Find all School-related info fast with the new School-Specific MBA Forum

It is currently 02 Oct 2014, 03:33

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Geometry 2

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 19 Aug 2009
Posts: 81
Followers: 1

Kudos [?]: 59 [0], given: 46

Geometry 2 [#permalink] New post 26 Sep 2009, 02:22
1
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (05:09) correct 0% (00:00) wrong based on 1 sessions
Q.
In a triangle ABC, point D is on side AB and pt E is on side AC, such that BCED is a trapezium. DE : BC = 3:5. Calculate the ratio of the triangle ADE and trapezium BCED.

1. 3:4
2. 9:16
3. 3:5
4. 9:25
5. 5:9
1 KUDOS received
Manager
Manager
avatar
Joined: 27 Oct 2008
Posts: 186
Followers: 1

Kudos [?]: 77 [1] , given: 3

Re: Geometry 2 [#permalink] New post 26 Sep 2009, 11:33
1
This post received
KUDOS
I think the answer is 9/16. OA pls.

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)
thus
area of tri(ADE)/area of tri(ABC) = 9/25
from this we have
area of tri(ABC)/area of tri(ADE) = 25/9
we also know that area of tri(ABC) = area of tri(ADE) + area of trap(BCDE)
so substituting

(area of tri(ADE) + area of trap(BCDE))/area of tri(ADE) = 25/9
rearranging
1 + area of trap(BCDE)/area of tri(ADE) = 25/9
taking 1 to other side
area of trap(BCDE)/area of tri(ADE) = (25/9) - 1 = 16/9

since question is asking for area of tri(ADE)/area of trap(BCDE) = 9/16
Manager
Manager
avatar
Joined: 15 Sep 2009
Posts: 139
Followers: 1

Kudos [?]: 15 [0], given: 2

Re: Geometry 2 [#permalink] New post 27 Sep 2009, 00:48
Certainly it shud be 9:16
Manager
Manager
avatar
Joined: 19 Aug 2009
Posts: 81
Followers: 1

Kudos [?]: 59 [0], given: 46

Re: Geometry 2 [#permalink] New post 28 Sep 2009, 13:32
Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step

srivas wrote:
I think the answer is 9/16. OA pls.

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)
thus
area of tri(ADE)/area of tri(ABC) = 9/25
from this we have
area of tri(ABC)/area of tri(ADE) = 25/9
we also know that area of tri(ABC) = area of tri(ADE) + area of trap(BCDE)
so substituting

(area of tri(ADE) + area of trap(BCDE))/area of tri(ADE) = 25/9
rearranging
1 + area of trap(BCDE)/area of tri(ADE) = 25/9
taking 1 to other side
area of trap(BCDE)/area of tri(ADE) = (25/9) - 1 = 16/9

since question is asking for area of tri(ADE)/area of trap(BCDE) = 9/16
1 KUDOS received
Manager
Manager
avatar
Joined: 04 Sep 2009
Posts: 53
WE 1: Real estate investment consulting
Followers: 1

Kudos [?]: 17 [1] , given: 7

Re: Geometry 2 [#permalink] New post 28 Sep 2009, 14:49
1
This post received
KUDOS
It is 9/16

I approached via algebra (this is for the ones not familiar with the properties of similar triangles :) ):

Let the height of ADE = x, height of BCDE = y

Then:

S(ADE) = 3x/2
S(BCDE) = (3+5)/2 *y = 4y
S(ABC) = 5*(x+y)/2

Since S(ABC)=S(ADE)+S(BCDE)
3x/2 + 4y = 5(x+y)/2
3x+8y=5x+5y
x=1.5y

Now the ratio of ADE to BCDE = 3x/2 / 4y
3*1.5y/2 * 1/4y
4.5y/8y
4.5/8
9/16
1 KUDOS received
Manager
Manager
avatar
Joined: 11 Sep 2009
Posts: 129
Followers: 4

Kudos [?]: 173 [1] , given: 6

Re: Geometry 2 [#permalink] New post 28 Sep 2009, 14:52
1
This post received
KUDOS
I agree that the answer is B: 9:16 .

Please note though that the triangles DO NOT have to be similar. However, since this case applies to all trapezium, triangle pairs, you can simplify the problem by assuming ADE is similar to ABC. I took a slightly different approach. Since there are no restrictions given on Triangle ABC, I decided to create one that simplifies the problem (see attached image).

I assume that ABC is a right triangle, and that DE is parallel to BC so that ADE is similar to ABC. Therefore,

Area(Triangle) = Area(ADE)
Area(Trapezium) = Area(ABC) - Area(ADE)

Area(ADE) = 3 * 3 / 2 = 9/2
Area(ABC) = 5 * 5 / 2 = 25/2

Area(Triangle) = 9/2
Area(Trapezium) = 25/2 - 9/2 = 16/2 = 8

Area(Triangle) : Area(Trapezium) = 9/2 : 8 = 9 : 16
Attachments

triangle.JPG
triangle.JPG [ 10.83 KiB | Viewed 1476 times ]

1 KUDOS received
Manager
Manager
User avatar
Joined: 25 Mar 2009
Posts: 56
Followers: 1

Kudos [?]: 10 [1] , given: 9

Re: Geometry 2 [#permalink] New post 28 Sep 2009, 21:00
1
This post received
KUDOS
[quote="virtualanimosity"]Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step

Attachment:
untitled.JPG
untitled.JPG [ 9.57 KiB | Viewed 1425 times ]



ADK & ABH are similar triangle too:
AD/AB=AK/AH

Tri ADE & ABC are similar triangle:
AD/AB=AE/AC=DE/BC

SADE/SABC=(AK*DE/2)/(AH*BC/2)
=AK*DE/AH*BC
=AK/AH * DE/BC
=AD/AB * DE/BC
=DE/BC * DE/BC
=DE^2/BC^2

Is it clear to U now Virtualanimosity?
Manager
Manager
avatar
Joined: 19 Aug 2009
Posts: 81
Followers: 1

Kudos [?]: 59 [0], given: 46

Re: Geometry 2 [#permalink] New post 02 Oct 2009, 07:38
Thanks man...crystal clear
great detailed explanation..really helpful!!

Mikko wrote:
virtualanimosity wrote:
Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step

Attachment:
untitled.JPG



ADK & ABH are similar triangle too:
AD/AB=AK/AH

Tri ADE & ABC are similar triangle:
AD/AB=AE/AC=DE/BC

SADE/SABC=(AK*DE/2)/(AH*BC/2)
=AK*DE/AH*BC
=AK/AH * DE/BC
=AD/AB * DE/BC
=DE/BC * DE/BC
=DE^2/BC^2

Is it clear to U now Virtualanimosity?
Re: Geometry 2   [#permalink] 02 Oct 2009, 07:38
    Similar topics Author Replies Last post
Similar
Topics:
1 Geometry Triangle #2 cleetus 1 19 Nov 2011, 08:19
2 GMATPrep2 - Geometry ps seofah 6 20 Jul 2009, 14:10
1 2 Geometry ques...stuck greatchap 8 24 Feb 2008, 03:19
DS -GMAT Prep2 - Geometry seofah 4 28 Apr 2007, 14:13
DS Geometry 2 laxieqv 9 07 Jul 2006, 05:41
Display posts from previous: Sort by

Geometry 2

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.