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

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In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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Updated on: 18 Oct 2013, 22:10
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In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR.

(2) The area of triangle ABC is 75% of the area of square SQRE.

Originally posted by AccipiterQ on 18 Oct 2013, 09:34.
Last edited by AccipiterQ on 18 Oct 2013, 22:10, edited 1 time in total.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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20 Oct 2013, 13:03
16
5

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR. Look at the diagram below:
Attachment:

Untitled.png [ 4.07 KiB | Viewed 14556 times ]
$$AB=AC=\sqrt{4^2+3^2}=5$$.

The perimeter of SQRE is 4*4=16.
The perimeter of ABC is 6+5+5=16.

The difference is 0. Sufficient.

(2) The area of triangle ABC is 75% of the area of square SQRE. The height of ABC equals to the side of the square, thus we have that 1/2*ER*CB=3/4*ER^2 --> 2CB=3ER --> 2(ER+2RB)=3ER --> 4RB=ER. The same info as above. Sufficient.

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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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18 Oct 2013, 09:36
1
I think because it's a square, once you know the size of one side, you can calculate the perimeter, than if you know it's relation to the base of the triangle (which A gives you) then you can calculate the height of the triangle as well, and then you can calculate perimeter of triangle based on base, and the area, somehow. So I chose A.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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18 Oct 2013, 22:06
C according to me !
Statement 1 tells us about the base of the triangle. Insufficient...
Statement 2 describes area, but base of triangle is still unknown...Insufficient

Combine both and you have the perimeters.

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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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19 Oct 2013, 04:51
AccipiterQ wrote:
In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR.

(2) The area of triangle ABC is 75% of the area of square SQRE.

I believe we need at least one dimension of either the square side or one side of the triangle. Unless we have that we can not get the absolute value of the difference in the perimeter values.

We will get an equation converted in one of the sides.

Can any one come up with the OE. Thanks!!!
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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11 Dec 2013, 10:56
1/2*ER*CB=3/4*ER^2 --> 2CB=3ER --> 2(ER+2RB)=3ER --> 4RB=ER.

How do I know to do that? How do I know to take those exact steps to get to that result? I came up with (2/3)CB = ER which I believe is a correct statement but obviously, it would make it more difficult to determine sufficiency in the 2 or so minutes we have. With the formula I got, I know the length of the base relative to the height and I can find the hypotenuse and obviously I know the perimeter of the square (the height of the triangle) Is that enough to determine sufficiency?
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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09 Feb 2014, 11:24
Quote:
AB=AC=\sqrt{4^2+3^2}=5

Hi Bunuel,

Can you please explain how did you know the length of the sides as the only information given in the questions is about how many times a side is of another and not the absolute values?
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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10 Feb 2014, 01:02
1
Rohan_Kanungo wrote:
Quote:
AB=AC=\sqrt{4^2+3^2}=5

Hi Bunuel,

Can you please explain how did you know the length of the sides as the only information given in the questions is about how many times a side is of another and not the absolute values?

Yes, we don't know the lengths. We only know the ratios. But if you write x, 2x, 2x, x, and 4x instead of 1, 2, 2, 1, and 4 you'll get the same result.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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10 Feb 2014, 22:10
good brainshaker after long break)
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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02 Jun 2014, 00:40
Bunuel wrote:

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR. Look at the diagram below:
Attachment:
Untitled.png
$$AB=AC=\sqrt{4^2+3^2}=5$$.

The perimeter of SQRE is 4*4=16.
The perimeter of ABC is 6+5+5=16.

The difference is 0. Sufficient.

(2) The area of triangle ABC is 75% of the area of square SQRE. The height of ABC equals to the side of the square, thus we have that 1/2*ER*CB=3/4*ER^2 --> 2CB=3ER --> 2(ER+2RB)=3ER --> 4RB=ER. The same info as above. Sufficient.

I get why (1) is sufficient , but in statement 2 we are taking CB=( ER + 2RB) which means we are taking CE= RB
but can anybody explain why in statement 2 we are able to take CE = RB, how do we know they are equal.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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02 Jun 2014, 00:56
1
1
qlx wrote:
Bunuel wrote:

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR. Look at the diagram below:

$$AB=AC=\sqrt{4^2+3^2}=5$$.

The perimeter of SQRE is 4*4=16.
The perimeter of ABC is 6+5+5=16.

The difference is 0. Sufficient.

(2) The area of triangle ABC is 75% of the area of square SQRE. The height of ABC equals to the side of the square, thus we have that 1/2*ER*CB=3/4*ER^2 --> 2CB=3ER --> 2(ER+2RB)=3ER --> 4RB=ER. The same info as above. Sufficient.

I get why (1) is sufficient , but in statement 2 we are taking CB=( ER + 2RB) which means we are taking CE= RB
but can anybody explain why in statement 2 we are able to take CE = RB, how do we know they are equal.

AS = AQ implies that A is the midpoint of SQ.
AB = AC implies that triangle ABC is isosceles, so it's symmetrical around the altitude. So, half of it lies to the left of the height and another identical half to the right of the height.

Therefore, CE = RB.

Hope it's clear.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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02 Jun 2014, 01:02

My, somewhat foolish, approach was that I was looking for actual measurements. As there were none I figured the ratio's wouldn't help me come up with a definitive answer of an absolute difference.

Thanks for clearing that out!
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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27 Jul 2014, 10:01
Bunuel wrote:
Rohan_Kanungo wrote:
Quote:
AB=AC=\sqrt{4^2+3^2}=5

Hi Bunuel,

Can you please explain how did you know the length of the sides as the only information given in the questions is about how many times a side is of another and not the absolute values?

Yes, we don't know the lengths. We only know the ratios. But if you write x, 2x, 2x, x, and 4x instead of 1, 2, 2, 1, and 4 you'll get the same result.

Hi Bunuel,

Thanks for this, I have tried it for a few values to confirm and it indeed does come out to be zero for all values of x.

Just wanted to understand how did you come up with this generalization without having to try a number of different values, is there some basic logic or rule that can come into play here? As it seems like the concept of "x, 2x, 2x, x, and 4x" was more of a generalization after having tried one value?

Sagar
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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27 Jul 2014, 15:08
itsworththepain wrote:
Bunuel wrote:

Yes, we don't know the lengths. We only know the ratios. But if you write x, 2x, 2x, x, and 4x instead of 1, 2, 2, 1, and 4 you'll get the same result.

Hi Bunuel,

Thanks for this, I have tried it for a few values to confirm and it indeed does come out to be zero for all values of x.

Just wanted to understand how did you come up with this generalization without having to try a number of different values, is there some basic logic or rule that can come into play here? As it seems like the concept of "x, 2x, 2x, x, and 4x" was more of a generalization after having tried one value?

Sagar

We are given only ratios in the question, so we can assume values according to them: in x, 2x, 2x, x, and 4x, if x=1, then the lengths come out to be 1, 2, 2, 1, and 4.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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30 Jun 2015, 10:30
Bunuel wrote:

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR. Look at the diagram below:
Attachment:
Untitled.png
$$AB=AC=\sqrt{4^2+3^2}=5$$.

The perimeter of SQRE is 4*4=16.
The perimeter of ABC is 6+5+5=16.

The difference is 0. Sufficient.

(2) The area of triangle ABC is 75% of the area of square SQRE. The height of ABC equals to the side of the square, thus we have that 1/2*ER*CB=3/4*ER^2 --> 2CB=3ER --> 2(ER+2RB)=3ER --> 4RB=ER. The same info as above. Sufficient.

OA is C but not D(As per the Manhattan source).
Please explain without taking values and also explain why have considered few assumptions that you have mentioned.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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30 Jun 2015, 10:40
Mechmeera wrote:
Bunuel wrote:

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR. Look at the diagram below:
Attachment:
Untitled.png
$$AB=AC=\sqrt{4^2+3^2}=5$$.

The perimeter of SQRE is 4*4=16.
The perimeter of ABC is 6+5+5=16.

The difference is 0. Sufficient.

(2) The area of triangle ABC is 75% of the area of square SQRE. The height of ABC equals to the side of the square, thus we have that 1/2*ER*CB=3/4*ER^2 --> 2CB=3ER --> 2(ER+2RB)=3ER --> 4RB=ER. The same info as above. Sufficient.

OA is C but not D(As per the Manhattan source).
Please explain without taking values and also explain why have considered few assumptions that you have mentioned.

The OA of this question is D, not C. And it's explained WHY it's D. Those are not the actual lengths but ratios, which is explained here: in-the-figure-above-sqre-is-a-square-ab-ac-and-as-aq-161814.html#p1330094

You are mixing this question with this one: in-the-figure-above-pqrs-is-a-square-and-ab-ac-is-the-area-of-tria-192330.html (OA is C there).

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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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30 Jun 2015, 10:47
Bunuel wrote:
Mechmeera wrote:
Bunuel wrote:

In the figure above, SQRE is a square, AB = AC, and AS = AQ. What is the difference between the perimeter of triangle ABC and the perimeter of square SQRE?

(1) The length of RE is 4 times the length of BR. Look at the diagram below:
Attachment:
Untitled.png
$$AB=AC=\sqrt{4^2+3^2}=5$$.

The perimeter of SQRE is 4*4=16.
The perimeter of ABC is 6+5+5=16.

The difference is 0. Sufficient.

(2) The area of triangle ABC is 75% of the area of square SQRE. The height of ABC equals to the side of the square, thus we have that 1/2*ER*CB=3/4*ER^2 --> 2CB=3ER --> 2(ER+2RB)=3ER --> 4RB=ER. The same info as above. Sufficient.

OA is C but not D(As per the Manhattan source).
Please explain without taking values and also explain why have considered few assumptions that you have mentioned.

The OA of this question is D, not C. And it's explained WHY it's D. Those are not the actual lengths but ratios, which is explained here: in-the-figure-above-sqre-is-a-square-ab-ac-and-as-aq-161814.html#p1330094

You are mixing this question with this one: in-the-figure-above-pqrs-is-a-square-and-ab-ac-is-the-area-of-tria-192330.html (OA is C there).

Sorry for the mistake. You can remove my post here as it is not relevant(Im not able to delete it).
in-the-figure-above-pqrs-is-a-square-and-ab-ac-is-the-area-of-tria-192330.html
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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19 Dec 2015, 17:42
Bunuel I am reviewing the other problem that is similar to this and I have two quick follow up questions.

(1) How are you able to assume in statement 1 that CE is also 1? It makes sense intuitively, but I'm not seeing that it has to be equal to segment RB.

(2) For the second question, it is not as clear how to pick smart numbers for statement 1 or 2.

I found myself guessing the right answer for this question but spending a minute and a half trying to prove why it was correct. Any advice for approaching this problem is helpful.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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06 Jul 2016, 15:24
Hi guys,

I know this is an old question, but I'm not sure about one thing.

Should I assume by looking at the figure that CE=RB? Because in the first statement it just say that RE is less than twice the length of BR, but how do I know that the triangle is "centered" with the square, by not looking the second statement?

thanks
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.  [#permalink]

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07 Jul 2016, 04:29
guireif wrote:
Hi guys,

I know this is an old question, but I'm not sure about one thing.

Should I assume by looking at the figure that CE=RB? Because in the first statement it just say that RE is less than twice the length of BR, but how do I know that the triangle is "centered" with the square, by not looking the second statement?

thanks

Since AB = AC (triangle ABC is isosceles), and AS = AQ, then CE must be equal to RB.
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Re: In the figure above, SQRE is a square, AB = AC, and AS = AQ.   [#permalink] 07 Jul 2016, 04:29

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