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# M07-22

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:35
1
8
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Difficulty:

65% (hard)

Question Stats:

56% (00:42) correct 44% (01:07) wrong based on 133 sessions

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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 $$DEF$$

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15 Sep 2014, 23:35
1
Official Solution:

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.

Statement (2) by itself is insufficient. All it tells us is that both triangles are similar or proportionate to each other, but nothing about their size.

Statements (1) and (2) combined are insufficient. Combining the two statements, we still cannot determine whether the triangles have small values of their sides that yield greater perimeters or large values that yield greater area measurements.

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13 Jan 2015, 11:10
Hi Bunnel ,

In stmt 1 where does triangle DEF come in to play , the solution only mentions ABC ?
Could you elaborate stmt 1 .

Thanks and Regards ,
Shelrod007
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06 Apr 2015, 00:11
Hello Bunuel,

I am confused on statement1, can you please explain it more clearly?

Thanks
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20 Apr 2015, 06:02
Hi, Bunuel. I answered correctly but I just have a question. Is the solution below a right application of the properties of a triangle? or was I just lucky to arrive at the correct answer?

My approach:

Statement 1: Area ABC < Perimeter of DEF
To maximize the area of ABC, let's assume that it's an equilateral triangle with side x. The area becomes [x^2(√3)]/4

To maximize the area given a perimeter, let's assume that the triangle DEF is also an equilateral triangle with side, y. The area DEF becomes [y^2(√3)]/4

Since we don't know the length of the sides, x and y, the statement is insufficient.

Statement 2: Angles of ABC = Angles of DEF
This only tells us that the triangles are similar.

Combining both:
Since we already assumed that both are equilateral triangles, and therefore similar, statement 2 doesn't add any additional information. We still don't know the length of the sides and therefore, the both statements taken together are insufficient.
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24 Jul 2015, 08:02
Bunuel wrote:
Official Solution:

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.

Statement (2) by itself is insufficient. All it tells us is that both triangles are similar or proportionate to each other, but nothing about their size.

Statements (1) and (2) combined are insufficient. Combining the two statements, we still cannot determine whether the triangles have small values of their sides that yield greater perimeters or large values that yield greater area measurements.

Hi Bunuel ,

I am little confused about explanation for stmnt 1. In your explanation, you are talking about ABC only but the given stmnt talks about comparison of triangles. for stmnt 1 to be insufficient, it shd lead to 2 contradictory results. can you pls elaborate a little.
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19 Aug 2015, 09:22
1
Hi guys,

Since we know that a triangle can have an area > perimeter or perimeter > area, we can infer two scenarios: while, in the first, the area of DEF is bigger than perimeter of DEF, so area of DEF is bigger than area ABC, in the second, the perimeter of DEF is bigger than area of DEF, so we can have area of DEF > area of ABC and area of ABC > area of DEF. Therefore, statement 1 is insufficient.
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13 Jun 2016, 02:24
Can someone please explain the solution since the official solution is a bit unclear as it only talks about the ABC triangle and does not mention DEF triangle. I am clear with the statement 2 but not sure about statement 1.
Manager
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10 Jul 2016, 23:52
from statement 1) assume perimeter of DEF is 30 ( so that we can find the max area of DEF for equilateral triangle)

As per statement area of ABC < perimeter of DEF

area ABC < 30

From DEF , perimeter = 30 ----> each side is 10 (equilateral)

area DEF = 100 * (1.73/4) = 25*1.73 = value is greater than 30.

now we know area of ABC < Max area of DEF .

can we find the Max area of ABC ..... NO we dont know the perimeter or any side .

Not sufficient .

Statement 2) only tells about 2 similar angels . we can't prove the triangles are congruent .

Not sufficient .

E
_________________

way to victory .....

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Joined: 05 Mar 2015
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14 Sep 2016, 03:38
3
goGMAT2015 wrote:
Hello Bunuel,

I am confused on statement1, can you please explain it more clearly?

Thanks

For statement (1) plz consider attached fig.
case 1 :- both areas equal
case 2:- DEF<ABC............not suff....

(2) angles in attached fig. is 90deg.

combined also insufff..

Ans E
>> !!!

You do not have the required permissions to view the files attached to this post.

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13 Dec 2016, 22:05
1
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i could'nt find the solution clear.please provide a better one
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Joined: 03 Feb 2016
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22 Jan 2017, 06:51
Consider for ABC, A1 is the area and P1 is the perimeter.

Similarly consider A2 and P2 for DEF.

Now statement 1 gives A1< P2

And asks us if A1> A2?

Is is not hard to imagine A1< P2< A2 and therefor A1< A2 (if area of one triangle is smaller than the perimeter of another, we can easily imagine the latter to be infinitely big).

But can we imagine A2< A1< P2 (or A2< A1)?

The official solutions says that for triangles with small sides, perimeter> area. So we take smaller triangles to see if we can find an example.

Try with smaller sets (1, 1, √2/ 2, 2, 2√2 and 1, 1, √2 / 0.5, 0.5, 0.5√2) and you will find that statement goes both ways. Hence, insufficient.
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19 Feb 2017, 14:05
I have a question how can we compare a perimeter with an area?
Math Expert
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19 Feb 2017, 21:24
lorenzo393 wrote:
I have a question how can we compare a perimeter with an area?

You can compare numerical value of perimeter with numerical value of area.
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11 May 2017, 03:23
4
This question can be solved easily by using only 2 cases :

Fact 1 - Area of ABC < Perimeter of DEF
Consider both ABC and DEF as right angled triangles with 3, 4, 5 combination
Area of ABC (3,4,5) = 1/2 * 3 * 4 = 6
Perimeter of DEF (3,4,5) = 3+4+5 = 12

So Is Ar. of ABC > Ar. of DEF ? [Is 6 > 6?] ........ NO

Consider both ABC and DEF as equi triangles
ABC sides 3,3,3 and DEF sides 2,2,2.
Area of ABC (3,3,3) = $$x^2$$ $$\sqrt{3}/4$$ = 9$$\sqrt{3}/4$$ ~ 4.25
Perimeter of DEF (2,2,2) = 2+2+2 = 6

So Is Ar. of ABC > Ar. of DEF ? ........ Clearly YES

Fact 2 - Angles of ABC = Angles of DEF
We already proved in Fact 1 that its INSUFFICIENT.

Combining Fact 1 and Fact 2 , we have nothing new. Hence E
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Sortem sternit fortem!

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08 Mar 2018, 12:21
1
The question can also be solved by applying the property of similar triangles, i.e,
When two triangles (lets say Triangle ABC and Triangle DEF) are similar, the the ratio of the area of the triangles is equal to the squares of their respective perimeter, sides, medians, heights etc.
So,
Area(ABC)/Area(DEF)=[Perimeter(ABC)]^2/[Perimeter(DEF)]^2
Now,
from St.(1) The value of area of ABC is less than that of perimeter of DEF---->We cannot conclude anything-->insufficient.
from St.(2) Angles of ABC = Angles of DEF----->Can only conclude that the triangles are similar(using the A-A-A property)--->Insufficient
from St.1 and St.2: Since ABC and DEF are similar triangles, thus, Area(ABC)/Area(DEF)=[Perimeter(ABC)]^2/[Perimeter(DEF)]^2---eqn1
Now, St. 1 says area of ABC is less than that of perimeter of DEF.Applying this logic on eqn1 nd taking some integer values, we can see that the Perimeter(DEF)>Perimeter (ABC); thus Area (ABC)<Area(DEF). [No]
On taking decimal values on eqn1, we get that the Perimeter(DEF)<Perimeter (ABC); thus Area (ABC)>Area(DEF)....[Yes]
Thus insufficient.
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Thanks,
Rnk

Please click on Kudos, if you think the post is helpful

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11 Aug 2018, 03:17
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.
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10 Sep 2018, 11:00
I solved this question like this.

Drew two equal right triangles ABC and DEF with sides 3, 4 and 5.
Area of each triangle is 6 and perimeter of each triangle is 12.

(1) says that value of area of ABC is less than value of perimeter of DEF.

If we change nothing, area of ABC will be equal to the area of DEF and (1) will be satisfied, i.e. 6 is less than 12. In this case answer to the question is no

However, if we multiply each side of DEF by 0.99999999 then, area of DEF will be less than area of ABC. After multiplying each side by 0.9999999 perimeter of DEF will be slightly less than 12. Therefore statement (1) will be satisfied, because value of perimeter will definitely be greater than 6. In this case answer to the question yes

insufficient

(2) above mentioned example is valid for statement 2 as well,

(1) and (2) together. Above mentioned example satisfies both statements, and we get both yes and no, therefore answer is E
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01 Dec 2018, 03:49
I don't think that this solution is correct! If we know that the angles of of triangles are equal, we can conclude that those are similar triangles. In the similar triangles the sides are proportionate to each other thus we are only left with scale. If one triangle has larger perimeter, we can conclude that each of its sides is longer than the appropriate sides of other triangle. Thus, h*(side of triangle)/2 will be greater!

The answer should be C.

Please let me know if you find any mistake in my logic. Thanks!

My mistake I didn't notice that the value of area is compared to the value of perimeter
M07-22 &nbs [#permalink] 01 Dec 2018, 03:49
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# M07-22

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