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geometry question

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by hemant_rajput » Wed Feb 27, 2013 9:13 am
In the figure below, AB=BC=CD=DE=EF=FG=GA. Then angle DAE is approximately?

a. 15
b. 20
c. 30
d. 25
e. 45

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[/img]
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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Source: — Problem Solving |

by Anurag@Gurome » Wed Feb 27, 2013 10:37 am
hemant_rajput wrote:In the figure below, AB=BC=CD=DE=EF=FG=GA. Then angle DAE is approximately?
Image[/img]

Let us assume, DAE = x
Triangle ABC is isosceles as AB = BC --> BCA = CAB = x
Hence, CBD = CAB + BCA = x + x = 2x .............. [External angle of triangle ABC]

Triangle BCD is isosceles as BC = CD --> CBD = CDB = 2x
Hence, DCE = DAE + CDA = x + 2x = 3x .............. [External angle of triangle ACD]

Triangle CDE is isosceles as CD = DE --> DCE = DEC = AED = 3x

Similarly, ADE = EFD = AEF + DAE = EGF + DAE = (DAE + GFA) + DAE = DAE + DAE + DAE = 3x

Hence, in triangle ADE, ADE + DAE + AED = 3x + x + 3x = 7x
Hence, 7x = 180 ---> x = 180/7 = 25.7.. ≈ 25

The correct answer is D.
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by hemant_rajput » Thu Feb 28, 2013 8:46 am
Anurag@Gurome wrote:
hemant_rajput wrote:In the figure below, AB=BC=CD=DE=EF=FG=GA. Then angle DAE is approximately?
Image[/img]

Let us assume, DAE = x
Triangle ABC is isosceles as AB = BC --> BCA = CAB = x
Hence, CBD = CAB + BCA = x + x = 2x .............. [External angle of triangle ABC]

Triangle BCD is isosceles as BC = CD --> CBD = CDB = 2x
Hence, DCE = DAE + CDA = x + 2x = 3x .............. [External angle of triangle ACD]

Triangle CDE is isosceles as CD = DE --> DCE = DEC = AED = 3x

Similarly, ADE = EFD = AEF + DAE = EGF + DAE = (DAE + GFA) + DAE = DAE + DAE + DAE = 3x

Hence, in triangle ADE, ADE + DAE + AED = 3x + x + 3x = 7x
Hence, 7x = 180 ---> x = 180/7 = 25.7.. ≈ 25

The correct answer is D.
How come GFA = DAE?
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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by Anurag@Gurome » Thu Feb 28, 2013 9:29 am
hemant_rajput wrote:How come GFA = DAE?
In triangle GAF, FG = GA ---> GAF = GFA
Now, GAF is nothing but DAE.

Hence, GFA = DAE

Hope that helps.
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by mariofelixpasku » Wed Mar 20, 2013 2:20 pm
can we expect such a question on the gmat ?
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by Brent@GMATPrepNow » Wed Mar 20, 2013 2:58 pm
mariofelixpasku wrote:can we expect such a question on the gmat ?
I'm going to say "no"

More than anything, it's too time-consuming. Simply drawing the diagram on scratch paper and labeling all of the equivalent lines (AB=BC=CD=DE=EF=FG=GA) will eat up a ton of time, and that's BEFORE any actual calculations begin.

IMPORTANT: On a Problem Solving question involving geometric figures, the diagram will be drawn to scale, unless it's stated otherwise. So, even if we don't know how to solve the question, we can probably eliminate one or two answer choices. In this case, the angle looks bigger than 15 degrees and less than 45 degrees. So, then we'd guess B, C or D.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by shivani susel » Thu Apr 17, 2014 7:25 pm
Is there any simple way to find ADE
after finding AED = 3x.
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