GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Aug 2018, 14:46

### 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

# In the figure above, is the area of triangular region ABC

Author Message
TAGS:

### Hide Tags

Manager
Joined: 26 Feb 2015
Posts: 118
In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

13 Apr 2015, 02:41
Is a valid solution to this to plug in numbers instead of X to avoid painful calculations?

I simply assigned "AC" to 6 --> (AC)^2 = 2(AD)^2 --> 36 = 2(AD)^2. --> 18 =(AD)^2. AD = $$3\sqrt{2}$$

Jumping straight to (1)+(2): So, they are Isoceles: 45-45-90 ratio aka $$6-6-6\sqrt{2}$$. Which means Area of ABC = 36/2 = 18

Area of DBA = $$6\sqrt{2} * 3\sqrt{2}$$ / 2 = 18

I mean... as long as Im conistent with keeping "AC" to 6, it shouldn't really matter since i'm just expressing AC in a different way, right? I just find it way easier to work with actual numbers than with letters
Manager
Joined: 26 Feb 2015
Posts: 118
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

14 Apr 2015, 01:31
1
EMPOWERgmatRichC wrote:
Hi erikvm,

Yes, TESTing VALUES on this question is a great way to deal with a situation that is really 'technical.' As long as you're labeling your work and being thorough, you'll find that this approach works on LOTS of Quant questions.

GMAT assassins aren't born, they're made,
Rich

Thanks Rich. I am a big fan of testing values, but got a bit sceptical when it comes to DS questions.
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12183
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

14 Apr 2015, 22:35
Hi erikvm,

TESTing VALUES is a useful approach on most DS questions; you just have to be sure to adjust the tactic a bit. On a PS question, you're (typically) looking for the one answer that matches your values. On a DS question, you're TESTing multiple VALUES to see if a pattern emerges (or if you can prove that a pattern does NOT exist). Since that strategy works in so many different ways on Test Day, it's a great approach to practice throughout your studies.

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Intern
Joined: 13 Nov 2014
Posts: 41
Location: United States
Concentration: Marketing, Finance
GMAT 1: 640 Q47 V30
GPA: 3.5
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

11 May 2015, 00:07
jfk wrote:
Attachment:
OG13DS79v2.png
In the figure above, is the area of triangular region ABC equal to the area of triangular region DBA ?

(2) ∆ABC is isosceles.

Source: OG13 DS79

Ans C.

Keep CB as x. From equation 1, we get (AC)^2=2(AD)^2 , which gives as AD = x/ root 2. - NS

From 2, we get both AC and CB are x. which implies AB is root 2 * x. - NS as no info about other triangle.

Combine. we can find the area of both the triangles and its equal. Keeping CB as x is the key here. Hope it helps
Director
Joined: 10 Mar 2013
Posts: 561
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

Updated on: 20 Nov 2015, 15:49
1
Spent too much time on this one.....
Let write down what is given: Area1=Area2 ? (AC*BC)/2=(AD*AB)/2 ? --->(it's easier to eliminate 2 from denominator + $$ac^2*bc^2=ad^2*ab^2$$
We have also Info about Hypotenuse: $$DB^2=AD^2+AB^2$$, $$AB^2=AC^2+CB^2$$
1) $$AC^2=2*AD^2$$ --> $$2*AD^2*bc^2$$=$$ad^2*ab^2$$ --> $$2*bc^2=ab^2=ac^2+bc^2 --> bc^2=ac^2 ?$$ Not Sufficient
2) ac=bc --> ac^4=ad^2*ab^2 ?... Not Sufficient

(1+2) in (1) bc^2=ac^2 ? Statement 2 answers this question directly - Yes Answer (C)

Hi Math Experts, I see many people here dealing with complex numbers and roots in this problem, cano not we just square the expression in the question as I did ?
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Originally posted by BrainLab on 11 Jul 2015, 05:07.
Last edited by BrainLab on 20 Nov 2015, 15:49, edited 8 times in total.
Intern
Joined: 20 Aug 2014
Posts: 1
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

20 Jul 2015, 21:19
ltkenny wrote:
I still don't get it.
To compare the Area of (ABC) and Area of (DBA):
Then look at the graph:
(AB) is the hypotenuse of triangle (ABC)
-> AB > CB (2)
(1) and (2) :
-> the answer is NO, their areas are not equal
Thus, the answer must be A.
Can anyone please explain my mistake in here? Thanks

I got the same answer first. Then, i thought,
AB^2=AC^2+BC^2
That means Area of triangle ABD is
=0.5*sqrt(AC^2+BC^2)*AC/sqrt(2) which maybe greater than 0.5*AC*BC .
Unless you know BC and Ac values, you cannot say if area is greater, because of the square root 2, you cannot say, it reduces the value to 0.707 of the value.
So, it may be or may not be greater. Hence, it's not A.
Intern
Joined: 15 Mar 2015
Posts: 24
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

28 Jul 2015, 22:04
My approach on this was:

Is Area ABC = Area DBA ?

Area ABC = (CB.AC)/2

So, we are reduced to: Is (CB.AC)/2 = (AB.AD)/2 ? even further: Is (CB.AC)= (AB.AD) ?

St1-
√2(CB)= (AB) ? Cannot affirm anything about it ---> NOT SUFF

St2-
Triangle ABC is isoceles, so (AC)=(CB), but nothing can be affirmed the other segments: (AB) and (AD) ---> NOT SUFF

St1/2-
We know until now:
√2(CB)= (AB)
(AC)=(CB)
(AC)^2 + (CB)^2 = (AB)^2 (Pythagorean Theorem)

(CB)^2 + (CB)^2 = (AB)^2
2(CB)^2 = (AB)^2
√2(CB)=(AB)

Going back on the first equation:
yes, the areas are equal

I think I could've stopped in that moment where I noticed with all the three equations in hand I could find the relation between the areas.

Does this make sense ?
Board of Directors
Joined: 17 Jul 2014
Posts: 2717
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

16 Feb 2016, 20:17
took me some time to solve it..
we know that both triangles are right angles.
we are asked for areas if are equal
area is bxh/2

AC*CB=AD*AB - this is our question.

we know that:
AC^2+CB^2=AB^2

doesn't tell much.

2. ABC is isosceles. AC=CB
ok, so AB^2=2AC^2 or AB=AC*sqrt(2)

area of ABC is AC^2/2

area of ABD is AB*AD/2. we know AB, but it doesn't help much, as we still have 2 unknowns - AC and AD.

1+2.

since we found out in 2 that we have 2 unknowns, 1 fills out the gap:

We know that AB^2=2AC^2. we also know that AC^2=2AD^2.

we see that the areas are equal.
Manager
Joined: 17 Aug 2015
Posts: 115
GMAT 1: 650 Q49 V29
question 79 DATA Sufficiency GMAT OFFICIAL 13  [#permalink]

### Show Tags

08 Mar 2016, 07:49

triangle ABC area = AC*BC/2 AND TRinagle ADB area = AD*AB/2= ROOT2 AC* AB/2

SO Triangle ADB area will be always greater than triangle ABC REASON AB is the diagonal of right angle triangle abc and AB will be always greater than BC
Attachments

File comment: diagram picture DS 79

IMG_5494.JPG [ 540.96 KiB | Viewed 1801 times ]

Current Student
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
question 79 DATA Sufficiency GMAT OFFICIAL 13  [#permalink]

### Show Tags

08 Mar 2016, 07:57
vijaisingh2001 wrote:

triangle ABC area = AC*BC/2 AND TRinagle ADB area = AD*AB/2= ROOT2 AC* AB/2

SO Triangle ADB area will be always greater than triangle ABC REASON AB is the diagonal of right angle triangle abc and AB will be always greater than BC

Follow posting guidelines (link in my signatures). Search for a question before you post, type in the complete question and do not use pictures for questions. Topics merged.

The list of OG2013 quant questions is at the-official-guide-quantitative-question-directory-143450.html

Do not waste your time in questioning the answers of the official guide. They are always correct.
Math Expert
Joined: 02 Sep 2009
Posts: 47946
Re: question 79 DATA Sufficiency GMAT OFFICIAL 13  [#permalink]

### Show Tags

08 Mar 2016, 07:57
1
1
vijaisingh2001 wrote:

triangle ABC area = AC*BC/2 AND TRinagle ADB area = AD*AB/2= ROOT2 AC* AB/2

SO Triangle ADB area will be always greater than triangle ABC REASON AB is the diagonal of right angle triangle abc and AB will be always greater than BC

All questions are correct in Official Guide.

All of them have their own discussion on the forum:
THE OFFICIAL GUIDE FOR GMAT REVIEW 2016 DIRECTORY
Quantitative Review, 2ND Edition Directory
The Official Guide Quantitative Question Directory (13th Edition)

This one is here: in-the-figure-above-is-the-area-of-triangular-region-abc-134270.html

Hope it helps.

Locking the topic.

_________________
Current Student
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

08 Mar 2016, 08:07
vijaisingh2001 wrote:

triangle ABC area = AC*BC/2 AND TRinagle ADB area = AD*AB/2= ROOT2 AC* AB/2

SO Triangle ADB area will be always greater than triangle ABC REASON AB is the diagonal of right angle triangle abc and AB will be always greater than BC

C is the correct answer for this question.

Your statement above is NOT correct.

Area (ABC) = 0.5*AC*AB = 0.5*$$\sqrt{2}$$*AD*BC = 0.7 * AD*BC

Now although, BC < AB , do you know for sure that 0.7*BC = (a fixed value)* AB , do you know the 'fixed value'? NO. This the reason why this statement is NOT sufficient. Be careful with your interpretations in DS questions. You must be absolutely sure of the final value.

Hope this helps.
SVP
Joined: 06 Nov 2014
Posts: 1888
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

22 Jun 2016, 22:09
Area (ABC) = 1/2*AC*BC
Area (DBA) = 1/2*DA*AB

Statement 1:
AC = √2* AD. But we do not know anything about the base
ISUFFICIENT

Statement 2:
We just have the information about ABC
INSUFFICIENT

Combining both statements:
On combining, AC = CB, and AC = √2*AD

We have the values of base and perpendicular of both triangles in terms of AD.
Hence we can calculate and compare the area
SUFFICIENT

Correct Option: C
Intern
Joined: 19 Sep 2012
Posts: 13
In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

01 Dec 2016, 03:50
My approach on this one was as follows:

I reworded the question, so if the question asks if the area of triangular region ABC is equal to the area of triangular region DBA, then it means that

$$( (AC) (CB) ) / 2 = ((AD) (AB)) /2$$, we can simplify it as $$(AC) (CB) = (AD) (AB)$$. The reworded question is then $$(AC) (CB) = (AD) (AB)$$ ?

1) $$(AC)^2 = 2(AD)^2$$

$$\sqrt[2]{(AC)^2}= \sqrt[2]{2(AD)^2}$$

$$AC = AD \sqrt[2]{2}$$

But we don't know anything about CB and AB so the stem is Insufficient.

2) ABC is isosceles.

We don't know anything about the measures of each triangle so, stem (2) is insufficient.

Together (1) and (2):

We know that the proportions between AC and AD from stem 1. That's to say,
$$AC = AD \sqrt[2]{2}$$

And we know that an isosceles triangle has two of the sides that are equal and they always follow this proportion: $$x:x:x\sqrt[2]{2}.$$

So AC and CB should be equal. That means, $$CB = AD \sqrt[2]{2}$$
and if $$x = AD \sqrt[2]{2}$$ then $$AB = AD \sqrt[2]{2} \sqrt[2]{2} = AD \sqrt[2]{4} = 2(AD)$$

Then area of ACB is $$((AD \sqrt[2]{2}) (AD \sqrt[2]{2}) ) / 2 = (2(AD)^2)/2$$
And area of DAB is $$((AD) 2(AD))/2 = (2(AD)^2)/2$$

That means, $$(2(AD)^2)/2 = (2(AD)^2)/2$$ so area of ACB and area of DAB are equal.
Attachments

unnamed.jpg [ 28 KiB | Viewed 1335 times ]

Intern
Joined: 27 May 2015
Posts: 12
Location: Venezuela
GMAT 1: 720 Q49 V40
GPA: 3.76
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

29 May 2017, 19:02
I think I found an effective way of solving this problem in under two minutes. Let me explain the logic of it:

For simplicity purposes, let's call $$(\overline{AC})=x; (\overline{CB})=y; (\overline{AD})=w$$; and $$(\overline{AB})=z$$.

The question is basically asking if $$xy=wz$$.

Let's analyze the statements:

Statement 1)

$$x^{2}=2w^{2}$$. You can solve for $$w$$ and see that $$w=\frac{x\sqrt{2}}{2}$$. This does not help us to answer the question, so Statement 1 is not sufficient.

Statement 2)

∆ABC being an isosceles right triangle means that the triangle is a $$x:x:x\sqrt{2}$$ triangle. In other words, that $$x=y$$, and consequently that $$\text{A}^{ABC}=\frac{x^{2}}{2} \rightarrow 2\text{A}^{ABC}=x^{2}$$. Again, useful, but not sufficient.

Statements 1 and 2)

From Statement 2 we know that $$2\text{A}^{ABC}=x^{2}$$, and from Statement 1 we know that $$x^{2}=2w^{2}$$. From these, we can say that:

$$2\text{A}^{ABC}=2w^{2} \rightarrow \text{A}^{ABC}=w^{2} \rightarrow \frac{xy}{2}=w^{2}$$. It's time to use some of the information we had from Statement 1, in this case regarding $$w$$:

$$\frac{xy}{2}=\left(\frac{x\sqrt{2}}{2}\right)^{2} \rightarrow \frac{xy}{2}=\frac{x^{2}}{2}$$.

Now, since from Statement 2 we know that $$x=y$$, then the equation will hold true, and therefore, the correct answer is C.

Hope it helps.
Director
Joined: 02 Sep 2016
Posts: 734
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

06 Jul 2017, 10:32
Area of triangle ABC= 1/2 * AC*CB

This just gives a relation between AC and AD. We can't estimate exact values.

2) AC=CB
Therefore it is a 45-45-90 triangle with sides in ratio 1:1: sq.root 2

Now our question becomes: Is AC^2= AD*AB

No information about any other side.

Both
We can say that:

Area of ABC= x^2/2
Area of DAB= x^2/2

The area of both the triangles is same.
_________________

Help me make my explanation better by providing a logical feedback.

If you liked the post, HIT KUDOS !!

Don't quit.............Do it.

Intern
Joined: 03 Aug 2017
Posts: 14
Re: In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

29 Apr 2018, 02:24
Hi the question is
is the area of triangular region ABC equal to the area of triangular region DBA ?

from Statement 1 we know
Which means AC is probably bigger than AD . however for the area of a triangle we need the relation to both Height and Base to draw a conclusion hence insufficient.
From Statement 2 we know

(2) ∆ABC is isosceles.
However we still have no idea about AD HENCE WE STILL CONT COMPARE AREAS OF BOTH THE TRIANGLES.

Now together we know
(2) ∆ABC is isosceles.

Using both, ratio of sides of ABC are 1:1:2 √
1:1:2
= AC:BC : AB
Area of ABC = 1/2*1*1 = 1/2

Area of DAB = 1/2∗AD∗AB=1/2∗1/2 √ ∗2 √ =1/2

Areas of both the triangles is the same
Intern
Joined: 31 Mar 2018
Posts: 3
Location: India
Concentration: Sustainability, General Management
Schools: Erasmus '20
GMAT 1: 550 Q44 V23
GMAT 2: 660 Q44 V38
GMAT 3: 710 Q47 V41
GPA: 3
WE: Architecture (Other)
In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

18 Jul 2018, 00:59
teal wrote:
does anyone have any other alternative method to solve?

We have to check if $$1/2 * AB * BC = 1/2* AD * AB$$
i.e, $$AB * BC = AD * AB$$

From (1), $$AC^2 = 2*AD^2$$

$$AC = \sqrt{2}AD$$ . Not Sufficient

From (2), $$\triangle ABC$$ is isosceles.

$$AC = BC$$

in $$\triangle ABC$$, $$AB^2 = AC^2 + BC^2$$
$$AB^2 = 2*BC^2$$ .
$$BC = (1/\sqrt{2})AB$$. Not Sufficient

From (1) and (2),
$$AC * BC = \sqrt{2}AD * (1/\sqrt{2})AB$$
$$AC * BC = AD * AB$$
Sufficient. so (C)
Manager
Joined: 11 Sep 2013
Posts: 156
Concentration: Finance, Finance
In the figure above, is the area of triangular region ABC  [#permalink]

### Show Tags

03 Aug 2018, 02:57
Logical Approach:

To find the Area of any right angle triangle, we need to know the value of at least two sides of that triangle.

St 1: NS---- Because we only know one side of each triangle.
St 2: NS---- Because we don't have any information about another triangle.

By combining----- We now have the value of two sides of triangle ABC and DBA. It doesn't matter what these values are. Because we just need to know that whether we can find the area of these two triangles. In this DS question, we don't have to prove anything. Either they will be equal or not equal.

Sufficient.

Ans C
In the figure above, is the area of triangular region ABC &nbs [#permalink] 03 Aug 2018, 02:57

Go to page   Previous    1   2   [ 40 posts ]

Display posts from previous: Sort by

# Events & Promotions

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