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Area of an overlapping section

by prachich1987 » Fri Jan 07, 2011 10:40 pm
Source : Kaplan

What is the area of the region in which squares ABCD and EFGH overlap?

STATEMENT 1:

EF bisects BC.

STATEMENT 2:

The distance from point C to point E is 2*sq root 2 and the distance from point C to point F is 2*sq root 2.
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by bblast » Fri Jan 07, 2011 11:37 pm
You forgot to complete the 2nd statement :P

anyways.

statement 1 is insufficient:
if the statement read - "EF is the perpendicular bisector of BC" then it would have been of any interest to us.
(in this case even that would have been insufficient as we do not know the position of point E and F. statement would have been suff if it said EF and BC bisect each other at a right angle )

statement 2 is sufficient.
if we know one diagonal length of the square we can calculate
A = (d^2)/2


Here we are given the diagonal length for 2 small squares which make up 2/7 th of the entire figure, hence area of 1/7th of the entire region(overlapping region) can be calculated.

IMO answer is B

@Anurag u might need to correct me on this :)
Last edited by bblast on Fri Jan 07, 2011 11:42 pm, edited 3 times in total.
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by Anurag@Gurome » Fri Jan 07, 2011 11:38 pm
I think we are missing something in statement 2.
prachich1987 wrote:STATEMENT 2:

The distance from point C to point E is something should be here and the distance from point C to point F is something should be here too.
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by bblast » Sat Jan 08, 2011 12:02 pm
prachich1987 wrote:Source : Kaplan

What is the area of the region in which squares ABCD and EFGH overlap?

STATEMENT 1:

EF bisects BC.

STATEMENT 2:

The distance from point C to point E is and the distance from point C to point F is .
any update on this one ?
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by prachich1987 » Sat Jan 08, 2011 12:05 pm
bblast wrote:
prachich1987 wrote:Source : Kaplan

What is the area of the region in which squares ABCD and EFGH overlap?

STATEMENT 1:

EF bisects BC.

STATEMENT 2:

The distance from point C to point E is and the distance from point C to point F is .
any update on this one ?
I edited the question
Sorry for the inconvenience
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by prachich1987 » Sat Jan 08, 2011 12:07 pm
Is it necessary that the overlapping portion has to be a square?
I don't know how to prove this :(
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by bblast » Sat Jan 08, 2011 12:16 pm
prachich1987 wrote:Is it necessary that the overlapping portion has to be a square?
I don't know how to prove this :(
i already gave u the explanation, it will be a square only if both diagonals given in statement 2 are equal.

its just an application for a simple formula for area of a square = (d^2)/2

statment 1 is insufficient for my said reasons,

maybe anurag provides a better explanation than mine tmrw :(
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by anshumishra » Sat Jan 08, 2011 1:35 pm
prachich1987 wrote:Source : Kaplan

What is the area of the region in which squares ABCD and EFGH overlap?

STATEMENT 1:

EF bisects BC.

STATEMENT 2:

The distance from point C to point E is 2*sq root 2 and the distance from point C to point F is 2*sq root 2.
Question : area of EXCY = ?

Statement 1 :
BX = CX
Not sufficient (As In this case EX can be equal to XF or not, and depending on that area of EXCY varies).

Statement 2 :
CE = CF = 2*sqrt2
Not Sufficient - As you can rotate EXCY, which changes the shape BUT NOT THE AREA ( still CE=CF = 2*sqrt2)
Please check the post on the next page which has the full OA Explanation.

I could verify by rotating it such that the shape is converted into a triangle with sides (2sqrt2 each).
Area = 1/2*2sqrt2*2sqrt2 = 4 = the same area when it was square of side length 2.
We can verify for intermediate shape also, but i would say move on. IT would be way too much time worth spending.

So, B
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by Anurag@Gurome » Sun Jan 09, 2011 12:35 am
anshumishra wrote:Although it is clear, however here is how you can prove that ECXY is a square with side length equal to 2.
The triangles EXC and CXF are congruent (EX = XF, XC = XC , CE = XF)
So, angle EXC = angle FXC = 90 degree
CX = 2, EC = 2*sqrt2; Hence EX = 2
CY = EX = 2
EY = XC = 2
Just want to make a small addition which explains why we can take BC is perpendicular bisector of EF.

Note that we cannot assume that EX = XF from the image, because if we can do so, then the question stem itself is necessary to solve the problem.

From statement we know that C is equidistant from E and F. This is only possible if C lies on the perpendicular bisector of the line segment EF. Hence BC is the perpendicular bisector of EF. Rest of the explanation is same as Anshu's explanation.
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by jaxis » Sun Jan 09, 2011 5:33 am
anshumishra wrote:
prachich1987 wrote:Source : Kaplan

What is the area of the region in which squares ABCD and EFGH overlap?

STATEMENT 1:

EF bisects BC.

STATEMENT 2:

The distance from point C to point E is 2*sq root 2 and the distance from point C to point F is 2*sq root 2.
Question : area of EXCY = ?

Statement 1 :
BX = CX
Not sufficient (As In this case EX can be equal to XF or not, and depending on that area of EXCY varies).

Statement 2 :
CE = CF = 2*sqrt2
EXCY is a square with side length = 2 -> Sufficient

So, B

Although it is clear, however here is how you can prove that ECXY is a square with side length equal to 2.
The triangles EXC and CXF are congruent (EX = XF, XC = XC , CE = XF)
So, angle EXC = angle FXC = 90 degree
CX = 2, EC = 2*sqrt2; Hence EX = 2
CY = EX = 2
EY = XC = 2
Hi Ashu,
I think I am missing some thing here.
I understood till So, angle EXC = angle FXC = 90 degree.
I couldn't understand the part where you took CX= 2.How can we reach to this conclusion.?


I think you have considered Angle ECX 90 degrees. But we cant prove it.Please correct me if im wrong.

@ anurag I understood that BC is perpencula bisec to of EF , but Point C has to be the center of square EFGH to make EXCY a square.

Please calrify my doubt.
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by anshumishra » Sun Jan 09, 2011 7:04 am
jaxis wrote:
anshumishra wrote:
prachich1987 wrote:Source : Kaplan

What is the area of the region in which squares ABCD and EFGH overlap?

STATEMENT 1:

EF bisects BC.

STATEMENT 2:

The distance from point C to point E is 2*sq root 2 and the distance from point C to point F is 2*sq root 2.
Question : area of EXCY = ?

Statement 1 :
BX = CX
Not sufficient (As In this case EX can be equal to XF or not, and depending on that area of EXCY varies).

Statement 2 :
CE = CF = 2*sqrt2
EXCY is a square with side length = 2 -> Sufficient

So, B

Although it is clear, however here is how you can prove that ECXY is a square with side length equal to 2.
The triangles EXC and CXF are congruent (EX = XF, XC = XC , CE = XF)
So, angle EXC = angle FXC = 90 degree
CX = 2, EC = 2*sqrt2; Hence EX = 2
CY = EX = 2
EY = XC = 2
Hi Ashu,
I think I am missing some thing here.
I understood till So, angle EXC = angle FXC = 90 degree.
I couldn't understand the part where you took CX= 2.How can we reach to this conclusion.?


I think you have considered Angle ECX 90 degrees. But we cant prove it.Please correct me if im wrong.

@ anurag I understood that BC is perpencula bisec to of EF , but Point C has to be the center of square EFGH to make EXCY a square.

Please calrify my doubt.
@jaxis,
If you agree on:
angle EXC = 90 degree
EX = XF = EF/2 = 4/2 = 2
EC = 2sqrt2
Then, CX = sqrt(8-4) = 2

However, I just thought something else, I am going to post next.
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by anshumishra » Sun Jan 09, 2011 7:14 am
Anurag@Gurome wrote:
anshumishra wrote:Although it is clear, however here is how you can prove that ECXY is a square with side length equal to 2.
The triangles EXC and CXF are congruent (EX = XF, XC = XC , CE = XF)
So, angle EXC = angle FXC = 90 degree
CX = 2, EC = 2*sqrt2; Hence EX = 2
CY = EX = 2
EY = XC = 2
Just want to make a small addition which explains why we can take BC is perpendicular bisector of EF.

Note that we cannot assume that EX = XF from the image, because if we can do so, then the question stem itself is necessary to solve the problem.

From statement we know that C is equidistant from E and F. This is only possible if C lies on the perpendicular bisector of the line segment EF. Hence BC is the perpendicular bisector of EF. Rest of the explanation is same as Anshu's explanation.
Anurag,

Consider EFGH as a frame, hinged at C (include CE and CF as part of the frame with length 2sqrt2). One can rotate it slightly in clockwise or anti-clockwise direction and still C will be equidistant from E and F.

If that is true, then we shouldn't assume anything based on the figure drawn, and consider B as insufficient.
And probably combining "A" and "B", we can make sure that the frame is not skewed and it is totally symmetric. Hence "C" could be the answer.

I haven't given too much thought on it, so let me know if you find any flaw with the reasoning provided.
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by gmatmachoman » Sun Jan 09, 2011 7:17 am
jaxis wrote:
anshumishra wrote:
prachich1987 wrote:Source : Kaplan

What is the area of the region in which squares ABCD and EFGH overlap?

STATEMENT 1:

EF bisects BC.

STATEMENT 2:

The distance from point C to point E is 2*sq root 2 and the distance from point C to point F is 2*sq root 2.
Question : area of EXCY = ?

Statement 1 :
BX = CX
Not sufficient (As In this case EX can be equal to XF or not, and depending on that area of EXCY varies).

Statement 2 :
CE = CF = 2*sqrt2
EXCY is a square with side length = 2 -> Sufficient

So, B

Although it is clear, however here is how you can prove that ECXY is a square with side length equal to 2.
The triangles EXC and CXF are congruent (EX = XF, XC = XC , CE = XF)
So, angle EXC = angle FXC = 90 degree
CX = 2, EC = 2*sqrt2; Hence EX = 2
CY = EX = 2
EY = XC = 2
Hi Ashu,
I think I am missing some thing here.
I understood till So, angle EXC = angle FXC = 90 degree.
I couldn't understand the part where you took CX= 2.How can we reach to this conclusion.?


I think you have considered Angle ECX 90 degrees. But we cant prove it.Please correct me if im wrong.

@ anurag I understood that BC is perpencula bisec to of EF , but Point C has to be the center of square EFGH to make EXCY a square.

Please calrify my doubt.
The original poster has asked why the "overlapping portion is a square"?

Here is my explanation !

angle (ECF) = 90 , CE = CF = 2V2. So this forms a Isoceles Right triangle( ECF)

Sin 45 = CX/CE

1/V2 = CX/2V2

CX= 2

tan 45 = EX/CX

1= EX/CX; so EX = 2

Since EX=CX=2, the overlapping figure is a square.
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by Anurag@Gurome » Sun Jan 09, 2011 7:25 am
anshumishra wrote:Anurag,

Consider EFGH as a frame, hinged at C (include CE and CF as part of the frame with length 2sqrt2). One can rotate it slightly in clockwise or anti-clockwise direction and still C will be equidistant from E and F.
Yes, and it'll be on the perpendicular bisector of EF. :)
If you have any doubt, I can give you the proof.
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by anshumishra » Sun Jan 09, 2011 7:35 am
Anurag@Gurome wrote:
anshumishra wrote:Anurag,

Consider EFGH as a frame, hinged at C (include CE and CF as part of the frame with length 2sqrt2). One can rotate it slightly in clockwise or anti-clockwise direction and still C will be equidistant from E and F.
Yes, and it'll be on the perpendicular bisector of EF. :)
If you have any doubt, I can give you the proof.
That is true. Also, C is the center of the square. Please see the other part of the previous post. Don't you think the answer based on that reasoning should be "C" ?
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