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# Equilateral Triangle Question

Author Message
Manager
Joined: 01 Aug 2008
Posts: 108

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07 Jun 2009, 00:30
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

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Pls help.

--== Message from GMAT Club Team ==--

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Do not answer without sharing the reasoning behind ur choice
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Working on my weakness : GMAT Verbal
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SVP
Joined: 04 May 2006
Posts: 1824
Schools: CBS, Kellogg

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07 Jun 2009, 02:58
mbaMission wrote:
Pls help.

D

The larger area =36square root3
the smaller area=9square root3

So the area of the border=27square root 3
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Manager
Joined: 10 May 2009
Posts: 63

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07 Jun 2009, 03:06
How did you get the length of bigger triangle?

sondenso wrote:
mbaMission wrote:
Pls help.

D

The larger area =36square root3
the smaller area=9square root3

So the area of the border=27square root 3
Manager
Joined: 01 Aug 2008
Posts: 108

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07 Jun 2009, 03:51
sondenso wrote:
mbaMission wrote:
Pls help.

D

The larger area =36square root3
the smaller area=9square root3

So the area of the border=27square root 3

How to calculate the larger area?
_________________

==============================================
Do not answer without sharing the reasoning behind ur choice
-----------------------------------------------------------
Working on my weakness : GMAT Verbal
------------------------------------------------------------
Why, What, How, When, Where, Who
==============================================

Director
Joined: 03 Jun 2009
Posts: 772
Location: New Delhi
WE 1: 5.5 yrs in IT

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07 Jun 2009, 04:10
2
KUDOS
Attachment:

solution.jpg [ 12.81 KiB | Viewed 1305 times ]

refer to the attachment.

For smaler triangle, s = 18 / 3 =6
Hieght = √(6^2 + 3^2) = 3√3
Area = 1/2 * 6 * 3√3 = 9√3

For larger traingle, given DQ = √3
Angle QBD = 30 degree (since QB will bisect angle CBA)
Angle QDA = 90 degree, and hence angle DQB = 60

sides of 30-60-90 triangle are always in ratio of 1:√3:2
since DQ = √3
=> DB = √3 * √3 = 3
and QB = 2√3 = AP

Hence, base BC = 6 + 3 + 3 = 12
Height = 3√3 + √3 + 2√3 = 6√3
Area = 1/2 * 12 * 6√3 = 36√3

Diff = 27√3

OA is D.

Please let me know if you need more explanation
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Manager
Joined: 01 Aug 2008
Posts: 108

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07 Jun 2009, 04:39
bigoyal wrote:
Attachment:
solution.jpg

refer to the attachment.

For smaler triangle, s = 18 / 3 =6
Hieght = √(6^2 + 3^2) = 3√3
Area = 1/2 * 6 * 3√3 = 9√3

For larger traingle, given DQ = √3
Angle QBD = 30 degree (since QB will bisect angle CBA)
Angle QDA = 90 degree, and hence angle DQB = 60

sides of 30-60-90 triangle are always in ratio of 1:√3:2
since DQ = √3
=> DB = √3 * √3 = 3
and QB = 2√3 = AP

Hence, base BC = 6 + 3 + 3 = 12
Height = 3√3 + √3 + 2√3 = 6√3
Area = 1/2 * 12 * 6√3 = 36√3

Diff = 27√3

OA is D.

Please let me know if you need more explanation

gr8 job ! thanks mate
_________________

==============================================
Do not answer without sharing the reasoning behind ur choice
-----------------------------------------------------------
Working on my weakness : GMAT Verbal
------------------------------------------------------------
Why, What, How, When, Where, Who
==============================================

Manager
Joined: 28 Jan 2004
Posts: 201
Location: India

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07 Jun 2009, 22:50
I think the height of the smaller traingle should be 5.
By using Pythagoras theorem....

(Height)^2 + (Base)^2 = (Hypot..)^2
h^2 + 3^2 = 6^2
or H=5.
How is it 3sqrt(3) ???????????????
Manager
Joined: 01 Aug 2008
Posts: 108

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07 Jun 2009, 23:36
mdfrahim wrote:
I think the height of the smaller traingle should be 5.
By using Pythagoras theorem....

(Height)^2 + (Base)^2 = (Hypot..)^2
h^2 + 3^2 = 6^2
or H=5.
How is it 3sqrt(3) ???????????????

Altitude of equilateral triangle is (sqr rt 3) / 2 X (length of side)
_________________

==============================================
Do not answer without sharing the reasoning behind ur choice
-----------------------------------------------------------
Working on my weakness : GMAT Verbal
------------------------------------------------------------
Why, What, How, When, Where, Who
==============================================

Director
Joined: 03 Jun 2009
Posts: 772
Location: New Delhi
WE 1: 5.5 yrs in IT

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08 Jun 2009, 02:23
mdfrahim wrote:
I think the height of the smaller traingle should be 5.
By using Pythagoras theorem....

(Height)^2 + (Base)^2 = (Hypot..)^2
h^2 + 3^2 = 6^2
or H=5.
How is it 3sqrt(3) ???????????????

You are calculating it incorrectly.
h^2 + 3^2 = 6^2
or, h^2 = 36 - 9 = 27
or, h = √27 = 3√3

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: Equilateral Triangle Question   [#permalink] 08 Jun 2009, 02:23
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