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hemant_rajput: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 22, 2012 7:13 am; Thanked: 46 times; Followed by:13 members; GMAT Score:700

geometry question

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

[/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: Beat The GMAT — Problem Solving | Wed Feb 27, 2013 9:13 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

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?

[/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.. â‰ˆ 25

The correct answer is D.

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hemant_rajput: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 22, 2012 7:13 am; Thanked: 46 times; Followed by:13 members; GMAT Score:700

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?
[/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.. â‰ˆ 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.

Quote

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

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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mariofelixpasku: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Sun Sep 23, 2012 4:26 am; Thanked: 2 times

by mariofelixpasku » Wed Mar 20, 2013 2:20 pm

can we expect such a question on the gmat ?

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

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