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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
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joseph0alexander wrote:
Attachment:
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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


Similar question to practice: in-the-figure-above-sqre-is-a-square-ab-ac-and-as-aq-161814.html
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
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Is Area of a Triangle > Area of the Square?
In other words is (RE*CB)/2>RE^2?

We can simplify the equation by dividing both sides by RE (which is always >0) and multiply by 2, we get: Is CB>2*RE?

(1) Insufficient. Tells us that RE<2*BR. If we would know that BR=EC, then we could answer our question, because CB would be equal to 2BR+RE. However, if BR is not equal to CE, then we cannot know the length of CB.

(2) Insufficient. Does not tell us anything about the values that we are looking for. The only information it gives is that CE=RE (note that we know that the triangle is equilateral).

(1)+(2) Sufficient. 2nd statement basically tells us that CB= RE+2*BR. So, our question then looks like that: Is RE+2*BR>2*RE ---> Is 2*BR>RE? And the 1st statement tells us explicitly that RE<2*BR.

Answer C
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
AdmitJA wrote:
Attachment:
Capture.PNG
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


Hi bunuel,

Could you please provide your comments for statement 2.

basically st2 AS = AQ telling us that CE and RB are symmetric then why this is not sufficient.

Please clarify.

Thanks
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
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AdmitJA wrote:
Attachment:
Capture.PNG
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


SOLUTION:

Area of triangle ABC=\(1/2*ER*BC\)

Area of Square SQRE=\((ER)^2\)

Area of Traingle:Area of Sqare=\(1/2*BC:ER\)=\(1/2*(BR+ER+CE):ER\)

Statement 1): We are given only RE<2BR , But we know nothing about relation between BR and CE.

Hence, Insufficient.

Statement 2): AS=AQ--->BR=CE, but we don't know anything about relation between ER and any of BR and CE.

Hence, Insufficient.

Statement 1 & Statement 2:
Let ER=1
Area of Traingle: Area of Sqare=\(1/2(1+more than 1):1\)=\(More than 1: 1\).

So,YES, Area of traingle ABC is greater than the area of sqare SQRE.

Ans :C
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
Bunuel same question here, but I am re-posting because this question is different than the other question, in my opinion.

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) It is not as clear how to pick smart numbers for statement 1 or 2. Can you provide an explanation for this question?

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, PQRS is a square and AB = AC. Is the area of tria [#permalink]
Why we don't need to know if the triangle is a right triangle? If it was a right triangle then the area=CA*CB/2 instead of CB*(side of the square)/2
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
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Gabyms89 wrote:
Why we don't need to know if the triangle is a right triangle? If it was a right triangle then the area=CA*CB/2 instead of CB*(side of the square)/2


The triangle cannot be a right triangle, right angled at C. If it were, then AB would be the hypotenuse. The hypotenuse is always larger than both legs so AC = AB would not be possible.
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
I was thinking that the right angle might be A not C. Why it cannot be a right triangle at A?
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
VeritasKarishma Bunuel
gmatbusters

It is already given that AB=AC and therefore any line drawn from vertex A will act as perpendicular bisector at point P to the line BC.

Now if we take two triangle why cannot we apply RHS rule and further conclude that both are congruent.

The Perpendicular AP will be common to both triangle APC and APB. The hypotenuse AC=AB is given.

So why do we need option B.

With this reasoning I marked A as answer.
Please tell me where am I going wrong.
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
Gabyms89 wrote:
I was thinking that the right angle might be A not C. Why it cannot be a right triangle at A?


Hi,

How do you conclude it's right angled at A?
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
warrior1991 wrote:
VeritasKarishma Bunuel
gmatbusters

It is already given that AB=AC and therefore any line drawn from vertex A will act as perpendicular bisector at point P to the line BC.

Now if we take two triangle why cannot we apply RHS rule and further conclude that both are congruent.

The Perpendicular AP will be common to both triangle APC and APB. The hypotenuse AC=AB is given.

So why do we need option B.

With this reasoning I marked A as answer.
Please tell me where am I going wrong.


VeritasKarishma

Need your inputs here as well.Please see my concern above


Sent from my iPhone using GMAT Club Forum mobile app
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
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warrior1991 wrote:
VeritasKarishma Bunuel
gmatbusters

It is already given that AB=AC and therefore any line drawn from vertex A will act as perpendicular bisector at point P to the line BC.

Now if we take two triangle why cannot we apply RHS rule and further conclude that both are congruent.

The Perpendicular AP will be common to both triangle APC and APB. The hypotenuse AC=AB is given.

So why do we need option B.

With this reasoning I marked A as answer.
Please tell me where am I going wrong.


You need stmnt 2 because of the way data is given in statement 1.

Note that area of the triangle depends on 2 things - altitude and base. Area = (1/2)*Altitude*Base
Area of the square is the side of the square (which is same an altitude SE/QR). Area = Altitude^2

So depending on what is greater: Base/2 or Altitude, we will know whether the triangle has more area or square.

(1) The length of RE is less than twice the length of BR.

RE (side of square) is the same length as altitude.

RE < 2*BR
Altitude < 2*BR
BR is just a part of the base. We don't know the relation BR has to the rest of the base.

(2) AS = AQ
With this info, now we know the relation of BR with rest of the base.
Base = BR + CE + Altitude = BR + BR + Altitude
Base = 2*BR + Altitude
Since 2*BR is greater than Altitude, Base > 2*Altitude
So Base/2 is greater and hence triangle has more area.
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
AdmitJA wrote:
Attachment:
The attachment Capture.PNG is no longer available
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

MANHATTAN PREP OFFICIAL EXPLANATION
To rephrase the question, it is helpful to see what connections exist between the triangle and the square and furthermore how it is possible to relate their areas. Draw a vertical line from point A straight down to the base of the triangle (and label the bottom point of this line Z). This new line, AZ (of length x), is the height of triangle ABC and has the same length as any side of the square. Label the diagram with the rest of the given information: AB = AC, implying that triangle ABC is isosceles and that angle B must equal angle C.
Attachment:
Capture.JPG
Capture.JPG [ 15.25 KiB | Viewed 17737 times ]


The area of the square can be expressed as \(x^2\) and the area of the triangle as \(1/2\)(BC)(x) (label all lengths equal to AZ as x)
Since BC is comprised of ER + RB + CE, and ER is also a side of the square or x, you can rewrite BC as
x + RB + CE. Thus the question becomes “Is \(1/2\)(x+RB+CE)(x) > \(x^2\)?"
If RB and CE together sum to the length of x, then the left side of the inequality will be identical to the right: \(1/2\)(x+x)(x) = \(x^2\)

For the left side of the inequality to be greater than the right, however, RB and CE must sum to more than x, thus the final rephrase can be stated as
“Is RB + CE > x?”

(1) NOT SUFFICIENT: The statement can be translated as ER < 2RB

x < 2RB .... Again replacing ER with the variable x
\(x/2\) < RB

To know if RB + CE > x, you would need to know something about CE as well. Can you assume that CE = RB? From the given, triangle ABC is isosceles and thus symmetrical. However, there is no guarantee that is symmetrically placed in the square. If the triangle is symmetrically placed, RB = CE and both are greater than \(x/2\) and would thus sum to greater than x (scenario II below).
Scenario III below would also satisfy the question, but scenario I would not necessarily.
Attachment:
Capture1.JPG
Capture1.JPG [ 34.93 KiB | Viewed 17755 times ]


(2) NOT SUFFICIENT: This statement establishes the symmetric placement of triangle ABC within the square SQRE. On its own it does not relate RB and CE to x.

(1) AND (2) TOGETHER SUFFICIENT: Given the symmetric placement of triangle ABC within the square SQRE from statement (2) along with the information from statement (1), scenario II above is established and RB + CE > x.

The correct answer is (C).
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
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I am a bit confused by this question and the OE is not helping. I feel that statement 1 is sufficient because:

Let's call RB = a

If x < 2a, x^2 (area of the square MUST be less then ((x + 2a) * x)/2 if we substitute x in for 2a.

It should not matter whether or not the triangle is centered in the square right? the area will remain constant since it is an isosceles triangle. Would really appreciate an expert's advice. Thank you
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Re: In the figure above, PQRS is a square and AB = AC. Is the area of tria [#permalink]
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