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

It is currently 24 May 2020, 21:55

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.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
B
Joined: 01 Oct 2009
Posts: 335
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36
WE: Business Development (Consulting)
GMAT ToolKit User Reviews Badge
ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post Updated on: 06 May 2020, 07:59
2
19
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

39% (02:19) correct 61% (02:40) wrong based on 165 sessions

HideShow timer Statistics

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

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

Attachment:
1.JPG
1.JPG [ 5.35 KiB | Viewed 16134 times ]

Originally posted by RaviChandra on 14 Sep 2010, 22:58.
Last edited by Bunuel on 06 May 2020, 07:59, edited 3 times in total.
Renamed the topic and edited the question.
Senior Manager
Senior Manager
avatar
B
Joined: 01 Oct 2009
Posts: 335
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36
WE: Business Development (Consulting)
GMAT ToolKit User Reviews Badge
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 14 Sep 2010, 22:59
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.PNG [ 2.36 KiB | Viewed 16027 times ]
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 64068
ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 14 Sep 2010, 23:54
2
RaviChandra wrote:
Image

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

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
7.PNG [ 5.36 KiB | Viewed 15985 times ]
Answer: D.

RaviChandra these are not GMAT questions, so I wouldn't worry about them too much and definitely wouldn't spend much time on them.
_________________
Intern
Intern
User avatar
Joined: 07 Sep 2010
Posts: 9
Location: Doha, Qatar
Schools: INSEAD, Princeton, HKUST, Oxford, ISB, SP Jain
WE 1: 5.5
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 15 Sep 2010, 00:39
Wht is answer of the 2nd question ???

24 ?
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 702
Location: London
GMAT ToolKit User Reviews Badge
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 15 Sep 2010, 01:03
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

_________________
Intern
Intern
User avatar
Joined: 07 Sep 2010
Posts: 9
Location: Doha, Qatar
Schools: INSEAD, Princeton, HKUST, Oxford, ISB, SP Jain
WE 1: 5.5
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 15 Sep 2010, 04:02
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
Manager
Manager
avatar
Joined: 24 Jul 2011
Posts: 160
Location: India
GMAT 1: 570 Q50 V19
GMAT 2: 650 Q49 V28
GMAT 3: 690 Q50 V34
WE: Information Technology (Investment Banking)
GMAT ToolKit User Reviews Badge
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 28 Sep 2015, 05:38
2
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!
Manager
Manager
avatar
S
Joined: 02 Jan 2020
Posts: 211
CAT Tests
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 06 May 2020, 02:18
MentorTutoring

Can you pls explain how we ascertain height of shaded triangle relative to triangle ABC in this question
Tutor
User avatar
D
Joined: 16 May 2019
Posts: 709
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 06 May 2020, 05:13
1
GDT wrote:
MentorTutoring

Can you pls explain how we ascertain height of shaded triangle relative to triangle ABC in this question

Thank you, GDT, for tagging me. This is a visual puzzle more than anything, and to be honest, I cannot better demonstrate how to approach this question than Bunuel has done above. If you rotate the figure and mark all the congruencies, you will see how the answer fits without knowing exact values. I will only add that a useful three-step approach to any Quant question is to ask yourself the following:

1) What is the question asking?

2) What information do I have to work with?

3) What is the core concept being tested?

In this instance, I think many test-takers would overlook the obvious in 3) above. The question is asking about area, and the shape is a triangle, so the equation

\(A=\frac{1}{2}*b*h\)

will come into play. To make any sort of meaningful comparison, we need to find a way to hold one of our two variables constant while we manipulate the other. In terms of a guessing strategy, only (C) or (D) make any sense, given the setup. It is really down to whether you can find a way to justify half of the area of the overall triangle or slightly more than half.

- Andrew
Manager
Manager
avatar
S
Joined: 02 Jan 2020
Posts: 211
CAT Tests
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 06 May 2020, 05:33
MentorTutoring wrote:
GDT wrote:
MentorTutoring

Can you pls explain how we ascertain height of shaded triangle relative to triangle ABC in this question

Thank you, GDT, for tagging me. This is a visual puzzle more than anything, and to be honest, I cannot better demonstrate how to approach this question than Bunuel has done above. If you rotate the figure and mark all the congruencies, you will see how the answer fits without knowing exact values. I will only add that a useful three-step approach to any Quant question is to ask yourself the following:

1) What is the question asking?

2) What information do I have to work with?

3) What is the core concept being tested?

In this instance, I think many test-takers would overlook the obvious in 3) above. The question is asking about area, and the shape is a triangle, so the equation

\(A=\frac{1}{2}*b*h\)

will come into play. To make any sort of meaningful comparison, we need to find a way to hold one of our two variables constant while we manipulate the other. In terms of a guessing strategy, only (C) or (D) make any sense, given the setup. It is really down to whether you can find a way to justify half of the area of the overall triangle or slightly more than half.

- Andrew


Hi MentorTutoring
I read the explanation given by Bunuel but could not understand how we get to know that height of shaded triangle is half of that of triangle ABC

Could you pls explain how we get to infer that from the figure

Thanks in advance!
Tutor
User avatar
D
Joined: 16 May 2019
Posts: 709
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 06 May 2020, 06:36
1
GDT wrote:

Hi MentorTutoring
I read the explanation given by Bunuel but could not understand how we get to know that height of shaded triangle is half of that of triangle ABC

Could you pls explain how we get to infer that from the figure

Thanks in advance!

If you look at the left and right sides of the largest figure in the diagram, you can see how each third is congruent on that side. Since the problem tells us that ED = FD, because that diagonal cuts through the central third, we can deduce that the "height" of triangle ADC comprises exactly half of the middle third plus all of the bottom third, or one and a half out of three thirds of the largest triangle. How can you tell those thirds are congruent? It has to do with parallel lines (or segments) and, again, the information provided in the problem. If BE is 1/3 of BC and you drew a segment from point E that was parallel to AC, you would have 1/3 of side AB; likewise, if you drew a segment from point F that was parallel to AC, you have have 1/3 of side BC, since AF is 1/3 of AB. That leaves each side (that we are not calling the base) of the largest figure carved up into equal thirds. Does it make more sense now?

- Andrew
Manager
Manager
avatar
S
Joined: 02 Jan 2020
Posts: 211
CAT Tests
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 06 May 2020, 07:29
MentorTutoring wrote:
GDT wrote:

Hi MentorTutoring
I read the explanation given by Bunuel but could not understand how we get to know that height of shaded triangle is half of that of triangle ABC

Could you pls explain how we get to infer that from the figure

Thanks in advance!

If you look at the left and right sides of the largest figure in the diagram, you can see how each third is congruent on that side. Since the problem tells us that ED = FD, because that diagonal cuts through the central third, we can deduce that the "height" of triangle ADC comprises exactly half of the middle third plus all of the bottom third, or one and a half out of three thirds of the largest triangle. How can you tell those thirds are congruent? It has to do with parallel lines (or segments) and, again, the information provided in the problem. If BE is 1/3 of BC and you drew a segment from point E that was parallel to AC, you would have 1/3 of side AB; likewise, if you drew a segment from point F that was parallel to AC, you have have 1/3 of side BC, since AF is 1/3 of AB. That leaves each side (that we are not calling the base) of the largest figure carved up into equal thirds. Does it make more sense now?

- Andrew


Is there any theorem that when a diagonal of a trapezium is getting bisected then at that point the ht. is divided into exactly 2 halves?
Tutor
User avatar
D
Joined: 16 May 2019
Posts: 709
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 06 May 2020, 09:26
2
GDT wrote:

Is there any theorem that when a diagonal of a trapezium is getting bisected then at that point the ht. is divided into exactly 2 halves?

I am in the middle of a lesson, so I must be brief. I think you are looking for a definition more than a theorem: midsegment. Here is a link you might find useful on the topic. I hope it helps.

- Andrew
Manager
Manager
avatar
S
Joined: 02 Jan 2020
Posts: 211
CAT Tests
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 06 May 2020, 23:01
MentorTutoring wrote:
GDT wrote:

Is there any theorem that when a diagonal of a trapezium is getting bisected then at that point the ht. is divided into exactly 2 halves?

I am in the middle of a lesson, so I must be brief. I think you are looking for a definition more than a theorem: midsegment. Here is a link you might find useful on the topic. I hope it helps.

- Andrew


Thank you! Now, it's clear
Tutor
User avatar
D
Joined: 16 May 2019
Posts: 709
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =  [#permalink]

Show Tags

New post 07 May 2020, 07:14
1
GDT wrote:
MentorTutoring wrote:
GDT wrote:

Is there any theorem that when a diagonal of a trapezium is getting bisected then at that point the ht. is divided into exactly 2 halves?

I am in the middle of a lesson, so I must be brief. I think you are looking for a definition more than a theorem: midsegment. Here is a link you might find useful on the topic. I hope it helps.

- Andrew


Thank you! Now, it's clear

That is music to my ears, GDT. Glad I could borrow a key "prop," as it were, from Bunuel. To reconstruct that same diagram in the way I know would take a long time. I will scroll back up to give that post a kudos right after this.

Good luck with your studies.

- Andrew
GMAT Club Bot
Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =   [#permalink] 07 May 2020, 07:14

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne