In the figure above, SQRE is a square and AB = AC.

[TagelS22]

In the figure above, SQRE is a square and AB = AC. Is the area of triangle ABC greater than the area of square SQRE?

(1) The length of RE is less than twice the length of BR.
(2) AS = AQ

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

Correct answer: (C)

My Question:
I thought option (E) is the correct answer.
It isn't written that the triangle must be outside the square.
Please tell me what am I missing in the next example:


(attached a picture)
i62.tinypic.com/2nltmhx.png

AS=AQ and the length of RE is less than twice of BR,
but I have 2 options:
if the triangle is inside the square -> the area of the triangle is less than the area of the square.
if the triangle is outside of the square (so CE=BR)-> the area of the triangle is more than the area of the square.

[RonPurewal]

What does the original picture look like?

If the original diagram shows the triangle inside the square, then ... well, the triangle is inside the square.

[RonPurewal]

Regardless of whether they are "to scale", diagrams NEVER lie about the placement or order of things.
Think about it for a second. If it were possible to rearrange the items in the diagram"”"”as it appears you're trying to do here"”"”then ALL diagrams would be completely useless, since it would become impossible for a diagram to depict any concrete relationship or relative placement.

If there are two or more possible orders/relative placements of the items in a diagram, then the problem MUST either ...
... (a) include two or more diagrams"”"”one per possible situation"”"”
or
... (b) omit the diagram altogether and describe the situation verbally (in a way that allows for the intended ambiguity).

In practice, you'll see (b) more often. In fact, if visual entities are described verbally but not actually drawn, your ability to visualize multiple possibilities is very likely to be the crux of the entire problem.

[RonPurewal]

More generally, NOTHING on the gmat exam will ever be intentionally misleading.

In other words, there are no "trick questions".

If you come up with a "tricky" interpretation of a straightforward diagram (or sentence), you are almost certainly over-thinking the issue"”"”and misunderstanding the fundamental nature of this exam.

[RonPurewal]

Also, you may want to read the basic rules part near the beginning of the OG, and/or read the instructions that precede the practice exams in the GMAT PREP software.
In those places, you'll see these things ("diagrams don't lie about placement/order", "all numbers are real; no imaginary numbers"; etc.) laid out in detail, just in case you want to see them explicitly.
But, "no trick questions" is a pretty apt summary of just about all of it.

[drtfyghujd403]

Alright, so, this question doesn't have an explanation still...
Can someone explain whats the right answer and why, please? :(

[jlucero]

Alright, so, this question doesn't have an explanation still...
Can someone explain whats the right answer and why, please? :(

The correct answer is D. The image that the OP posted was one from the exam (correct) and a separate picture that he/she thought as a possibility. The main point is to not imagine alternative shapes that could work while changing the main idea of that picture. You can't assume angles are perfectly made, but you can assume the picture as a whole should be accurate.

The explanation is that the triangle will have a base of x + 4x + x, and a height of 4x. Using those two values, you can create a right triangle with a base of 3x, height of 4x, inferring the third side is 5x. The square has 4 sides of 4x = 16x. The triangle has a base of 6x and two sids of 5x = 16x. Therefore, no matter how big or small the figures are, they have the same perimeter.

Same idea with statement 2. Since the height of the two figures are the same, we can create an equation:

1/2bh = 3/4 h * h
2 bh = 3 hh
2b = 3h
h = 2/3 b

Using this you can find that same ratio 3/4/5 ratio for the right triangle and use that to find perimeter.

[RonPurewal]

Alright, so, this question doesn't have an explanation still...
Can someone explain whats the right answer and why, please? :(

In the future, please ask at least one specific question about the problem.

Problems from our exams all come with answer keys. So, in response to something like "What's the right answer and why?", our best response is, basically, "Check out the answer key."

The forum is intended to be a resource for specific questions, beyond what's offered in the answer key. It is not meant to become a collection of answer keys.

Thanks.

[AlessadraS228]

According to the CAT answer D is not correct. Answer C is.

Why is this? I cannot figure it out and cannot find an adequate explanation.

[RonPurewal]

According to the CAT answer D is not correct. Answer C is.

Why is this? I cannot figure it out and cannot find an adequate explanation.

this seems to imply that you think (d) should be the answer.

• what work have you already done on this problem?

• what has led you to think that's the answer?

[RonPurewal]

also, all of our problems come with answer keys. if you just say something like "i don't know why the answer is C", then the best answer is "Check out the answer key".

so...
• have you checked out the answer key?
• if so, what parts of it don't you understand? (please quote those parts, along with enough context to understand them)

[Meerak869]

Going for statement 2
if AS = AQ then it places the triangle in symmetric position in the center.
From this we can conclude that RB=CE as the triangle is an isosceles triangle.
I'm not really sure why this option is wrong. It is very confusing for me.

[RonPurewal]

Going for statement 2
if AS = AQ then it places the triangle in symmetric position in the center.
From this we can conclude that RB=CE as the triangle is an isosceles triangle.
I'm not really sure why this option is wrong. It is very confusing for me.

...yes, that's all correct...

...but...

...but...

...but are you answering the actual question?
the question is about whether the triangle is bigger than the square.

even with the whole isosceles thing, RB and CE could be really tiny, or they could be huge. try drawing those possibilities, and look at the relative areas of the square and triangle.

[ajaym8]

Alright, so, this question doesn't have an explanation still...
Can someone explain whats the right answer and why, please?

The correct answer is D. The image that the OP posted was one from the exam (correct) and a separate picture that he/she thought as a possibility. The main point is to not imagine alternative shapes that could work while changing the main idea of that picture. You can't assume angles are perfectly made, but you can assume the picture as a whole should be accurate.

The explanation is that the triangle will have a base of x + 4x + x, and a height of 4x. Using those two values, you can create a right triangle with a base of 3x, height of 4x, inferring the third side is 5x. The square has 4 sides of 4x = 16x. The triangle has a base of 6x and two sids of 5x = 16x. Therefore, no matter how big or small the figures are, they have the same perimeter.

Same idea with statement 2. Since the height of the two figures are the same, we can create an equation:

1/2bh = 3/4 h * h
2 bh = 3 hh
2b = 3h
h = 2/3 b

Using this you can find that same ratio 3/4/5 ratio for the right triangle and use that to find perimeter.

Hi Jamie,
I could reduce the question to - is CE+BR > l ? (where l= side of square = RE= a part of the base of the triangle.)
Few questions .:

1.

The explanation is that the triangle will have a base of x + 4x + x, and a height of 4x.

I guess you have labeled CE= x, ER = 4x, BR=x. (Please let me know if I need to attach the diagram).
How are you assuming that CE=BR =x ? What I gather from statement 1 is l < 2BR. (where l= side of square = RE= a part of the base of the triangle.)
When I solved this, I labeled statement 1 as sufficient because I wrongly used AAA as a congruence criterion for 2 small triangles formed with sides CE & BR ( I am referring to 2 small triangles that lie outside the square). That means I can now say that st.1 is INSUFFICIENT.

2.

Same idea with statement 2. Since the height of the two figures are the same, we can create an equation:
1/2bh = 3/4 h * h

I understand 1/2bh. That is area of the triangle. How does 3/4 h * h come ? and how are these 2 expressions equal ?
When I solved this, I labeled statement 2 as insufficient. If AS=AQ, how does this answer the question - is CE+BR > l ?

So, in my view the answer should be E. Please tell me where am I going wrong.
NOTE.: MGMAT CAT shows the answer to this question as C. I do not agree with the explanation given there for statement 2. First, it says that because AS=AQ, triangle is placed symmetrically in the square. I agree with this. Then it goes on to say because of this symmetric placement, alongwith statement 1, we can say that CE+BR > l. Wait a minute, are we saying that due to symmetric placement of the triangle, CE=BR ? How !!??

Many doubts as described above. Please help
Thanks,

[RonPurewal]

hi,
we're going to need to see the figure to address these questions. can you please post the figure (and/or post it on an image-hosting website and put a link here)?

thank you.