GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 23 Jan 2020, 22:51

### 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

# ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 02 Oct 2009
Posts: 355
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36
ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

### Show Tags

Updated on: 09 Jul 2013, 06:50
2
12
00:00

Difficulty:

95% (hard)

Question Stats:

43% (02:10) correct 57% (02:42) wrong based on 176 sessions

### HideShow timer Statistics

ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure.
Attachment:

1.JPG [ 5.35 KiB | Viewed 14501 times ]

A. 1/3
B. 13/16
C. 5/9
D. 1/2

Originally posted by RaviChandra on 14 Sep 2010, 23:58.
Last edited by Bunuel on 09 Jul 2013, 06:50, edited 1 time in total.
Renamed the topic and edited the question.
Retired Moderator
Joined: 02 Sep 2010
Posts: 714
Location: London

### Show Tags

15 Sep 2010, 02:03
2
RaviChandra wrote:
In triangle ABC, D and E are points on AC and AB such DE || BC and length of DE is one-third of BC. If the area of triangle ABC is 216 square units, find the area of the shaded triangle.
a. 12
b. 18
c. 24
d. 16

Attachment:
2.PNG

Note that AED is similar to ABC with each side as 1/3rd of the original
=> Area(AED) = 1/9 * Area(ABC) =24

Area(ABD) = Area(ACE) = 1/3 * Area(ABC) = 72
[same height, 1/3rd base]

Area(ABD) = Area(ACE)
=> Area(ABD) - Area(AEDF) = Area(ACE) - Area(AEDF)
=> Area(BEF) = Area(CDF) = x (say)

Finally note that DEF is similar to BFC with one side in the ratio of 1:3
Thus area(DEF) = 1/9 * area(BFC) = y (say)

Area(EDC) = Area(EDF)+Area(DFC) = x + y
ALSO
Area(EDC) = Area(AEC) - Area(AED) =72 - 24 = 48
So x+y = 48

Area(BEDC) = Area(BEF) + Area(CDF) + Area(EDF) + Area(BFC) = x+x+y+9y
ALSO
Area(BEDC) = Area(ABC)-Area(AED)=216-24=192
So x+5y=96

Subtracting the 2 equations
4y=48
y=12

Hence, Area(DEF)=12

_________________
Senior Manager
Joined: 02 Oct 2009
Posts: 355
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36

### Show Tags

14 Sep 2010, 23:59
1
In triangle ABC, D and E are points on AC and AB such DE || BC and length of DE is one-third of BC. If the area of triangle ABC is 216 square units, find the area of the shaded triangle.
a. 12
b. 18
c. 24
d. 16

Attachment:

2.PNG [ 2.36 KiB | Viewed 14428 times ]
Math Expert
Joined: 02 Sep 2009
Posts: 60627

### Show Tags

15 Sep 2010, 00:54
1
RaviChandra wrote:
ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure.
Attachment:
The attachment 1.JPG is no longer available

a. 1/3
b. 13/16
c. 5/9
d. 1/2

Rotate the triangle so that CA to be the base. Now, triangle CDA has the same base as CBA and half of its height ("read" the diagram for explanation), which means that area of CDA will be half of the area of CBA, so 1/2.
Attachment:

7.PNG [ 5.36 KiB | Viewed 14386 times ]

RaviChandra these are not GMAT questions, so I wouldn't worry about them too much and definitely wouldn't spend much time on them.
_________________
Manager
Joined: 24 Jul 2011
Posts: 162
Location: India
GMAT 1: 570 Q50 V19
GMAT 2: 650 Q49 V28
GMAT 3: 690 Q50 V34
WE: Information Technology (Investment Banking)
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

### Show Tags

28 Sep 2015, 06:38
1
Whoa!!! A tough one

A small trick that helped me to solve it in 1:37

Imagine that the triangle ABC is an isosceles triangle with AB = BC. (Reason: All the mentioned conditions are satisfied so area ratio should also hold for Isosceles triangle)

Redraw it in your notepad, you will find symmetry. D is just in middle of AF and DE. You will figure out that D is a mid-point so its height should be half. Now if height of triangle is reduced to half, area of triangle will also reduce to half of its original value (Area = 0.5* base * height).

Hence D.
_________________
Middle of nowhere!
Intern
Joined: 07 Sep 2010
Posts: 9
Location: Doha, Qatar
Schools: INSEAD, Princeton, HKUST, Oxford, ISB, SP Jain
WE 1: 5.5

### Show Tags

15 Sep 2010, 01:39
Wht is answer of the 2nd question ???

24 ?
Intern
Joined: 07 Sep 2010
Posts: 9
Location: Doha, Qatar
Schools: INSEAD, Princeton, HKUST, Oxford, ISB, SP Jain
WE 1: 5.5

### Show Tags

15 Sep 2010, 05:02
Wow... thks a ton.

I had guessed it to be 24 - Thinking T. EFD should be of area 1/9th of main traingle.

If I knew T. AED = 1/9 T. ABC, 12 would have been my guess, as it (T. EFD) looks half the size of T. AED
Non-Human User
Joined: 09 Sep 2013
Posts: 14004
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

### Show Tags

08 Jun 2018, 00:04
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =   [#permalink] 08 Jun 2018, 00:04
Display posts from previous: Sort by