It is currently 20 Oct 2017, 13:25

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Equilateral Triangle Question

Author Message
Manager
Joined: 01 Aug 2008
Posts: 117

Kudos [?]: 161 [0], given: 2

### Show Tags

07 Jun 2009, 00:30
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Pls help.
Attachments

File comment: Equilateral Triangle Question

EquiTriQue.JPG [ 19.25 KiB | Viewed 1276 times ]

_________________

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

Kudos [?]: 161 [0], given: 2

SVP
Joined: 04 May 2006
Posts: 1881

Kudos [?]: 1399 [0], given: 1

Schools: CBS, Kellogg

### Show Tags

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
_________________

Kudos [?]: 1399 [0], given: 1

Manager
Joined: 10 May 2009
Posts: 65

Kudos [?]: 117 [0], given: 11

### Show Tags

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

Kudos [?]: 117 [0], given: 11

Manager
Joined: 01 Aug 2008
Posts: 117

Kudos [?]: 161 [0], given: 2

### Show Tags

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
==============================================

Kudos [?]: 161 [0], given: 2

Director
Joined: 03 Jun 2009
Posts: 782

Kudos [?]: 902 [2], given: 56

Location: New Delhi
WE 1: 5.5 yrs in IT

### Show Tags

07 Jun 2009, 04:10
2
KUDOS
Attachment:

solution.jpg [ 12.81 KiB | Viewed 1251 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
_________________

Kudos [?]: 902 [2], given: 56

Manager
Joined: 01 Aug 2008
Posts: 117

Kudos [?]: 161 [0], given: 2

### Show Tags

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
==============================================

Kudos [?]: 161 [0], given: 2

Manager
Joined: 28 Jan 2004
Posts: 202

Kudos [?]: 28 [0], given: 4

Location: India

### Show Tags

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) ???????????????

Kudos [?]: 28 [0], given: 4

Manager
Joined: 01 Aug 2008
Posts: 117

Kudos [?]: 161 [0], given: 2

### Show Tags

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
==============================================

Kudos [?]: 161 [0], given: 2

Director
Joined: 03 Jun 2009
Posts: 782

Kudos [?]: 902 [0], given: 56

Location: New Delhi
WE 1: 5.5 yrs in IT

### Show Tags

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
_________________

Kudos [?]: 902 [0], given: 56

Re: Equilateral Triangle Question   [#permalink] 08 Jun 2009, 02:23
Display posts from previous: Sort by