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Gmatprep3 Problem

Expert replies
by faraz_jeddah » Sun Sep 22, 2013 12:48 am
No Idea how to solve this.

Image
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Source: — Problem Solving |

by theCodeToGMAT » Sun Sep 22, 2013 5:25 am
Edited:
Area of the BCE = √2/4

Answer [spoiler][/spoiler]

Sorry, earlier I made a wrong diagram which resulted in answer 1/2
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Last edited by theCodeToGMAT on Sun Sep 22, 2013 7:24 am, edited 1 time in total.
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by faraz_jeddah » Sun Sep 22, 2013 5:33 am
I am not sure what the logic you have used but the answer is incorrect.

OA is B
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by theCodeToGMAT » Sun Sep 22, 2013 5:53 am
faraz_jeddah wrote:I am not sure what the logic you have used but the answer is incorrect.

OA is B
Really??????? Can you please share the explanation.. I was very convinced that I was correct :|
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by theCodeToGMAT » Sun Sep 22, 2013 7:03 am
faraz_jeddah wrote:I am not sure what the logic you have used but the answer is incorrect.

OA is B
yes, correct.. the answer is indeed .. small mistake :)

I have edited my post which is just below the question.
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by runningguy » Sun Sep 22, 2013 7:25 am
I'm confused as to why the answer is not 1/2 (A=1/2(b)(h)).

b=1
h=1

A=1/2(1)(1)=1/2
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by theCodeToGMAT » Sun Sep 22, 2013 7:32 am
runningguy wrote:I'm confused as to why the answer is not 1/2 (A=1/2(b)(h)).

b=1
h=1

A=1/2(1)(1)=1/2
How did you find that base and height are 1?? that's not possible

if you say that you drew a perpendicular on BE from C and that weighted 1.. then the triangle is not possible as BC & CE are 1.. so perpendicular to BE from C must be less than 1.

if you say that you extended the C and dropped a perpendicular from E on the extended C line.. then also similar rule applies..

Refer my solution just below the question..
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by [email protected] » Sun Sep 22, 2013 12:33 pm
Hi All,

You might find this approach to be a bit easier; it's based on spotting a "hidden" 45/45/90 triangle...


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by runningguy » Sun Sep 22, 2013 3:27 pm
theCodeToGMAT wrote:
runningguy wrote:I'm confused as to why the answer is not 1/2 (A=1/2(b)(h)).

b=1
h=1

A=1/2(1)(1)=1/2
How did you find that base and height are 1?? that's not possible

if you say that you drew a perpendicular on BE from C and that weighted 1.. then the triangle is not possible as BC & CE are 1.. so perpendicular to BE from C must be less than 1.

if you say that you extended the C and dropped a perpendicular from E on the extended C line.. then also similar rule applies..

Refer my solution just below the question..
I had incorrectly assumed that one of the legs that was equal to 1 was the height. After looking at my flash cards, I see that for an isosceles triangle, you can divide the hypotenuse by sqrt2 to find the length. This was a very tricky question for me.
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by GMATGuruNY » Mon Sep 23, 2013 2:41 am
Image

In square ABCD, since s=1, BD = √2 and CF = √2/2.
It is given that CE=1.
Thus, EF = √2/2 + 1.

∆BED = (1/2)(BD)(EF) = (1/2)(√2)(√2/2 + 1) = (1/2)(1 + √2) = 1/2 + √2/2.

∆BCD = (1/2)(square ABCD) = 1/2.

Quadrilateral BEDC = ∆BED - ∆BCD = (1/2 + √2/2) - 1/2 = √2/2.

∆BCE = (1/2)(quadrilateral BEDC) = (1/2)(√2/2) = √2/4.

The correct answer is B.
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by faraz_jeddah » Mon Sep 23, 2013 2:51 am
GMATGuruNY wrote:Image

In square ABCD, since s=1, BD = √2 and CF = √2/2.
It is given that CE=1.
Thus, EF = √2/2 + 1.

∆BED = (1/2)(BD)(EF) = (1/2)(√2)(√2/2 + 1) = (1/2)(1 + √2) = 1/2 + √2/2.

∆BCD = (1/2)(square ABCD) = 1/2.

Quadrilateral BEDC = ∆BED - ∆BCD = (1/2 + √2/2) - 1/2 = √2/2.

∆BCE = (1/2)(quadrilateral BEDC) = (1/2)(√2/2) = √2/4.

The correct answer is B.
Thanks Mitch! I must say it would be hard to figure this question out in less than 3 minutes.
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by shinys » Tue Sep 24, 2013 6:19 pm
Rich,
Can you please help me understand how you figured out that 135 deg angle?
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by [email protected] » Tue Sep 24, 2013 6:36 pm
Hi shinys,

We're told that the sides BC=CD=CE=1 and we're told that BE=DE. This info tells us that the 2 triangles are IDENTICAL.

Since they're identical, their two "big angles" + the corner of the square = 360 degrees

2(Big angle) + 90 = 360
2(Big angle) = 270
Big angle = 135

Be on the lookout for the "special" right triangles in GMAT Quant questions; sometimes they're obvious, sometimes they're hidden.

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Rich
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by Resp007 » Thu Sep 26, 2013 4:38 am
HI Rich,
I used GMATGuruNY approach and solved it correctly but as pointed its bit time consuming,
So looked into yours and couldn't understand how you found H.
How you were able to conclude its a 45-45-90 triangle please?

[email protected] wrote:Hi All,

You might find this approach to be a bit easier; it's based on spotting a "hidden" 45/45/90 triangle...


Image


GMAT assassins aren't born, they're made,
Rich
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by Resp007 » Thu Sep 26, 2013 4:39 am
repeat post...
Last edited by Resp007 on Thu Sep 26, 2013 4:42 am, edited 1 time in total.
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