ . The perimeter is much greater than the area of a triangle with these values. However if the sides of $$ABC$$ are 10, 10, and $$10\sqrt{2}$$ ; then the perimeter is $$20+10\sqrt{2}$$ and the area is 50. The perimeter is much smaller than the area.

The highlight portion of the OE contradicts the ST1 condition. Thus, its invalid. Can someone please explain the problem? Thanks.
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Re: m07 #22 [#permalink]  18 Nov 2013, 05:37
Lets consider right angled triangles for the sake of simplicity:

I name them, a,b,c with area = 1/2*a*b and a',b',c' with area = 1/2*a'*b'.

Now, using statement 1, we have 1/2*a*b < a'+b'+c'.
Using this, we need to see if we can arrive at some declarative result for 1/2*a*b>1/2*a'*b' (Yes or No). However, if you observe carefully, we can never be sure if the perimeter of a triangle is always less than or greater than its area. Hence option A and D are ruled out!

Now, using statement 2, we have all the angles to be equal, or the triangles to be similar. This has nothing to do with 1/2*a*b > 1/2*a'*b'. Hence, option B ruled out.

Using both the statements together we get, 1/2*a*b<a'+b'+c'.
divide by 1/2*a'*b' on both sides. This will give us, a*b/a'*b' < (1/b' + 1/c' + c'/a'*b').

Now this value - (1/b' + 1/c' + c'/a'*b') can vary from less than 1 to more than 1, giving us more than one answers. Hence Option C is ruled out. Therefore, Option E is the right answer.
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Re: m07 #22 [#permalink]  12 May 2014, 08:33
is it really a good question? I'm still unable to judge what ability it is trying to test here?
And is it possible to solve this sum under 2 mins or even 3 mins?
Re: m07 #22   [#permalink] 12 May 2014, 08:33
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# m07 #22

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