BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Geometry

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Tue Nov 23, 2010 12:28 am
Let BM = x. So MC = 2x.
Or BC = x + 2x = 3x = AD.
Or BM/AD is x/3x = 1/3.
Now triangle BGM and AGD are similar.
So (Area of BGM)/(Area of AGD) = (BM)^2/(AD)^2 = 1/9.
Or Area of AGD is (9* area of BGM) = 9*1 = 9.
Next consider triangles ABM and BDC.
Now their heights will be same because they are between same two parallel lines.
So their areas will correspond to their bases which are in the ratio 1 : 3.
Hence the areas of ABM and BDC will be in the ratio 1 : 3.
Let the area of triangle ABG be m and the area of quadrilateral GMCD be n.
So m+1 = area of ABM and n+1 = area of BDC.
Or m+1 = 1/3*(n+1). (equation 1)
Also since area of ABD = area of BDC, m+9 = n+1. (equation 2).
From the two equations we get m = 3 and n = 11.
So area of parallelogram is 3+9+11+1 = 24.
The correct answer is D.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Nov 23, 2010 1:34 am
Image


Rahul Thanks for the Above explanation, Can we do the problem Like this,

We know the Area of Traingle BGM is 1

= 1/2 ( BM * Height of traingle )

Now BM * Height of Traingle Must be equal to 2

Only then the Area of Triangle will be 1 .

IF we assume that Length of BM is 2 and Height Lets say GX is 1

We know Triangles BGM and ADG is similar

BM / GX = AD / GY { Y is the point of line AD }

2/1 = 6 / GY

GY = 3

We know formula of parallelogram is Base and Height

Total Height is 1 + 3 = 4
Base is 6

answer is 24

I am not sure can we do it this way for Not Because i assumed everything from like value of BM , Value of GX But Got the same answer ...
Please tell me Can we do it this way............
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Nov 23, 2010 1:41 am
Rahul@gurome wrote:Let BM = x. So MC = 2x.
Or BC = x + 2x = 3x = AD.
Or BM/AD is x/3x = 1/3.
Now triangle BGM and AGD are similar.
So (Area of BGM)/(Area of AGD) = (BM)^2/(AD)^2 = 1/9. { why squaring }

Or Area of AGD is (9* area of BGM) = 9*1 = 9.

Next consider triangles ABM and BDC.

Now their heights will be same because they are between same two parallel lines.

So their areas will correspond to their bases which are in the ratio 1 : 3.

Hence the areas of ABM and BDC will be in the ratio 1 : 3.

Let the area of triangle ABG be m and the area of quadrilateral GMCD be n.

So m+1 = area of ABM and n+1 = area of BDC.
Or m+1 = 1/3*(n+1). (equation 1)
Also since area of ABD = area of BDC, m+9 = n+1. (equation 2).
From the two equations we get m = 3 and n = 11.
So area of parallelogram is 3+9+11+1 = 24.
The correct answer is D.
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 537
Joined: Fri Jan 22, 2010 10:06 pmFollowed by: 1 member
Thanked: 14 times

by frank1 » Tue Nov 23, 2010 2:28 am
goyalsau wrote:
BM / GX = AD / GY { Y is the point of line AD }

2/1 = 6 / GY
i understood every thing except this
we have considered BM=1 and MC=2
thats mean BC=3
as it is parallelogram AD should be 3 how is it 6? (i dont see any other relations)
Bit confused...

thanks
GMAT score is equally counted as your GPA and 78 clicks can change you life.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Tue Nov 23, 2010 2:54 am
In similar triangles, the ratio of areas of two triangles is the square of the ratio of corresponiding sides.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Nov 23, 2010 4:29 am
frank1 wrote:
goyalsau wrote:
BM / GX = AD / GY { Y is the point of line AD }

2/1 = 6 / GY
i understood every thing except this
we have considered BM=1 and MC=2
thats mean BC=3
as it is parallelogram AD should be 3 how is it 6? (i dont see any other relations)
Bit confused...

thanks
I am considering BM is equal to 2 and MC is equal to 4

Then AD is equal to 6
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Tue Nov 23, 2010 4:32 am
Rahul@gurome wrote:In similar triangles, the ratio of areas of two triangles is the square of the ratio of corresponiding sides.
This means If we have two corresponding sides of similar triangles we can determine the ratio of the Areas of two triangles

and this will not be the actual Area it will only be the ratio of AREAS.
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Tue Nov 23, 2010 5:12 am
goyalsau wrote:Image
I solved this very quickly by guessing and checking.

Let's try BM=1. Then MC = 2 and BC=3. This means that AD=3.
Triangles BMG and AGD are similar because all their corresponding angles are equal.
Since BM=1 and AD=3, the ratio of the heights of triangles BMG and AGD is 1:3.
Since the area of triangle BMG=1 and BM=1, h = area/(1/2*b) = 1(1/2*1) = 2.
Since the ratio of the heights of the two triangles is 1:3, the height of triangle AGD = 3*2 = 6.
Thus, the height of parallelogram ABCD = height of BMG + height of AGD = 2+6 = 8.
Area of parallelogram ABCD = bh = 3*8 = 24.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1