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Re: M07-22 [#permalink]
Hello Bunuel,

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

Thanks
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Re: M07-22 [#permalink]
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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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Re: M07-22 [#permalink]
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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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Re: M07-22 [#permalink]
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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 ? [Is 9\(\sqrt{3}/4\) > 4\(\sqrt{3}/4\)?]........ 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 :-D

Originally posted by Sash143 on 11 May 2017, 04:23.
Last edited by Sash143 on 20 Jul 2022, 03:21, edited 1 time in total.
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Re: M07-22 [#permalink]
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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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Re: M07-22 [#permalink]
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.


Answer: E


VeritasKarishma

Can you pls explain why in explanation of statement 1 we are comparing area and perimeter of ABC?
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Re: M07-22 [#permalink]
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GDT wrote:
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.


Answer: E


VeritasKarishma

Can you pls explain why in explanation of statement 1 we are comparing area and perimeter of ABC?


What we are trying to show in statement 1 is that numerical values of area and perimeter are not comparable even with the same triangle. In some cases, area value might be higher, in other cases perimeter value. So knowing that perimeter of one triangle is more than area of another gives us no information about their relative sizes and areas.

Take @Bunuel's numbers:

Say ABC is 1, 1, sqrt(2).
Area < < Perimeter

Say DEF is just like ABC but slightly smaller.
Then, area DEF < area ABC

Say DEF is just like ABC but slightly larger.
Then area of DEF > area ABC.

But in both cases, area ABC < < perimeter of DEF

Hence, this is not enough.
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Re: M07-22 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M07-22 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re M07-22 [#permalink]
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